Math in Focus

This handy Math in Focus Grade 8 Workbook Answer Key Chapter 11 Lesson 11.3 Independent Events detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 11 Lesson 11.3 Answer Key Independent Events

Math in Focus Grade 8 Chapter 11 Lesson 11.3 Guided Practice Answer Key

Solve. Show your work.

Question 1.
A game is played with a bag of 6 color tokens and a bag of 6 letter tiles. The 6 tokens consist of 2 green tokens, 1 yellow token, and 3 red tokens. The 6 letter tiles consist of 4 tiles of letter A and 2 tiles of letter B. To win the game, you need to randomly get a yellow token and a tile of letter B from a random selection in each bag.
a) Copy and complete the tree diagram.

Answer:

b) Use the multiplication rule of probability to find the probability of winning the game in one try.
P(winning the game) = P(Y, B)
= P(Y) • P(B)
= •
=
The probability of winning the game is .
Answer:
P(winning the game) = P(Y, B)
= P(Y) • P(B)
P(Y) = 1/6 and P(B) = 2/6
= 1/6 • 2/6
= 2/36
= 1/18
The probability of winning the game is 1/18.

Technology Activity

Materials:
spreadsheet software

SIMULATE RANDOMNESS

Work in pairs.

Background
Two fair six-sided number dice are thrown. Using a spreadsheet, data will be generated to investigate how frequently the outcome of doubles (1 and 1, 2 and 2, … , 6 and 6) occurs.
STEP 1: Label your spreadsheet as shown.

STEP 2: To generate a random integer between 1 and 6, in cell A2, enter the formula = INT(RAND()*6 + 1) to simulate rolling a die. A random number from 1 to 6 should appear in the cell.

STEP 3: To model 100 rolls, select cells A2 to A101 and choose Fill Down from the Edit menu.

STEP 4: Repeat STEP 2 and STEP 3 for cells B2 to B101.

STEP 5: In cell C2, enter the formula = A2 – B2. Select cells C2 to C101 and choose Fill Down from the Edit menu. This column serves as a check to see if the random numbers generated in columns A and B are the same. If the numbers are the same, their difference is 0. A zero difference indicates doubles outcome.

STEP 6: To see how many times the data shows double occurring, in cell E1, enter the formula = COUNTIF(C2:C101,0).

STEP 7: Find the experimental probability of the occurrence of two number dice showing the same number by dividing the number you get in cell E1 by the total, 100 rolls.

Math Journal Find the theoretical probability of rolling doubles with 2 fair number dice. Compare this theoretical probability with the experimental probability you obtained in the spreadsheet simulation. Are these two values the same?

Answer:

Question 2.
In a bag, there are 9 magenta balls and 1 orange ball. Two balls are randomly drawn, one at a time with replacement.
a) Find the probability of drawing two magenta balls.

P(M, M) = P(M) • P(M)
= •
=
The probability drawing two magenta balls is .
Answer:

P(M) = 9/10
P(M, M) = P(M) • P(M)
= 9/10 • 9/10
= 81/100
The probability drawing two magenta balls is 81/100.

b) Find the probability of drawing an orange ball followed by a magenta ball.
P(O, M) = P(O) • P(M)
= •
=
The probability drawing an orange ball followed by a magenta ball is .
Answer:
P(O) = 1/10
P(M) = 9/10
P(O, M) = P(O) • P(M)
= 1/10 • 9/10
= 9/100
The probability drawing an orange ball followed by a magenta ball is .

c) Find the probability of drawing both orange balls.
P(O, O) = P(O) • P(O)
= •
=
The probability drawing both orange balls is .
Answer:
P(O) = 1/10
P(O, O) = P(O) • P(O)
= 1/10 • 1/10
= 1/100
The probability drawing both orange balls is 1/100.

Question 3.
On weekends, Carli either jogs (J) or plays tennis (T) each day, but never both. The probability of her playing tennis is 0.75.
a) Find the probability that Carli jogs on both days.
Because J and T are complementary,
P(J) = 1 – P(T)
= 1 –
=

P(J, J) = P(J) • P(J)
P(J, J) = •
=
The probability that Carli jogs both days is .
Answer:
P(J) = 1 – P(T)
= 1 – 0.75
= 0.25

P(J, J) = P(J) • P(J)
P(J, J) = 0.25 • 0.25
= 0.0625
The probability that Carli jogs both days is 0.0625.

b) Find the probability that Carli jogs on exactly one of the days.
Using the addition rule of probability:
P(J, J) + P(T, J) = P(J) • P(T) + P(T) • P(J)
= • + •
=
The probability that Carli jogs on exactly one of the days is .
Answer:
Using the addition rule of probability:
P(J, J) + P(T, J) = P(J) • P(T) + P(T) • P(J)
= 0.25 • 0.75 + 0.75 • 0.25
= 0.125 + 0.125
= 0.25
The probability that Carli jogs on exactly one of the days is 0.25.

Math in Focus Course 3B Practice 11.3 Answer Key

Draw a tree diagram to represent each compound event.

Question 1.
Tossing a fair coin followed by drawing a marble from a bag of 3 marbles:
1 yellow, 1 green, and 1 blue.
Answer:

H represents heads
T represents  tails
Y represents yellow
G represents green
B represents blue

Question 2.
Drawing two balls randomly with replacement from a bag with 1 green ball and 1 purple ball.
Answer:

B represents ball
G represents green ball
P represents purple ball

Question 3.
Drawing a ball randomly from a bag containing 1 red ball and 1 blue ball, followed by tossing a fair six-sided number die.
Answer:

R represents red ball
B represents blue ball

Question 4.
Tossing a fair coin twice.
Answer:
If we toss a fair coin twice, we have the following possible outcomes, or events: {(H,H), (H,T),(T,H), (T,T). The total number of possible outcomes is therefore 4 and the number of outcomes where the result is two heads is 1. The probability of getting two heads in tossing a fair coin twice is therefore 1/4.

Question 5.
Reading or playing on each day of a weekend.
Answer:

R represents reading
P represents playing

Question 6.
On time or tardy for school for two consecutive days.
Answer:

Solve. Show your work.

Question 7.
Mindy is playing a game that uses the spinner shown below and a fair coin. An outcome of 3 on the spinner and heads on the coin wins the game.

a) Draw a tree diagram to represent the possible outcomes of this game.
Answer:

H represents Heads
T represents Tails

b) Find the probability of winning the game in one try.
Answer:
1/2 × 1/3 = 1/6
Thus the probability of winning the game in one try is 1/6.

c) Find the probability of losing the game in one try.
Answer:
1/2 + 1/3
= 3/6 + 2/6
= 5/6
Thus the probability of losing the game in one try is 5/6.

Question 8.
There are 2 blue balls and 4 yellow balls in a bag. A ball is randomly drawn from the bag, and it is replaced before a second ball is randomly drawn.
a) Draw a tree diagram to represent the possible outcomes.
Answer:

b) Find the probability that a yellow ball is drawn first, followed by another yellow ball.
Answer:
The probability of obtaining 2 yellow balls is 4/9.
Explanation:
P(Y,Y) = P(Y) x P(Y)
= 4/6 x 4/6
= 16/36
= 4/9

c) Find the probability that a yellow ball is drawn after a blue ball is drawn first.
Answer:
The probability of picking a blue than a yellow ball is 2/9.
Explanation:
P(B,Y) = P(B) x P(Y)
= 2/6 x 4/6
= 8/36
= 2/9

Question 9.
Jasmine has 3 blue pens and 2 green pens in her pencil case. She randomly selects a pen from her pencil case, and replaces it before she randomly selects again.
a) Draw a tree diagram to represent the possible outcomes.
Answer:

B represents blue
G represents green

b) Find the probability that she selects 2 blue pens.
Answer:
P(B) =3/5
P(B) × P(B) = 3/5 × 3/5 = 9/25

c) Find the probability that she selects 2 green pens.
Answer:
P(G) = 2/5
P(G) × P(G) = 2/5 × 2/5 = 4/25
Thus the probability that she selects 2 green pens is 4/25

d) Find the probability that she selects 2 pens of the same color.
Answer:
P(B, B) + P(G, G) = 9/25 + 4/25 = 13/25
Therefore the probability that she selects 2 pens of the same color is 13/25

Question 10.
Henry has 4 fiction books, 6 non-fiction books, and 1 Spanish book on his bookshelf. He randomly selects two books with replacement.

a) Draw a tree diagram to represent the possible outcomes.
Answer:

b) Find the probability that he selects a fiction book twice.
Answer:
The probability of obtaining 2 fictional books is 16/121 or around 0.13
Explanation:
P(F,F) = P(F) x P(F)
= 4/11 x 4/11
= 16/121
~ 0.13

c) Find the probability that he first selects a non-fiction book, and then a Spanish book.
Answer:
The probability of obtaining a non-fiction book and a spanish book is 6/121 or around 0.05
Explanation:
P(N,S) = P(N) x P(S)
= 6/11 x 1/11
= 6/121
~ 0.05

d) Find the probability that he first selects a fiction book, and then a non-fiction book.
Answer:
The probability of obtaining a fiction book and a non-fiction book is 24/121 or around 0.20
P(F,N) = P(F) x P(N)
= 4/11 x 6/11
= 24/121
~ 0.20

Question 11.
Andy tosses a fair six-sided number die twice. What is the probability of tossing an even number on the first toss and a prime number on the second toss?

Answer:

There are 36 total possible outcomes and 12 of them are favorable outcomes
P(E,P) = favorable outcomes/total outcomes
= 12/36
=1/3
Therefore, the probability of obtaining an even number and a prime number on the 2 tosses is 1/3.

Question 12.
The probability that Fiona wakes up before 8 A.M. when she does not need to set her alarm is \(\frac{4}{10}\). On any two consecutive days that Fiona does not need to set her alarm, what is the probability of her waking up before 8 A.M. for at least one of the days?
Answer:
The probability of waking up earlier than 8 AM at least once is 0.64.
Explanation:
P(E) + P(L) = 1
P(L) = 1 – P(E)
P(L) = 1 – 2/5
P(L) = 3/5
Therefore, the probability of that she will wakeup late is 3/5.
The probability of Fiona waking up late two days consecutively is 0.36
P(L,L) = P(L) x P(L)
= 3/5 x 3/5
= 9/25
= 0.36
The probability of waking up earlier than 8 AM at least once
A = 1 – P(L,L)
A = 1 – 0.36
A = 0.64

Question 13.
A globe is spinning on a globe stand. The globe surface is painted with 30% yellow, 10% green, and the rest is painted blue. Two times Danny randomly points to a spot on the globe while it spins. The color he points to each time is recorded.
a) What is the probability that he points to the same color on both spins?
Answer:
The probability of getting to the same color on both spins is 0.46
Explanation:
P(M) = P(Y,Y) + P(G,G) + P(B,B)
= P(Y) x P(Y) + P(G) x P(G) + P(B) x P(B)
= 0.3 x 0.3 + 0.1 x 0.1 + 0.6 x 0.6
= 0.09 + 0.01 + 0.36
= 0.46

b) What is the probability that he points to yellow at least one time?

Answer:
The probability of selecting yellow at least once is 0.51
Explanation:
P(NY) + P(Y) = 1
P(NY) = 1 – P(Y)
P(NY) = 1 – 0.3
P(NY) = 0.7
Therefore, the probability of not selecting yellow is 0.7
P(NY,NY) = P(NY) x P(NY)
P(NY,NY) = 0.7 x 0.7
P(NY,NY) = 0.49
Therefore, the probability of selecting twice is 0.49
P(A) + P(NY, NY) = 1
P(A) = 1 – 0.49
P(A) = 0.51
Therefore, the probability of selecting yellow at least once is 0.51.

Question 14.
Math Journal Sally thinks that for two independent events, because the occurrence of one event will not have any impact on the probability of the other event, they are also mutually exclusive. Do you agree with her? Explain your reasoning using an example.
Answer:
In an independent event, the existence of one does not affect the others. For example, eating chocolates and watching TV as they both are not mutually exclusive. Mutually exclusive events are two (or more) events that cannot be done at the same time. Thus independent events are not mutually exclusive.

Question 15.
A game is designed so that a player wins when the game piece lands on the letter A. The game piece begins on letter G. A fair six-sided number die is tossed. If the number tossed is odd, the game piece moves one step counterclockwise. If the number tossed is even, the game piece moves one step clockwise.
a) What is the probability that a player will win after tossing the number die once?
Answer:
Winning with one roll of die means getting an even number, say P(E). A number die has 6 possible results and 3 of these are even so the probability of getting even number die is the same as the probability of winning in one roll of the number die, which is 3/6 or 1/2.

b) What is the probability that a player will win after tossing the number die twice?

Answer:
The probability of arriving on A with two rolls of the dice is 1/4.
Explanation:
P(O,O) = P(O) x P(O)
P(O,O) = 1/2 x 1/2
P(O,O) = 1/4
Therefore, the probability of arriving on A with two rolls of the dice is 1/4.

This handy Math in Focus Grade 8 Workbook Answer Key Chapter 11 Lesson 11.2 Probability of Compound Events detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 11 Lesson 11.2 Answer Key Probability of Compound Events

Math in Focus Grade 8 Chapter 11 Lesson 11.2 Guided Practice Answer Key

Use a possibility diagram to find each probability.

Question 1.
Two fair four-sided number dice, each numbered 1 to 4 are rolled together. The result recorded is the number facing down. Find the probability that the product of the two numbers is divisible by 2.
Answer:
Given,
Two fair four-sided number dice, each numbered 1 to 4 are rolled together.
The result recorded is the number facing down.

3/4

Question 2.
One colored disc is randomly drawn from each of two bags. Both bags each have 5 colored discs: 1 red, 1 green, 1 blue, 1 yellow, and 1 white. Find the probability of drawing a blue or yellow disc.
Answer:
Given,
One colored disc is randomly drawn from each of two bags. Both bags each have 5 colored discs: 1 red, 1 green, 1 blue, 1 yellow, and 1 white.

R represents red
G represents green
B represents blue
Y represents yellow
W represents white
16/25

Question 3.
A box has 1 black, 1 green, 1 red, and 1 yellow marble. Another box has 1 white, 1 green, and 1 red marble. A marble is taken at random from each box. Find the probability that a red marble is not drawn.
Answer:
Given,
A box has 1 black, 1 green, 1 red, and 1 yellow marble.
Another box has 1 white, 1 green, and 1 red marble.
Marble is taken at random from each box.

B represents black
R represents red
G represents green
Y represents yellow
W represents white
1/2

Solve. Show your work.

Question 4.
Three fair coins are tossed together.
a) Draw a tree diagram to represent the possible outcomes.
Answer:

H represents Heads
T represents tails

b) Using your answer in a), find the probability of getting all heads.
Answer:
Total outcomes = 24
Number of getting all heads = 3
3/24 = 1/8
The probability of getting all heads = 1/8

c) Using your answer in a), find the probability of getting at least two tails.
Answer:
Total outcomes = 24
1/2
The probability of getting at least two tails = 1/2

Math in Focus Course 3B Practice 11.2 Answer Key

Solve. Show your work

Question 1.
A bag contains 2 blue balls and 1 red ball. Winnie randomly draws a ball from the bag and replaces it before she draws a second ball. Use a possibility diagram to find the probability that the balls drawn are different colors.
Answer:
Given,
A bag contains 2 blue balls and 1 red ball = 3
P(red) = 1/3
P(blue) = 2/3
possible outcomes = 4
Thus the probability that the balls drawn are different colors = 4/6

Question 2.
A letter is randomly chosen from the word FOOD, followed by randomly choosing a letter from the word DOG. Draw a tree diagram to find the probability that both letters chosen are the same.
Answer:

There are 12 possible outcomes and 3 among them are favorable outcomes, then
Probability = favorable outcomes/total outcomes
= 3/12
= 1/4
Therefore, the probability of getting two similar letters is 1/4.

Question 3.
Three pebbles are placed in a bag: 1 blue, 1 green, and 1 yellow. First a pebble is randomly drawn from the bag. Then a fair four-sided number die labeled from 1 to 4 is rolled. The result recorded is the number facing down. Use a possibility diagram to find the probability of drawing a yellow pebble and getting a 4.
Answer:

As there are 12 possible outcomes and 1 favorable outcome, then
Probability = favorable outcomes/total outcomes
= 1/12
Therefore, the probability of getting (4,y) is 1/12.

Question 4.
Thomas rolled a fair six-sided number die and a fair four-sided number die labeled 1 to 4 together. Use a possibility diagram to find the probability of rolling the number 3 on both.

Answer:

As there are total of 24 possible outcomes and 1 favorable outcome, then
Probability = favorable outcome/total outcomes
= 1/24
Therefore, the probability of getting (3,3) is 1/24.

Question 5.
At a bike shop, there are 3 bikes with 20-speed gears and 2 bikes with 18-speed gears. The bike shop also sells 1 blue helmet and 1 yellow helmet. Use a possibility diagram to find the probability of getting an 18-speed bicycle and a blue helmet if randomly selecting one bike and one helmet from among these.

Answer:

As there are 10 possible outcomes and 2 of them are favorable outcomes, then
Probability = favorable outcomes/total outcomes
= 2/10
= 1/5
Therefore, the probability of getting (B,18) is 1/5.

Question 6.
Susan randomly draws a card from three number cards: 1, 3, and 6. After replacing the card, Susan randomly draws another number card. The product of the two numbers drawn is recorded.

a) Use a possibility diagram to represent the possible outcomes.
Answer:

b) Using your answer in a), what is the probability of forming a number larger than 10 but less than 30?
Answer:

There are 9 possible outcomes and 2 of them are favorable outcomes,then
Probability(>10<30) = favorable outcomes/total outcomes
= 2/9
Therefore, the probability of getting greater than 10 and less than 30 is 2/9.

Question 7.
Jane and Jill watch television together for 2 hours. Jane selects the channel for the first hour, and Jill selects the channel for the second hour. Jane’s remote control randomly selects from Channels A, B, and C. Jill’s remote control randomly selects from Channels C, D, and E.
a) Use a possibility diagram to represent the possible outcomes for the channels they watch on television for 2 hours.
Answer:

b) Using your answer in a), what is the probability of watching the same channel both hours?
Answer:

As there are 9 possible outcomes and 1 favorable outcome, then
Probability = favorable outcomes/total outcomes
= 1/9

Question 8.
A color disc is randomly drawn from a bag that contains the following discs.

After a disc is drawn, a fair coin is tossed. Use a possibility diagram to find the probability of drawing a red disc and landing on heads.
Answer:

As there are 10 possible outcomes and 2 of them are favorable outcomes, then
Probability = favorable outcomes/total outcomes
= 2/10
= 1/5
Therefore, the probability of getting a red disc and head is 1/5.

Question 9.
Karen tosses a fair coin three times. Draw a tree diagram to find the probability of getting the same result in all three tosses.
Answer:

As there are total of 8 possible outcomes and 2 favorable outcomes, then
Probability = favorable outcomes/total outcomes
= 2/8
= 1/4
Therefore, the probability of acquiring a (H,H’,H”) and (T,T’,T”) is 1/4.

Question 10.
Mrs. Bridget’s recipes require her to put in some fine maize flour in bowl 1, followed by wheat flour in bowl 2, and rice flour in bowl 3. However, the jars of flour are not labeled, so she randomly guesses which flour to put in which bowl.
a) Draw a tree diagram to represent the possible outcomes.
Answer:

b) Using your answer in a), find the probability of getting the correct flour in the correct order.

Answer:
There are total of 6 possible outcomes and 1 favorable outcomes, then
Probability = favorable outcomes/total outcomes
= 1/6
Therefore, the probability of selecting a cornmeal flour and low fat milk is 1/6.

This handy Math in Focus Grade 8 Workbook Answer Key Chapter 11 Lesson 11.1 Compound Events detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 11 Lesson 11.1 Answer Key Compound Events

Math in Focus Grade 8 Chapter 11 Lesson 11.1 Guided Practice Answer Key

Tell whether the outcomes described are from a simple or compound event. If it is a compound event, identify the simple events that form the compound event.

Question 1.
Obtaining two heads when two coins are tossed
Answer:
There are two simple events each consisting of tossing a coin.
Thus it is a compound event

Question 2.
Winning a football game
Answer:
Simple event

Question 3.
Getting a number less than 4 or getting a number greater than 5 when a fair six-sided number die is rolled
Answer: Simple event

Question 4.
Rolling two fair six-sided number dice and obtaining a sum of 10 from the throws
Answer: There are two simple events each consisting of rolling a fair number die.

Represent and tell the number of possible outcomes for each compound event described.

Question 5.
Two fair coins are tossed together.
Answer:
The possibilities of tossing two coins are {(HH), (TT), (H,T), (T, H)}
Thus there are 4 outcomes of tossing two fair coins.

Question 6.
The results of rolling two fair six-sided number dice are multiplied.
Answer:
(1, 1), (1, 2) (1, 3), (1, 4) (1, 5) (1, 6)
(2, 1), (2, 2) (2, 3), (2, 4) (2, 5) (2, 6)
(3, 1), (3, 2) (3, 3), (3, 4) (3, 5) (3, 6)
(4, 1), (4, 2) (4, 3), (, 4) (4, 5) (4, 6)
(5, 1), (5, 2) (5, 3), (5, 4) (5, 5) (2, 6)
(6, 1), (6, 2) (6, 3), (6, 4) (6, 5) (6, 6)
The possible outcomes are 36.

Question 7.
A fair six-sided number die and a fair four-sided number die labeled 1 to 4 are rolled. The results that face down on both number dice are recorded.

Answer:

For each compound event, draw a tree diagram to represent the possible outcomes. Then tell the number of possible outcomes.

Question 8.
Joshua has two bags. In the first bag, there are 2 blue beads and 1 green bead. In the second bag, there are 3 lettered cards with the letters P, Q, and R. Joshua randomly takes an item from the first bag, and then from the second bag.
Answer:

The possible outcomes are 9
Where,
B represents blue
G represents Green
P represents letter P
Q represents letter Q
R represents letter R

Question 9.
A fair coin is tossed then a fair four-sided color die with faces painted yellow, green, blue, and black is rolled. The color facing down is the result recorded.
Answer:

The possible outcomes are 8.
Where,
H represents heads
T represents tails
Y represents yellow
G represents green
B represents blue
B represents black

Math in Focus Course 3B Practice 11.1 Answer Key

Tell whether each statement is True or False.

Question 1.
Selecting letter A from the word PROBABILITY is a compound event.
Answer:
P – 1
R – 1
O – 1
B – 2
A – 1
I – 2
L – 1
T – 1
Y – 1
So, the statement is false.

Question 2.
Selecting letter B from the word BASEBALL and ABLE is a simple event.
Answer:
B – 2
A – 2
S – 1
E – 1
L – 2
The statement is false
ABLE is a simple event

Question 3.
Tossing a fair six-sided number die to get either an even number or a five is a compound event.
Answer:
Rolling an even number: {2, 4, 6}
Rolling a 5: {5}
The statement “Tossing a fair six-sided number die to get either an even number or a five is a compound event” is false.

Question 4.
Umberto has 3 red cards and 4 blue cards. Drawing two red cards in a row, without replacing the first card before drawing the second card, is a compound event.
Answer: true

Tell whether the outcomes described are from a simple or compound event. If it is a compound event, identify the simple events that form the compound event.

Question 5.
Getting a 6 when a fair six-sided number die is rolled.
Answer:
Rolling a 6: {6}
A simple event is an event with only one outcome.

Question 6.
Rolling three fair six-sided number dice and obtaining a sum of 18 from the throws.
Answer:
There are 3 possible outcomes.
A compound event is an event with more than one outcome.

Question 7.
Getting an eighteen when a fair twenty-sided number die is rolled.

Answer:
There is one possible outcome thus it is a simple event.

Question 8.
Susan has 3 red cards and 4 blue cards. She first draws a blue card. Without replacing the first card, she then draws another blue card.
Answer: Compound event

Solve. Show your work.

Question 9.
In the top drawer, there are two battery operated flash lights, one red and one yellow. In the second drawer, there are three packages of batteries: small, medium, and large. A flashlight and a package of batteries are randomly selected.
a) Use a possibility diagram to represent the possible outcomes.
Answer:

R represents red
Y represents yellow
S represents small
M represents medium
L represents large

b) How many possible outcomes are there?
Answer: 6 outcomes

Question 10.
Two electronic spinners, A and B, are spun by pressing a button. Spinner A has four sectors labeled 1 to 4, while B has three sectors, labeled 1 to 3. Spinner B, due to technical error, will never land on number 2 if spinner A lands on a 4.
a) Use a possibility diagram to represent the possible outcomes.
Answer:
The target of this task is to analyze the events of turning two spinners labelled t 2, 3, and 4, and 1, 2, and 3 then make a diagram that shows all its possibLe outcomes then determine the number of its possible outcomes given a condition of not being able to get (4, 2).

Since there are two spinners A and B, marked from 1 – 4 and 1 – 3 then the illustration for the possible outcomes would be as depicted in the table.

b) How many possible outcomes are there?
Answer:
The possible outcomes are (1, 1), (2, 1), (3, 1), (4, 1), (1, 2), (2, 2), (3, 2), (1, 3), (2, 3), (3, 3). and (4, 3)Therefore, there are 11 possible outcomes.

Question 11.
Winston has two boxes. The first box has 3 black pens and 1 red pen. The second box has 1 green ball and 1 yellow ball. Use a tree diagram to represent the possible outcomes for randomly drawing a pen and a ball. Then tell the number of possible outcomes.

Answer:

8 outcomes
R represents red
Y represents yellow
G represents green
B represents black

Question 12.
Seraphina first tosses a fair six-sided number die. She then tosses a fair coin. Use a tree diagram to represent the possible outcomes.
Answer:
The objective of this task is to make a tree diagram for the possibLe outcomes for tossing a number die and flipping a coin.

Observe that rolling a die has 6 possible outcomes, then start the tree diagram with 6 branches. Now, at the end of these branches put the possible outcomes: 1, 2, 3, 4, 5 and 6. Next in tossing a coin, since there are 2 possible outcomes then for each number, make another 2 branches. At the end of these branches put the possibLe outcomes: H for head and T for tail. Finally, conclude the possible results. The tree diagram for this event would be as depicted below.

Question 13.
A game was designed such that a participant needs to accomplish 2 rounds to be considered the overall winner. The first round is to roll a 4 from a fair four-sided number die labeled 1 to 4. The result recorded is the number facing down. The second round is to randomly draw a red ball from a box of 2 differently colored balls.
a) Draw a tree diagram to represent the possible outcomes.
Answer:

R represents red
R’ represents not red

b) How many possible outcomes are there?
Answer: 8 outcomes

c) Math Journal If the participant first draws the colored ball and then rolls the four-sided number die, will the number of possible outcomes be the same? Use a diagram to explain your reasoning.
Answer: Yes, there are still 8 outcomes

R represents red
R’ represents not red

Question 14.
Zoe first rolls a fair four-sided number die labeled 1 to 4. Then she rolls another fair four-sided number die labeled 2 to 5. The result recorded is the number facing down.

a) Use a possibility diagram to find the number of favorable outcomes for an odd sum.
Answer:
Note that there are 4 possible outcomes for born toss of the die, add these results, then the outcomes in tossing a die twice and adding their result would be as depicted in the table. Mark the sum that are odd.

Observe that there are 16 possib[e outcomes and 8 are odd then

Therefore, the probability of obtaining an odd sum is \(\frac{1}{2}\).

b) Use a possibility diagram to find the number of favorable outcomes for a difference greater than 2.
Answer:
Since there are 4 possible outcomes for both toss of the die, subtract these results, then the outcomes in tossing a die twice and subtracting their result would be as depicted in the table. Mark the difference that are bigger than 2.

Notice that there are 16 possible outcomes and only 1 is larger than 2 then

Therefore, the probability of obtaining a difference greater than 2 is \(\frac{1}{16}\).

This handy Math in Focus Grade 8 Workbook Answer Key Chapter 11 Probability detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 11 Answer Key Probability

Math in Focus Grade 8 Chapter 11 Quick Check Answer Key

Solve. Show your work.

A box has 2 black balls, 7 red balls, and 3 green balls. A ball is randomly chosen from the box.

Question 1.
What is the probability of choosing a green ball?
Answer:
Given,
A box has 2 black balls, 7 red balls, and 3 green balls.
Total number of balls = 2 + 7 + 3 = 12
We have to find the probability of choosing a green ball.
Number of green balls = 3
Probability of choosing green balls = 3/12 = 1/4
Thus the Probability of choosing green balls is 1/4.

Question 2.
What is the probability of choosing a black ball?
Answer:
Given,
A box has 2 black balls, 7 red balls, and 3 green balls.
Total number of balls = 2 + 7 + 3 = 12
We have to find the probability of choosing a black ball.
Number of green balls = 2
Probability of choosing green balls = 2/12 = 1/6
Thus the Probability of choosing green balls is 1/6.

Question 3.
What is the probability of choosing a blue ball?
Answer:
Given,
A box has 2 black balls, 7 red balls, and 3 green balls.
Total number of balls = 2 + 7 + 3 = 12
We have to find the probability of choosing a black ball.
Number of blue balls = 0
Thus the Probability of choosing blue balls is 0.

Question 4.
What is the probability of choosing a ball that is not red?
Answer:
A box has 2 black balls, 7 red balls, and 3 green balls.
Total number of balls = 2 + 7 + 3 = 12
We have to find the probability of choosing a ball that is not red.
Number of red balls = 7
The number of balls that are not red is 5
So, the probability of choosing a ball that is not red is 5/12.

Question 5.
What is the probability of choosing a red or green ball?
Answer:
Given that,
A box has 2 black balls, 7 red balls, and 3 green balls.
Total number of balls = 2 + 7 + 3 = 12
Find the probability of choosing a red or green ball.
Number of green balls = 2
Probability of choosing green balls = 2/12 = 1/6
Thus the Probability of choosing green balls is 1/6.
Number of red balls = 7
Probability of choosing green balls = 7/12
The probability of choosing a red or green ball is 9/12 = 3/4.

Tell whether the events X and Y are mutually exclusive events.

Question 6.
A fair coin and a fair six-sided number die are tossed. X is the event that a head is obtained. Y is the event that a six is obtained.
Answer:
Given,
A fair coin and a fair six-sided number die are tossed.
X is the event that a head is obtained.
The probability of getting heads is p(A) = 1
Y is the event that a six is obtained.
The probability of getting six is p(B) = 1
Yes the events X and Y are not mutually exclusive events.

Question 7.
A fair six-sided number die is rolled. X is the event of obtaining a three. Y is the event of obtaining a five.
Answer:
A fair six-sided number die is rolled.
X is the event of obtaining a three.
Y is the event of obtaining a five.
Events 3 and 5 are mutually exclusive because we cannot get 3 and 5 at the same time.

Question 8.
Two fair six-sided number dice are tossed. X is the event that the sum of the score is six. Y is the event that the sum of the score is 10.
Answer:
Given,
Two fair six-sided number dice are tossed.
X is the event that the sum of the score is six.
Y is the event that the sum of the score is 10.
Total number of possible results from two six-sided dice is 6 × 6 = 36.
The possibility of the sum of the score is six is (1, 5) (2, 4) (3, 3) (4, 1) (5, 1) = 5/36
The sum of the score is 10 is (4, 6) (5, 5) (6, 4) = 3/36 = 1/12
Thus they are mutually exclusive events.

Question 9.
X is the event consisting of the factors of 24. Y is the event consisting of multiples of 6 less than 20.
Answer:
Factors of 24 is (1, 24), (2, 12) (3, 8) and (4, 6).
Multiples of 6 less than 20 is 6, 12, 18.
Thus they are mutually exclusive events.

Go through the Math in Focus Grade 4 Workbook Answer Key Chapter 1 Place Value of Whole Numbers to finish your assignments.

Math in Focus Grade 4 Chapter 1 Answer Key Place Value of Whole Numbers

Math Journal

Question 1.
Kim wrote these statements about the three numbers shown here.
Do you agree? Explain why or why not.
3,869 is less than 85,945.
85,691 is greater than 85,945.

Answer:
The above-given statements are:
1. 3,869 is less than 85,945.
2. 85,691 is greater than 85,945.
The first statement is correct.
The second statement is wrong.
The first statement is correct because we are comparing numbers with their place values.
In Mathematics, place value charts help us to make sure that the digits are in the correct places. To identify the positional values of numbers accurately, first, write the digits in the place value chart and then write the numbers in the usual and the standard form.
Rules for comparing numbers:
There are certain rules, based on which it becomes easier to compare numbers. These rules are:
* Numbers with more digits
* Numbers starting with a larger digit
Rule 1: Numbers with more digits
– When we compare numbers, then check if both the numbers are having the same number of digits or not. If a number has more digits, then it is greater than the other number.
Examples:
– 33 > 3
– 400 > 39
– 5555 > 555
– 10000 > 9999
Rule 2: Numbers starting with a larger digit
This rule is applicable when two numbers are having the same number of digits. In such cases, we need to check the digit at the leftmost place, whichever is greater. Therefore, the number with a greater digit at the leftmost place of the number is greater than the other number.
Examples:
* 323>232 [323 is greater than 232]
* 343<434 [343 is less than 434]
Thus, in the above examples, we can see that, when we compare the two numbers, though the number of digits is the same, one number is greater than/less than the other number.
Now according to the above rules, we check statement 1:
1. 3,869 is less than 85,945.
– In this statement, two numbers are given. The first number is having fewer digits than the second number so here rule 1 is applicable.
absolutely, 3,869 < 85,945
This can be written in another way:
85,945 > 3,869.
Now we check for statement 2:
2. 85,691 is greater than 85,945.
Here the same number of digits are having so rule 2 is applicable.
– 85,691 and 85,945 both the numbers have an equal number of digits, therefore, we will compare the left-most digit of both the numbers.
– As we can see the first two digits are the same for both the numbers, thus we need to compare the next left-most digit of both numbers.
6 < 9
Here it is a wrong statement because 6 is not greater than 9. So the second statement is wrong.

Question 2.
Sam continued this number pattern.
Do you agree? Explain why or why not.

Answer:

For number patterns there are certain rules:
To create a complete pattern, there are a set of rules to be considered. To apply the rule, we need to understand the nature of the sequence and the difference between the two successive numbers. It takes some amount of guesswork and checking to see whether the rule works throughout the whole series.
There are two basic divisions to find out the rules in number patterns:
* When the numbers in the given pattern get larger, they are said to be in ascending order. These patterns usually involve addition or multiplication.
* When the numbers in the given pattern get smaller, they are said to be in descending order. These patterns usually involve subtraction or division.
So the number pattern is correct.

Question 3.
Read the example. Then write your own 5-digit number and clues. Ask a friend or family member to solve your puzzle.

Example
45,870
1. The digit 5 is in the thousands place.
2. The value of the digit 7 is 70.
3. The digit in the hundreds place is 10 – 2.
4. The digit in the ten thousands place is 1 less than the digit in the thousands place.
5. The digit in the ones place is 0.

Number: _______
Clues: ______
Answer:
– According to the clues, the number is 15,870
– Now in line number 4, the digit in the ten thousand places is 1 less than hundreds of places. So, 1 < 5. 5 is in the thousand places so I placed 5 in the thousands place.
– In line 3 the hundreds place is 8. In the above-example 10-2 is given which is 8. So I placed 8 in the hundreds place.
– In line 2 the number 70 is given, 70 is nothing but 7 tens so I placed 7 in the tens place.
– In line 5 directly given that one’s place is 0.

Put on Your Thinking Cap!

Challenging Practice

Complete.

A 5-digit number is made up of different digits that are all odd numbers.

Question 1.
What is the greatest possible number? ______
Answer: 9,99,99
Explanation:
The largest 5 digit number that you can have when all digits are odd would be 99999.
that’s with no restrictions.
any other number you choose would have to be smaller.
the restrictions are:
the ten thousand digits are 3 times the one’s digit.
since the ten-thousands digit is fixed at 9, then the one’s digit has to be 3 because 3 * 3 = 9.
the thousands digit is 2 more than the hundreds digit.
since the thousands digit is fixed at 9, then the hundreds digit has to be 7 because 7 + 2 = 9.
that makes your number equal to 99793.
any other arrangement would force the ten-thousands digit to be something less than 9 or the thousands digit to be less than 9 which would result in a smaller number.

Question 2.
What is the value of the digit in the hundreds place? _____
Answer: 9
Explanation:
The largest 5 digit number that you can have when all digits are odd would be 99999.
The place values of 99999 are:
9 is in the hundred thousand place.
9 is in the ten thousand place
9 is in the thousand place
9 is in the hundreds place
9 is in the tens place
9 is in the one’s place.

Continue the pattern.

Question 3.

Answer: 487  502
Explanation:
A list of numbers that follow a certain sequence is known as patterns or number patterns. The different types of number patterns are algebraic or arithmetic patterns, geometric patterns, Fibonacci patterns and so on.
The above-given numbers are in the arithmetic pattern.
The arithmetic pattern is also known as the algebraic pattern. In an arithmetic pattern, the sequences are based on the addition or subtraction of the terms. If two or more terms in the sequence are given, we can use addition or subtraction to find the arithmetic pattern.
The numbers are 412  427  442  457  472  –  -. Now, we need to find the missing term in the sequence.
Here, we can use the addition process to figure out the missing terms in the patterns.
In the pattern, the rule used is “Add 15 to the previous term to get the next term”.
In the numbers given above, take the second term (427). If we add “15” to the second term (427), we get the third term 442.
Similarly, we can find the unknown terms in the sequence.
First missing term: The previous term is 472. Therefore, 472+15 = 487.
Second missing term: The previous term is 487. So, 487+15 = 502
Hence, the complete arithmetic pattern is 412  427   442   457   472   487   502.

Fill in the blanks.

Question 4.
What is 3 ten thousands + 14 tens + 16 ones? ______
Answer:30,156
Explanation:
In a number, the place (local) value of a non-zero digit is the value of this digit according to its position.
A number can have many digits, and each digit has a special place and value
ten thousand can be written as 10,000
tens can be written as 10
ones can be written as 1
3 x 10,000 = 30,000, 14 x 10 = 140 and 16 x 1 = 16
30 000 + 140 + 16 = 30 156.

Question 5.
7 thousands = _____ hundreds 10 tens
Answer:
7 thousand=7 hundreds 10 tens
This can be written as:
7000=700*10
7 thousands=70 hundreds
7 thousand=700 tens

Answer these questions.

Question 6.
what is the value of the digit 5 in the ? ____
Answer: 5000
Explanation:
In Mathematics, place value charts help us to make sure that the digits are in the correct places. To identify the positional values of numbers accurately, first, write the digits in the place value chart and then write the numbers in the usual and the standard form.
According to the above solution, 5 is in the thousands place.
5 thousands is nothing but 5000
So it can be written as 5000.

Question 7.
what is the value of the digit 5 in the ? ____
Answer:50
Explanation:
In Mathematics, place value charts help us to make sure that the digits are in the correct places. To identify the positional values of numbers accurately, first, write the digits in the place value chart and then write the numbers in the usual and the standard form.
According to the above solution, 5 is in the tens place.
5 tens are nothing but 50
So it can be written as 50.

Question 8.
what is the difference between the answers in Exercises 6 and 7? ____
Answer: Place values.
Explanation:
The place value is the position of each digit in a number. The place value of digits is determined as ones, tens, hundreds, thousands,ten-thousands and so on, based on their position in the number. For example, the place value of 1 in 1002 is thousands, i.e.1000.
The above-given number 75,859
7 is in the ten thousand place
5 is in the thousands of place
8 is in the hundreds place
5 is in the tens place
9 is in the one’s place.

Question 9.
, the difference between the values of the digits in the and the is 8,930. What is the digit in the ?
Answer:9
The above-given number 5 X,278
We need to find out the X.
The difference between and =8,930
Already we know the value in the square that is 7
X-7=8,930
If we subtract 59,278 and 50,348 then we get 8,930
so 9 is in the ten thousand places.
Check which number we subtract from the given number to get 8,930.
If we kept 9 in the X place and subtract with the assumed number that is 50,348 then we get the given number 8,930. Likewise we need to think and process the problem.

Put On Your Thinking cap!

Problem Solving

The button and the button do not work.
Justin wants to enter the number 82,365.
Explain what he can do to key in the number. Give two solutions.

Answer:
The above-given number is 82,365
But the 2 and 6 buttons are not working.
Here given clue we can add or subtract
First solution:
For example, if we add two 1’s means 1+1=2
if we add two 3’s or we can add 5+1=6 and 3+3=6.
Second solution:
If we subtract 5-3=2
if we subtract 7-1=6
These are the chances to enter the right values.

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 5 Systems of Linear Equations to score better marks in the exam.

Math in Focus Grade 8 Course 3 A Chapter 5 Answer Key Systems of Linear Equations

Math in Focus Grade 8 Chapter 5 Quick Check Answer Key

Complete the table of values and graph each linear equation.

Question 1.
y = 5x

Answer:
When x = 0, y=0,
x =1 , y = 5, x = 2, y = 10, when x = 3 y =15,

Explanation:

Question 2.
y = -x + 2

Answer:
x =0, y = 2, x= 1, y =1,
x= 2, y = 0, x =3,y =-1,

Explanation:

Solve. Show your work.

Question 3.
Samuel bought 30 books. The hardcover books cost $20 each while the rest,
which are paperbacks, cost $8 each. If he spent a total of $480, how many paperbacks did he buy?
Answer:
Number of paperbacks he buy are 10,

Let x be the no. of hard cover books and y be the no. of paper back books.
x + y = 30,
x = 30 – y,
20x + 8y = 480,
20(30 – y) + 8y = 480,
600 – 20y + 8y = 480,
-12y = – 120,
y = 120/12, y = 10,
x = 30-10 = 20,
No.of hard cover books = x = 20,
No.of paper back books = y = 10.

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 5 Lesson 5.1 Introduction to Systems of Linear Equations to score better marks in the exam.

Math in Focus Grade 7 Course 3 A Chapter 5 Lesson 5.1 Answer Key Introduction to Systems of Linear Equations

Math in Focus Grade 8 Chapter 5 Lesson 5.1 Guided Practice Answer Key

Solve the system of linear equations by copying and completing the tables of values.
The values x and y are positive integers.

Question 1.
A bottle of water and a taco cost $3. The cost of 3 bottles of water is $1 more than the cost of a taco.
Let x be the price of a bottle of water and y be the price of a taco in dollars.
The related system of equations and tables of values are:
3x – y = 1
x + y = 3

Answer:
The cost of a bottle of water is $2 and the cost of a taco is $1,

Explanation:

Solve each system of equations by making tables of values, x and y are positive integers.

Question 2.
x + y = 6
x + 2y = 8
Answer:

Explanation:
Solved each system of equations by making tables of values, x and y are positive integers.
For equation x+y = 6, when x =1 then y =5 and when x =2 then y=4,
For equation x +2y =8, when x =1 then y =7/2 and when x = 2 then x =3.

Question 3.
x + y = 8
x – 3y = -8
Answer:

Explanation:
Solved each system of equations by making tables of values, x and y are positive integers.
For equation x+y = 6, when x =1 then y =5 and when x =2 then y=4,
For equation x +2y =8, when x =1 then y =7/2 and when x = 2 then x =3.

For each linear equation, list in a table enough values for x and y to obtain a solution.
Remember that they must be positive integers.

Technology Activity

Materials:

• graphing calculator

Use Tables On A Graphing Calculator To Solve A System Of Equations

Work in pairs.

You can use a graphing calculator to create tables of values and solve systems of equations.
Use the steps below to solve this system:

8x + y = 38
x – 4y = 13

Step 1.
Solve each equation for y in terms of x. Input the two resulting expressions for y into the equation screen.

Caution
Use parentheses around fractional coefficients and the key for negative coefficients

Step 2.
Set the table function to use values of x starting at 0, with increments of 1.

Step 3.
Display the table. It will be in three columns as shown.

Step 4.
Find the row where the two y-values are the same. This y-value and the corresponding x-value will be the solutión to the equations.
The solution to the system of equations is given by x = and y = .

Math Journal How can you tell from the two columns of y-values that there is only one row where the y-values are the same?

Math in Focus Course 3A Practice 5.1 Answer Key

Solve each system of linear equations by making tables of values.
Each variable x is a positive integer less than 6.

Question 1.
2x + y = 5
x – y = -2
Answer:

Explanation:
Solved each system of linear equations by making tables of values.
Each variable x is a positive integer less than 6.

Question 2.
x + 2y = 4
x = 2y
Answer:

Explanation:
Solved each system of linear equations by making tables of values.
Each variable x is a positive integer less than 6.

Question 3.
3x + 2y = 10
5x – 2y = 6
Answer:

Explanation:
Solved each system of linear equations by making tables of values.
Each variable x is a positive integer less than 6.

Question 4.
x – 2y = -5
x = y
Answer:

Explanation:
Solved each system of linear equations by making tables of values.
Each variable x is a positive integer less than 6.

Question 5.
2y – x = -2
x + y = 2
Answer:

Explanation:
Solved each system of linear equations by making tables of values.
Each variable x is a positive integer less than 6.

Question 6.
2x + y = 3
x + y = 1
Answer:

Explanation:
Solved each system of linear equations by making tables of values.
Each variable x is a positive integer less than 6.

Question 7.
x + 2y = 1
x – 2y = 5
Answer:

Explanation:
Solved each system of linear equations by making tables of values.
Each variable x is a positive integer less than 6.

Question 8.
2x – y = 5
2y + x = -1
Answer:

Explanation:
Solved each system of linear equations by making tables of values.
Each variable x is a positive integer less than 6.

Question 9.
2x + y = -1
x + y = 1
Answer:

Explanation:
Solved each system of linear equations by making tables of values.
Each variable x is a positive integer less than 6.

Solve by making a table of values. The values x and y are integers.

Question 10.
A shop sells a party hat at x dollars and a mask at y dollars.
On a particular morning, 10 hats and 20 masks were sold for $30.
In the afternoon, 8 hats and 10 masks were sold for $18. The related system of linear equations is:
10x + 20y =30
8x + 10y = 18
Solve the system of linear equations. Then find the cost of each hat and each mask.
Answer:
The cost of each hat is $1 and each mask is$1,

Explanation:
Given A shop sells a party hat at x dollars and a mask at y dollars.
On a particular morning, 10 hats and 20 masks were sold for $30.
In the afternoon, 8 hats and 10 masks were sold for $18.
The related system of linear equations is:
10x + 20y =30—(1)
8x + 10y = 18—-(2) dividing equation 1 by 10 we get
x +2y = 3,
x = 3-2y substituting in equation 2 we get
8(3-2y) +10y =18,
24 – 16y+10y =18,
24 -6y =18,
6y = 24 -18,
6y =6, y =1 we have x = 3-2X 1 = 3-2 = 1.

Question 11.
Alicia is x years old and her cousin is y years old.
Alicia is 2 times as old as her cousin.
Three years later, their combined age will be 27 years.
The related system of linear equations is:
x = 2y
x + y = 27
Solve the system of linear equations. Then find Alicia’s age and her cousin’s age.
Answer:
Alicia’s age is 18 and her cousin’s age is 9 years old,

Explanation:
GivenAlicia is x years old and her cousin is y years old.
Alicia is 2 times as old as her cousin.
Three years later, their combined age will be 27 years.
The related system of linear equations is: x = 2y,
x + y = 27 solving the equations  2y + y = 27,
3y = 27, y = 27/3 =9, so x = 2X 9 = 18, Alicia’s age is 18 and her
cousin’s age is 9 years old.

Question 12.
Steve and Alex start driving at the same time from Boston to Paterson.
The journey is d kilometers. Steve drives at 100 kilometers per hour and
takes t hours to complete the journey. Alex, who drives at 80 kilometers per hour is
60 kilometers away from Paterson when Steve reaches Paterson.
The related system of linear equations is:
100t = d
80t = d – 60

Solve the system of linear equations by making tables of values.
Then find the distance between Boston and Paterson.
Answer:
300 kilometers is the distance between Boston and paterson,

Explanation:
Given Steve and Alex start driving at the same time from Boston to Paterson.
The journey is d kilometers. Steve drives at 100 kilometers per hour and
takes t hours to complete the journey. Alex, who drives at 80 kilometers per hour is
60 kilometers away from Paterson when Steve reaches Paterson.
The related system of linear equations is:
100t = d
80t = d – 60 substituting 80t = 100t -60,
100t-80t = 60,
20t = 60, t = 60/20, t =3 so distance is 100 X 3= 300 kilometers.

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 5 Lesson 5.2 Solving Systems of Linear Equations Using Algebraic Methods to score better marks in the exam.

Math in Focus Grade 7 Course 3 A Chapter 5 Lesson 5.2 Answer Key Solving Systems of Linear Equations Using Algebraic Methods

Math in Focus Grade 8 Chapter 5 Lesson 5.2 Guided Practice Answer Key

Solve each system of linear equations using the elimination method.

Question 1.
2a+ 3b= 29 —Equation 1
2a — b = 17 — Equation 2
Subtract Equation 2 from Equation 1:

Answer:
The solution to the system of linear equations is a = 23/2 and b =6,

Explanation:

Question 2.
2x – y = 2
3x + y = 13
Answer:
The solution to the system of linear equations is  x = 3 and y =4.

Explanation:
Given 2x -y =2 -Equation 1,
3x + y = 13- Equation 2 Adding Equation 2 and 1,
2x – y =2
3x + y =13
5x = 15,
x = 15/5 = 3, substituting x =3 in Equation 1
y = 2x -2= 2 X 3 -2 = 6-2 =4.

Question 3.
x + 6 = 1
x + y = 6
Answer:
The solution to the system of linear equations is x = -5 and y = 11,

Explanation:
Given x + 6 =1 means x = 1-6 = -5,
substituting x = -5 in Equation 2 – x + y  = 6 as x = -5,
-5 + y = 6, y = 6 +5 = 11.
Therefore x = -5 and y = 11.

Solve each system of linear equations using the elimination method.

Question 4.
7m + 2n = -8
2m = 3n – 13
Answer:
The solution to the system of linear equations is m = -2 and n =3,

Explanation:
Given 7m +2n = -8 – Equation 1 and 2m = 3n – 13- Equation 2,
to make elmination we need common values we multiply
Equation 1 by 3 and Equation 2 by 2 so that n becomes common,
3 X(7m +2n )= 3 X -8 = 21m + 6n = -24 and
2 X 2m = 2 X (3n – 13)= 4m = 6n – 26 means 6n = 4m + 26,
so 21m + 4m + 26 = -24, 25m = -24-26,
25m = -50,
m = -50/25 = -2, so substituting m= -2 in 2m = 3n -13,
2(-2)+13 = 3n,
-4+13 = 3n,
3n = 9, n = 9/3 = 3.

Question 5.
3x – 2y = 24
5x + 4y = -4
Answer:
The solution to the system of linear equations is x = 4 and y = -6,

Explanation:
Given Equation 1 as 3x -2y = 24 and Equation 2 as 5x +4y = -4,
to make elmination we need common values we multiply
Equation 1 by 2  so that y can be elminated,
2X(3x-2y =24)= 6x -4y = 48 adding with Equation 2
6x -4y =48
(+)5x +4y = -4
11x = 44,
x = 44/11 = 4, we substitute x = 4 in Equation 1 ,
3 X 4 – 2y = 24,
12- 2y = 24,
12-24 = 2y,
2y = -12, y = -12/2 = -6.

Question 6.
2x + 7y = -32
4x – 5y = 12
Answer:
The solution to the system of linear equations is y = -4 and x = -2,

Explanation:
Given 2x +7y = -32 Equation 1 and 4x -5y = 12 Equation 2,
to make elmination we need common values we multiply
Equation 1 by -2  so that x can be elminated,
-2 X(2x +7y=-32) = -4x – 14y = 64 now we add eqaution 1 with
equation 2 as -4x – 14y = 64
(+) 4x- 5y =12
                                 -19y = 76, y = -76/19 =-4, substituting y =-4
4x -5(-4)=12, 4x +20 = 12, 4x = 12-20 = -8, 4x = -8 , x =-8/4 = -2.

Solve each system of linear equations by using the substitution method.

Question 7.
2x + y = 5 — Equation 1
y = 4x – 7 — Equation 2
Substitute Equation 2 into Equation 1:

Answer:
The solution to the system of linear equations is x =2 and y =1,

Explanation:

Question 8.
4x + 3y = 23 — Equation 1
5x + y = 15 — Equation 2
Use Equation 2 to express y in terms of x:
5x + y = 15

Answer:
The solution to the system of linear equations is x =2 and y = 5,

Explanation:

Question 9.
3x – y = 8
2x + 3y = 9
Answer:
The solution to the system of linear equations is x = 3 and y = 1,

Explanation:
Given 3x – y =8 Equation 1 and 2x +3y = 9 Equation 2,
we can write Equation 1 as y = 3x -8,
substituting y in Equation 2 as
2x +3 (3x -8) = 9,
2x + 9x – 24 =9,
11x = 9+ 24 = 33, x = 33/11 = 3,so y = 3 X 3 – 8 = 9-8 = 1.

Question 10.
7m + 2n = -8
2m = 3n – 13
Answer:
The solution to the system of linear equations is m = -2 and n =3,

Explanation:
Given 7m +2n = -8  Equation 1 and 2m = 3n – 13 Equation 2,
to make elmination we need common values we multiply
Equation 1 by 3 and Equation 2 by 2 so that n can be elminated as
3(7m +2n =-8) = 21m +6n = -24 – equation 3and
2(2m = 3n -13) = 4m = 6n – 26 means 6n = 4m +26 substituting 6n in eqaution 3
21m +4m + 26 = -24,
25m = -24 -26 = -50 , m = -50/25 = -2, now
6n = 4(-2)+ 26 = -8 +26 = 18,
6n = 18, n = 18/6 = 3.

Solve each system of linear equations using elimination method or
substitution method. Explain why you choose each method.

Question 11.
2x + 3y = 29
2x – 17 = y
Answer:
The solution to the system of linear equations is x = 10 and y = 3,
used substitution method as already y is given in Equation 2,

Explanation:
Given 2x+3y = 29 as equation 1 and y already there ,
so we use substitution method as y = 2x -17  in equation 1
2x + 3(2x – 17) = 29,
2x + 6x – 51= 29,
8x = 29 +51= 80,
x = 80/8 = 10,
y = 2 X 10 -17 = 20 -17 = 3.

Question 12.
3a – 2b = 5
2a – 5b = 51
Answer:
The solution to the system of linear equations is a= -7 and b = -13,
Used elmination method because of no common values of a and b,

Explanation:
Given 3a- 2b =5 Equation 1 and 2a – 5b = 51 in Equation 2 in both
equations a or b is not common so we use elmination and we multiply
equation 1 by 2 2(3a-2b =5) = 6a – 4b = 10 and equation 2 by 3
3(2a -5b = 51) = 6a – 15b = 153 subtracting eqaution 2 from eqaution 1
6a-4b =10
(-) -6a +15b =-153
11 b = -143,
b = – 143/11 = -13, substituting in eqaution 1
3a -2(-13)=5,
3a +26 =5,
3a = 5 -26 = -21,
a = -21/3 = -7.

Math in Focus Course 3A Practice 5.2 Answer Key

Solve each system of linear equations using the elimination method.

Question 1.
2j + k = 6
j – k = 8
Answer:
The solution to the system of linear equations is j = 14/3 and k = -10/3,

Explanation:
Given 2j +k =6 equation 1,
j – k = 8 equation 2 adding 1 and 2
2j+j = 6 +8,
3j = 14,
j = 14/3,
k = j – 8 = 14/3 – 8 = 14- 24/3 = -10/3,.

Question 2.
2j + 3k = 11
2j – 5k = 3
Answer:
The solution to the system of linear equations is j =4 and k =1,

Explanation:
Given 2j +3k = 11 – Equation -1, 2j – 5k =3- Equation -2
subtracting eqaution 2 from 1
2j +3k =11
-2j +5k =  -3
8k = 8, k = 8/8 = 1, substuting k =1 in equtation 2
2j – 5 = 3, 2j = 3+5 = 8, j = 8/2 = 4.

Question 3.
3m + n = 30
2m – n = 20
Answer:
The solution to the system of linear equations is m = 10, n =0.

Explanation:
Given 3m + n = 30 Equation 1, 2m -n = 20 Equation 2
adding 1 and 2 we get
3m +n =30
2m – n =20
5m = 50, m = 50/5 = 10, substituting m = 10 in equation2
2 X 10 – n =20,
n = 20 – 20 = 0.

Question 4.
3x – y = 9
2x – y = 7
Answer:
The solution to the system of linear equations is x =2 and y = -3,

Explanation:
Given 3x -y = 9 Equation 1 and 2x -y =7 Equation 2,
subtracting eqaution 2 from 1
3x -y =9
-2x +y =-7
x = 2, substituting x =2 in 2(2)-y = 7,
y = 4 – 7 = -3,

Question 5.
5s – t = 12
3s + t = 12
Answer:
The solution to the system of linear equations is s = 3 and t =3,

Explanation:
Given 5s -t =12 as equation 1 and 3s +t = 12 equation 2 adding1 & 2
5s -t =12,
(+)3s+t =12
8s = 24, s = 24/8 = 3, substituting s = 3 in eqaution 2
t = 12 – 3s,
t = 12 – 3 X 3 = 12-9 = 3.

Question 6.
2b + c = 10
2b – c = 6
Answer:
The solution to the system of linear equations is b = 4 and c = 2,
Explanation:
Given 2b + c = 10 as equation 1 and
2b – c =6 as equation 2 adding euqtion 1 & 2
2b +c =10
(+) 2b -c =6
4b = 16, b = 16/4 = 4, substituting b = 4 in equation 2
2X4 -6 = c,
c = 8-6 = 2.

Question 7.
3m – n = 7
21m + 6n = -29
Answer:
The solution to the system of linear equations is m = 1/3 and n = -6,

Explanation:
Given 3m – n = 7 as equation 1 means n = 3m -7,
21m +6n = -29 is equation 2 substituting n = 3m -7 in  equation 2
21m +6(3m -7) = -29,
21m +18m – 42 = -29,
39m = -29+42,
39m = 13,
m = 13/39 = 1/3, substituting n = 3(1/3)-7 = 1-7 = -6.

Question 8.
7a + b = 10
2a + 3b = -8
Answer:
The solution to the system of linear equations is

Explanation:
Given 7a + b = 10 as equation 1 we get b = 10-7a,
we have 2a + 3b = -8 as equation 2 substituting b = 10-7a in
equation 2 –2a +3(10-7a) =-8,
2a + 30 – 21a = -8,
-19a = -8 -30
a = 38/19= 2, substituting a=2 in b = 10 -7 X2 = 10 – 14 = -4.

Question 9.
2p + 5q = 4
7p + 15q = 9
Answer:
The solution to the system of linear equations is p= -3 and q = 2,

Explanation:
Given 2p +5q = 4 as eqaution 1 and 7p +15 q = 9 as equation 2,
if we multiply eqauation 1 with 3 we get 15 q so that we can easily
subtract from equation 2 so
3 X (2p +5q = 4)= equation 1 becomes
6p + 15q = 12 now we subtract equation 2
(-) 7p -15q = -9
-p = 3, p = -3 substituting p =-3 in 2p+5q = 4,
2(-3)+5q = 4,
5q = 4+6 = 10, q = 10/5 = 2.

Solve each system of linear equations using the substitution method.

Question 10.
2j + k = 3
k = j – 9
Answer:
The solution to the system of linear equations is j = 4 and k = -5,

Explanation:
Given 2j +k = 3 as equation 1 and wehave k = j -9 substituting in
equation 1 2j+(j -9) = 3,
3j = 3 + 9 = 12,
j = 12/3 = 4, therefore k = 4 – 9 = -5.

Question 11.
2h + 3k = 13
h = 2k – 4
Answer:
The solution to the system of linear equations is h =2 and k = 3,

Explanation:
Given 2h +3k = 13 as equation 1 and h = 2k-4
substituting h in equation 1 as
2 (2k-4) +3k = 13,
4k – 8 +3k = 13,
7k = 13+8= 21,
k = 21/7 = 3,So h = 2 X 3 – 4 = 6-4 =2.

Question 12.
3m + b = 23
m – b = 5
Answer:
The solution to the system of linear equations is b = 2 and m =7,

Explanation:
Given 3m+ b = 23 as equation 1 and as m-b = 5 ,
b = m-5 substituting in equation 1  3m + m -5 = 23,
4m = 23+5, 4m = 28, m= 28/4 = 7,
b = 7 – 5= 2.

Question 13.
3h – k = 10
h – k = 2
Answer:
The solution to the system of linear equations is h =4 and k =2,

Explanation:
Given 3h – k =10 as equation 1 and h -k = 2 ,
k = h -2 substituting in eqaution 1 k,
3h -h +2 = 10,
2h = 10-2 = 8,
h = 8/2 = 4 therefore k = h -2 = 4-2 =2.

Question 14.
3s – t = 5
s + 2t = 4
Answer:
The solution to the system of linear equations is s =2 and t = 1,

Explanation:
Given 3s – t = 5 as equation 1 and s +2t = 4 means
s = 4-2t substituting s in equation 1 we get
3(4-2t)- t =5,
12 – 6t – t =5,
12 -7t =5,
7t = 12-5 = 7,
t =7/7 = 1, therefore s = 4 – 2X 1 = 4-2 =2.

Question 15.
2x + y = 20
3x + 4y = 40
Answer:
The solution to the system of linear equations is x = 8 and y = 4,

Explanation:
Given 2x +y = 20 as equation 1 and 3x +4y = 40 equation2,
Equation 1 can be written as  y = 20-2x and
sustituted in equation 2 as 3x +4 (20-2x) = 40,
3x + 80 – 8x = 40,
-5x = 40 -80= -40,
5x = 40,
x = 40/5 = 8, so y = 20 -2 X 8 = 20 -16 =4.

Question 16.
3x + 2y = 0
5x – 2y = 32
Answer:
The solution to the system of linear equations is x = 4 and y = -6,

Explanation:
Given 3x+ 2y = 0 as equation 1 and 5x -2y = 32 as equation 2,
adding equation 1 and 2 we get
3x +2y =0,
5x -2y = 32
8x = 32,
x = 32/8 = 4 substituting x = 4 in equation 1
3(4) +2y = 0,
2y = -12, y = -12/2 = -6.

Question 17.
5x – y = 20
4x + 3y = 16
Answer:
The solution to the system of linear equations is x = 4 and y =0,

Explanation:
Given 5x – y = 20 as equation 1 and 4x +3y = 16 as equation 2
From equation 1 we have y = 5x -20,
substituting y in equation 2
4x + 3(5x -20) = 16,
4x + 15x – 60 = 16,
19x = 16 +60,
19 x = 76,
x = 76/19 = 4, so y = 5 x 4 – 20 = 0.

Question 18.
3p + 4q = 3
\(\frac{1}{2}\) + q = 3p
Answer:
The solution to the system of linear equations is p =\(\frac{1}{3}\) and
q = \(\frac{1}{2}\),

Explanation:
Given 3p +4q = 3 as equation 1 and \(\frac{1}{2}\) + q = 3p as equation 2
q = 3p – (1/2) substituting in equation 1
3p+4(3p – 1/2) =3,
3p +12 p – 2 = 3,
15p = 3+2 = 5,
p = 5/15 = 1/3, now q = 3 X 1/3  -1/2 = 1 – 1/2 = 1/2,
so p =\(\frac{1}{3}\) and q = \(\frac{1}{2}\).

Solve each system of linear equations using the elimination method or
substitution method. Explain why you choose each method.

Question 19.
2x + 7y = 32
4x – 5y = -12
Answer:
The solution to the system of linear equations is x= 2 and y =4,
used elimination method to elminate x,

Explanation:
Given 2x +7y = 32 as equation 1 and we have
4x – 5y = -12 as equation 2 if we multiply equation 1 by 2 and
subtarct equation 2 we can eliminate x
2X(2x +7y) = 64,
4x +14 y = 64
-4x +5y = 12
19y = 76, y = 76/19 = 4, now 4x -5 X 4= -12,
4x – 20 = -12,
4x = 8, x = 8/4 = 2.

Question 20.
3x + 3y = 22
3x – 2y = 7
Answer:
The solution to the system of linear equations is x = 13/3 and
y =3 used elmination method to elminate x,

Explanation:
Given 3x +3y = 22 as equation 1 and 3x – 2y = 7 as
equation 2 subtracting equation 2 from 1 we can eliminate x
3x + 3y = 22,
-3x +2y = -7
5y = 15,  y = 15/5 = 3, substituting y =3 in eqaution 2
3x – 2 X 3 = 7,
3x = 7+6 = 13,
x = 13/3.

Question 21.
7m + 2n = 20
2m = 3n – 5
Answer:
The solution to the system of linear equations is m = 2 and n =3,
used subtitution method,

Explanation:
Given 7m + 2n = 20 as equation 1 and 2m = 3n -5,
so using substitution m = (3n-5)/2 in equation 1,
7 (3n- 5)/2 + 2n = 20
7 (3n -5) + 4n = 40,
21n – 35 + 4n = 40,
25n = 40 +35 = 75,
25n = 75, n = 75/25 = 3,
n = 3, so m = (3 X3 -5)/2 = (9-5)/2 = 4/2 = 2.

Question 22.
3h – 4k = 35
k = 2h – 20
Answer:
The solution to the system of linear equations is h = 15 and
k =10 used substitution method,

Explanation:
Given 3h – 4k = 35 as equation 1 and k =2h – 20,
substituting k in equation 1
3h – 4(2h – 20) = 35,
3h – 6h + 80  = 35,
– 3h + 80 = 35,
3h = 80-35 = 45,
3h = 45, h = 45/3 = 15, so k = 2 X 15 – 20 = 30-20 = 10.

Question 23.
2h + 7k = 32
3h – 2k = -2
Answer:
The solution to the system of linear equations is h = 2 and k = 4,
used elmination method,

Explanation:
Given 2h +7k = 32 as equation 1 and 3h – 2k = -2 as equation 2
multiplying equation 1 by 3 and eqaution 2 by 2 to elminate h,
3(2h+ 7k= 32) becomes 6h + 21 k = 96 and
2(3h – 2k=-2) becomes 6h – 4k = -4 substracting 2 from 1,
6h +21k = 96
-6h +4k = 4
25 k = 100,
k = 100/25 = 4, substituting k =4 in eqaution 2
3h – 2 X 4= -2,
3h = -2 + 8 = 6,
h = 6/3 = 2.

Question 24.
2m+4=3n
5m – 3n = -1
Answer:
The solution to the system of linear equations m = -11 and n -6,
used substitution method,

Explanation:
Given 2m+ 4 = 3n as equation 1 and 5m – 3n = -1 as equation 2,
substituting 3n in equation 2 as
5m -3(2m +4) =-1,
5m – 6m -12 = -1,
-m = -1 + 12 = 11, m = -11, therefore 3n = 2 (-11) + 4,
3n = -22 + 4 = -18,
n = -18/3 = -6.

Solve.

Question 25.
Math Journal Sam solves the following system of linear equations by the elimination method,
without using calculator.

He can multiply the first equation by 3 and the second equation by 2 in order to eliminate x.
Or he can eliminate y by multiplying the first equation by 17 and the second equation by 3.
Which way should Sam choose? Explain.
Answer:
Sam can either choose any of the methods by
using method 1 he can elminate x or
by using method 2 he can elminate y and find solutions,

Explanation:
Given in method 1 sam can multiply the first equation by 3 and the
second equation by 2 in order to eliminate x  and in method 2
he can eliminate y by multiplying the first equation by 17 and
the second equation by 3 therefore Sam can either choose any of the method by
using method 1 he can elminate x or
by using method 2 he can elminate y and find solutions,
Method 1 :
3 X (2x + 3y = 1) and 2 X (3x – 17 y =23)
6x + 9y =3  and
(-)6x – 34y = 46
43 y = -43, y = -43/43, y = -1, substuting y = -1 in 2x +3y = 1,
2x + 3X -1 = 1,
2x – 3 = 1,
2x = 1+3 = 4, x = 4/2 = 2.
Method 2:
17 X (2X +3y = 1) and 3 ( 3x – 17y =23),
34x + 51 y = 17
9x – 51 y = 69
43x = 86, x = 86/43 = 2, x = 2 substituting  2 X 2 + 3y = 1,
4 +3y = 1, 3y = 1-4 = -3, 3y = -3, y = -3/3 = -1,
therefore by both methods we got x =2 and y = -1 ,
Sam can either choose any of the methods.

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 5 Lesson 5.3 Real-World Problems: Systems of Linear Equations to score better marks in the exam.

Math in Focus Grade 7 Course 3 A Chapter 5 Lesson 5.3 Answer Key Real-World Problems: Systems of Linear Equations

Math in Focus Grade 8 Chapter 5 Lesson 5.3 Guided Practice Answer Key

Solve using systems of linear equations.

Question 1.
Two bowls and one cup have a mass of 800 grams. One bowl and two cups have a mass of 700 grams. Find the mass of a bowl and the mass of a cup.
Let the mass of a bowl be b grams and the mass of a cup be c grams.
Mass of two bowls and on cup:  =  – Equation 1

Mass of one bowl and two cups:  =  – Equation 2

Use Equation 1 to express c in terms of b:

The mass of a bowl grams and the mass of a cup is grams.
Answer:
The mass of a bowl is 300 grams and
the mass of a cup is 200 grams,

Explanation:

Solve using systems of linear equations.

Question 2.
Elizabeth is thinking of a two-digit number. When the tens digit is subtracted
from the ones digit, the difference is 2. One-fifth of the number is 1 less than
the sum of the digits. What is the number?
Answer:
Let the tens digit be c and the ones digit be d.
Tens digit subtracted from ones digit:

If the tens digit is c and the ones digit is d, then the value of the number is 10c + d.

Substitute 3 for c into Equation 3:
d = 5
=
The number is .
Answer:

Question 3.
Christopher is thinking of two positive integers.
The sum of the integers is 27. When twice the first integer is added to
half of the second integer, the sums 24. Find i the integers.

Answer:
The integers are x =7 and y = 20,

Explanation:
Given Christopher is thinking of two positive integers.
Let numbers be x and y , x+y = 27,
When twice the first integer is added to
half of the second integer, the sums 24,
2x +y/2 = 24,
x = 27-y,
2(27-y) + y/2 = 24,
4(27-y)+y = 48,
108-4y +y = 48,
108-3y = 48,
3y = 108-48,
3y = 60,
y = 60/3 = 20 and x = 27-20 = 7.

Math in Focus Course 3A Practice 5.3 Answer Key

Solve using systems of linear equations.

Question 1.
Jean stocked her aquarium with 36 fresh-water fish, which cost $212.
The male fish cost $5 each, while the female fish cost $7 each.
Find the number of male fish and the number of female fish.

Answer:
The number of male fish is 20 and the number of female fish is 16,

Explanation:
Given Jean stocked her aquarium with 36 fresh-water fish, which cost $212.
The male fish cost $5 each, while the female fish cost $7 each.
Lets take male fish as M and female fish as F,
M + F = 36, M = 36-F and 5M +7F = 212,
5(36-F) +7F = 212,
180- 5F +7F = 212,
2F = 212-180= 32,
F = 32/2 = 16, So M = 36 – 16 = 20.
The number of male fish is 20 and the number of female fish is 16.

Question 2.
Seventy concert tickets were sold for $550. Each adult ticket cost
$9 and each children’s ticket cost $5.
Find the number of adult tickets and the number of children’s tickets sold.
Answer:
50 is the number of adult tickets and 20 is the number of children’s tickets sold,

Explanation:
Given Seventy concert tickets were sold for $550. Each adult ticket cost
$9 and each children’s ticket cost $5. Let number of adult tickets be a and
number of children tickets is c, a + c = 70, a = 70 -c, 9a +5c = 550,
9(70-c) + 5c = 550,
630 – 9c +5c = 550,
630-550 = 4c,
80 = 4c,
c = 80/4 = 20,
so a= 70 – 20 = 50 tickets.
50 is the number of adult tickets and 20 is the number of children’s tickets sold.

Question 3.
George paid $2.75 for 4 granola bars and 1 apple.
Addison paid $2.25for 2 granola bars and 3 apples.
Find the cost of each granola bar and each apple.
Answer:
The cost of each granola is $0.6 and each apple is $0.35,

Explanation:
Given George paid $2.75 for 4 granola bars and 1 apple.
Addison paid $2.25 for 2 granola bars and 3 apples.
Lets take granola as g and apple as a
4g+ a = 2.75, a = 2.75 -4g
2g +3a = 2.25, substituting
2g +3 (2.75 -4g)= 2.25,
2g +8.25 – 12g = 2.25,
10g = 8.25-2.25,
10g = 6,
g = 6/10= $0.6,
a = 2.75 – 4 X 0.6 = 2.75 – 2.4 = $0.35,
The cost of each granola is $0.6 and each apple is $0.35.

Question 4.
4 thumb drives and 1 compact disk have a total capacity of 9 gigabytes.
5 thumb drives have 9 gigabytes more capacity than 1 compact disk.
Find the capacity of 1 thumb drive and the capacity of 1 compact disk.
Answer:
The capacity of 1 thumb drive is 2GB and
the capacity of 1 compact disk is 1 GB,

Question 5.
A book cover has the length and width (in inches) shown in the diagram.

a) Find the values of a and b.
Answer:
a=4in and b = 2in,

Explanation:
We have equations  (3a-b) = 2+a+2b,
3a – a = 2+2b +b,
2a = 2 + 3b,
2a – 3b = 2  equation 1
and 3b = 2+a means  2a -2-a =2,
a =4 so 3b = 2+4 = 6, b= 6/3 = 2. ,
therefore a = 4in and b = 2in.

b) Find the perimeter of the book cover.
Answer:
Peimeter of the book cover is 32in,

Explanation:
Given l = 3a-b = 3 x 4 – 2 = 12-2 = 10, width = 3 b = 3 X 2 = 6,
perimeter = 2(l+w) = 2(10+ 6)= 2(16)= 32 in.

Question 6.
Eileen saves dimes and quarters. She has 40 coins,
which totaled $6.55, in her bank. How many of each coin does she have?
Answer:
Eileen has 23 dimes and 17 quarters,

Aiden is a percussionist in his school band. One instrument he
plays is in the shape of an equilateral triangle shown below.
The side lengths are in inches. Find x and y.

Answer:
x = -1 and y = -5,
Explanation:
Given Aiden is a percussionist in his school band. One instrument he
plays is in the shape of an equilateral triangle so
3x – 2y = 7,
3x = 2y +7,
x = (2y +7)/3,
8x – 3y = 7 substituting x
8(2y +7)/3 = 7+3y,
16y +56 = 3(7 +3y)
16y +56 = 21 +9y
16y -9y = 21-56,
7y = -35, y = -35/7 = -5,x=(2X -5+7)/3 = (-10+7)/3 = -3/3 = -1.

Question 8.
Mrs. Green gave three riddles for her class to solve.

a) The sum of the digits of a two-digit number is 11.
Twice the tens digit plus 2 equals ten times the ones digit. What is the number?
Answer:
The number is 92,

Explanation:
Given the sum of the digits of a two-digit number is 11.
x + y = 11,y = 11-x
Twice the tens digit plus 2 equals ten times the ones digit
2x + 2 = 10y,
2x + 2 = 10(11-x),
2x +2 = 110 -10x,
2x +10x = 110-2,
12x = 108, x = 108/12 = 9, so y = 11-9 = 2,
the number is 92.

b) There are two numbers. The first number minus the second number is 15.
One-third of the sum of the numbers is one quarter of the first number.
What are the two numbers?
Answer:
The two numbers are 12 and -3,

Explanation:
Let x and y be the numbers ,
The first number minus the second number is 15. x – y = 15,
(x +y)/3 = x/4,
4x + 4y = 3x,
x = -4y, substituting  -4y – y = 15,
-5y = 15, y = -15/5 = -3,
x = – 4(-3) = 12.
The two numbers are 12 and -3.

c) A two-digit number is 1 more than eight times the sum of the digits.
The ones digit is 3 less than the tens digit. What is the number?
Answer:
The number is 41,

Explanation:
Given A two-digit number is 1 more than eight times the sum of the digits.
10x + y = 8(x+y)+1
The ones digit is 3 less than the tens digit means x = y +3,
10(y +3)+y = 8(y+3) +8y +1,
10 y + 30 +y = 8y + 24 +8y +1,
11y + 30 = 16 y +25,
16y – 11y = 30 -25,
5y = 5, y = 5/5 = 1, x = 1+ 3 = 4,
The number is 41.

Question 9.

On Saturday, $585 was collected from the sale of 55 tickets for a performance.
The table below shows the information about the sale of the tickets.
Find the number of adult tickets and the number of student’s tickets sold on that day.

Answer:
30 number of adult tickets and 25 number of student’s tickets sold on that day,

Explanation:
Given Students tickets cost 12 X .25 = 3, 12-3 =9 is the cost for student tickets,
Let a be the number of adult tickets and s the number of
student’s tickets sold on that day, a + s = 55 , s = 55-a and
12a +9s = 585,
12a+9(55 -a) = 585,
12a + 495 -9a = 585,
3a = 585- 495= 90,
a= 90/3 = 30, s = 55-30 = 25.
30 number of adult tickets and 25 number of student’s tickets sold on that day.

Question 10.
Eight years ago, Mr. Fontana was six times as old as his son.
In twelve years’ time, he will be twice as old as his son. How old are they now?

Answer:
Fathers age is 38 and sons age is 13 years old,
Explanation:
Let f be fathers age now and s be sons age now,
Eight years ago, Mr. Fontana was six times as old as his son.
f – 8 = 6(s -8), In twelve years’ time, he will be twice as old as his son,
f + 12 = 2(s+12),
f = 2s +24-12 = 2s+12,
2s +12 – 8 = 6s – 48,
2s +4 = 6s -48,
6s-2s = 48+4,
4s = 52,
s = 52/4 = 13, f = 2 X 13 +12 = 26 +12 = 38.
Fathers age is 38 and sons age is 13 years old.

Question 11.
A restaurant sells four combo meals. Jolly Meal, which cost $12.60,
consists of 2 yogurt cups and 1 sandwich. The $pecial Meal, which is
made up of 2 sandwiches and 1 yogurt cup, cost $13.50.
Calculate the cost of the following combo meals if the charge for
sandwiches and yogurt cups are the same for all combo meals.
a) Children’s Meal: 1 sandwich and 1 yogurt cup
Answer:
Children’s Meal: 1 sandwich and 1 yogurt cup is $8.7,

Explanation:
Given a restaurant sells four combo meals.
Jolly Meal, which cost $12.60, consists of 2 yogurt cups and 1 sandwich.
The $pecial Meal, which is made up of 2 sandwiches and 1 yogurt cup, cost $13.50.
Let sandwiches is s and yogurt is y
Jolly Meal 2y +s = $12.60, s = 12.60 – 2y,
2s +y = $13.50,
2 (12.60 -2y)+y =13.50,
25.2 – 4y +y = 13.50,
25.2-13.50 = 3y,
3y = 11.7,y = 11.7/3 = 3.9,
so s = 12.60 – 2(3.9) = 12.60 – 7.8 = 4.8,
therefore 1 sandwich and 1 yogurt cup is $4.8 + $3.9 = $8.7.

b) Family Meal: 2 sandwiches and 3 yogurt cups
Answer:
Family Meal: 2 sandwiches and 3 yogurt cups is $21.3,

Explanation:
We got sandwiches s = $4.8, yogurt y = $3.9,
so for 2 sandwiches and 3 yogurt cups it is
2X $4.8 and 3 X $3.9 = $9.6 + $11.7 = $21.3.

Question 12.
In a boat race, Jenny’s team rowed their boat from point A to point 8 and back to point A.
Points A and 8 are 30 miles apart. During the race,
there was a constant current flowing from A to 8.
She took 2 hours to travel from A to 8 and 2.5 hours to travel from 8 to A.

a) Calculate the speed of the boat from A to 6 and the speed from 8 to A.
Answer:
The effective speed of the boat downstream is 15 mph,
The effective speed of the boat upstream is 12 mph,
Explanation:
Given one way distance is 30 miles,
Down stream travel took 2 hours,
and Up stream travel took is 2.5 hours,
The effective speed downstream was 30/2 = 15 miles per hour,
The effective speed upstream was 30/2.5 = 12 miles per hour,
The effective speed downstream is u + v, where u is the speed of the
boat in the still water and v is the current speed,
The effective speed up stream is u – v,
u + v = 15, u – v = 12, v = u -12,
u + (u -12) = 15,
2u = 15+ 12 = 27, u = 27/2 = 13.5 mph,
v = 13.5 – 12 = 1.5 mph

b) Find the speed of the boat from A to B if there was no current.
Answer:
The speed of the boat in still water is 13.5 mph

c) Find the speed of the current.
Answer:
The speed of the current is 1.5 mph

Question 13.
Alex deposited $3,500 in two banks. The first bank paid 2% simple interest
per year while the second bank paid 3%. At the end of one year,
the difference in the interest amounts was $30. Find the amount Alex deposited in each bank.
Answer:
Alex deposited $2,700 in one bank and $800 in another bank,

Explanation:
Given Alex deposited $3,500 in two banks,
x + y = 3500,so y = 3500-x
The first bank paid 2% simple interest
per year while the second bank paid 3% and
At the end of one year,
the difference in the interest amounts was $30.
I = pRT interest of x is x X 0.02,
Interest of y is y X 0.03,
0.02x-0.03 y = 30 multiplying both sides by 100,
2x – 3y = 3000,substituting x
2x – 3(3500 – x) = 3000,
2x – 10,500  +3x = 3000,
5x = 10,500 + 3,000= 13,500,
x = 13,500/5 = $2700, y = 3500 – 2700 = $800,
Alex deposited $2,700 in one bank and $800 in another bank.

Question 14.
Greg’s favorite activities are bike riding and hip-hop dancing.
He wants to try to do these activities for 6 hours each week.
He would also like to try to burn off about 2,100 calories a week.
He finds that biking uses about 425 calories each hour and
hip-hop dancing uses about 325 calories each hour.
He wants to do bike riding for x hours and hip-hop dancing for y hours each week.

a) Copy and complete the following table.

Answer:

Explanation:
Copied and completed the table as shown above.

b) How much time should he spend doing each activity to accomplish his goal?
Answer:
Greg’s wants to do bike riding for 3/2 or 1.5 hours and
hip-hop dancing for 9/2 or 4.5 hours eqch week,

Explanation:
Given Greg’s favorite activities are bike riding and hip-hop dancing.
He wants to try to do these activities for 6 hours each week.
He would also like to try to burn off about 2,100 calories a week.
He finds that biking uses about 425 calories each hour and
hip-hop dancing uses about 325 calories each hour.
Let x hours is bike riding and y hours br hip hop dancing so
x + y = 6, y = 6 – x and 425x + 325y = 2,100 substituting y
425x +325(6-x) =2100,
425x + 1950 – 325x = 2100,
100x = 2100-1950=150,
x = 150/100 = 3/2, y = 6-3/2 = 9/2,
So Greg’s wants to do bike riding for 3/2 or 1.5 hours and
hip-hop dancing for 9/2 or 4.5 hours each week.

Question 15.
In an experiment, Matthew was given two saline (salt) solutions,
whose concentrations are shown below. He had to prepare 10 fluid ounces of
a new saline solution with a 27% concentration from the two solutions.
Calculate the volume of each solution used to prepare the new solution.

Answer:
The volume of each solution used to prepare the new solution is
x as 7.5 fl oz and y as 2.5 fl oz,

Explanation:
Given Matthew was given two saline (salt) solutions,
whose concentrations are 30 and 18. He had to prepare 10 fluid ounces of
a new saline solution with a 27% concentration from the two solutions.
The volume of each solution used to prepare the new solution is
30x +18 y = 270 and x + y = 10, y = 10 -x,
30 x+18(10-x)=270,
30x +180 – 18x = 270,
12 x = 270-180,
12x = 90, x= 90/12 = 7.5 and y= 10-7.5 = 2.5, therefore
the volume of each solution used to prepare the new solution is
x as 7.5 fl oz and y as 2.5 fl oz.

Practice the problems of Math in Focus Grade 8 Workbook Answer Key Chapter 5 Lesson 5.4 Solve Systems of Linear Equations by Graphing to score better marks in the exam.

Math in Focus Grade 7 Course 3 A Chapter 5 Lesson 5.4 Answer Key Solve Systems of Linear Equations by Graphing

Math in Focus Grade 8 Chapter 5 Lesson 5.4 Guided Practice Answer Key

Technology Activity

Materials:

• graphing calculator

Explore The Graphical Method

Work in pairs.

Step 1.
Solve the system of linear equations using the elimination method or substitution method.
x + 2y = 4
x – y = 1

Step 2.
To solve this system of linear equations using a graphing calculator, solve each equation for y and enter each expression for y into the calculator.

Step 3.
Press the key. Use the function and select 5:lntersect to find
where the two graphs intersect.

Caution
Be sure to use parentheses around any fractional coefficients, and use the key if the coefficient is negative.

Step 4.
Repeat step 1 to step 3 for the system of linear equations
6x – 5y = -3,
x + y = 5.

Math Journal How is the solution you found in step 1 related to the coordinates of the point of intersection in step 3? Why do you think this happens?

Solve using the graphical method. Copy and complete the tables of values. Graph the system of linear equations on the same coordinate plane. Use 1 grid square on both axes to represent 1 unit for the x interval from -1 to 3 and the y interval from -1 to 5.

Question 1.
2x + y = 5
x – y = -2

2x + y = 5

x – y = -2

From the graph, the slope of 2x + y = 5 is and the y-intercept is . The slope of x – y = -2 is and its y-intercept is .
The point of intersection is (, ). Therefore, the solution to the system of equations is x = and y = .

Solve using the graphical method. Graph each system of linear equations on the same coordinate plane. Use 1 grid square on both axes to represent 1 unit for the Interval from -4 to 4.
Answer:
2x + y = 5
x – y = -2
We are given the system of equations:

For the equation 2x + y = 5 we build the table of values:

For the equation x – y = -2 we build the table of values:

We graph the two lines:
From the graph, the slope of 2x + y = 5 is -2 and the y-Intercept is 5.
The slope of x – y = -2 is 1 and the y-Intercept is 2.
The point of intersection is (1, 3). Therefore the solution of the system of equations is x = 1 and y = 3.
x = 1; y = 3

Question 2.
x + 2y = 5
x + y = 2
Answer:
x + 2y = 5
x + y = 2
We are given the system of equations:

For the equation x + 2y = 5 we build the table of values:

For the equation x + y = 2 we build the table of values:

We graph the two lines:
From the graph, the slope of x + 2y = 5 is -0.5 and the y-Intercept is 2.5. The slope of x + y = 2 is -1 and the y-Intercept is 2.
The point of intersection is (-1, 3). Therefore the solution of the system of equations is x = -1 and y = 3.
x = -1; y = 3

Question 3.
3x + 2y = 8
x – y = 1
Answer:
3x + 2y = 8
x – y = 1
We are given the system of equations:

For the equation 3x + 2y = 8 we build the table of values:

For the equation x – y = 1 we build the table of values:

We graph the two lines:
From the graph, the slope of 3x + 2y = 8 is -1.5 and the y-Intercept is 4. The slope of x – y = 1 is 1 and the y-Intercept is -1.
The point of intersection is (2, 1). Therefore the solution of the system of equations is x = 2 and y = 1.
x = 2; y = 1

Use graph paper. Use 1 grid square on both axis to represent 10 seconds for the t interval from 110 to 150 seconds, and 10°F for the T interval from 300 to 350°F.

Question 4.
Two ovens are being heated. Their temperatures are represented by the equations
Oven 1: T = t + 200
Oven 2: T = 2t + 80
where T is the temperature in °F of the oven and t is the time in seconds. Solve the system of linear equations graphically. When will the temperatures of the ovens be the same?
Answer:
T = t + 200
T = 2t + 80
We are given the system of equations:
For the equation T = t + 200 the slope is 1 and the y-intercept is 200.
For the equation T = 2t + 80 the slope is 2 and the y-intercept is 80.

We sketch the graph of the two equations using the slope and the y-irtercept values:
The point of intersection is (120, 320). Therefore the solution of the system of equations is t = 120 and T = 320.
t = 120 seconds; T = 320° F

Math in Focus Course 3A Practice 5.4 Answer Key

For this practice, unless otherwise stated, use 1 grid square to represent 1 unit on both axes for the interval from —8 to 8. Solve each system of linear equations using the graphical method.

Question 1.
x + y = 6
2x + y = 8
Answer:
x + y = 6
2x + y = 8
We are given the system of equations:

a) Copy and complete the tables of values for the system of linear equations, x + y = 6

Answer:
For the equation x + y = 6 we build the table of values:

For the equation 2x + y = 8 we build the table of values:

b) Graph x + y = 6 and 2x + y = 8 on the same coordinate plane. Find the point of intersection.
Answer:
We graph the two lines:
From the graph, the slope of x + y = 6 is -1 and the y-intercept is 6
The slope of 2x + y = 8 is -2 and the y-intercept is 8.
The point of intersection is (2, 4).

c) Use the graph in b) to solve the system of linear equations.
Answer:
Therefore the solution of the system or equations is x = 2 and y = 4.
x = 2; y = 4

Question 2.
x + y = 5
x – y = 2

a) Copy and complete the tables of values for the system of linear equations, x + y = 5

Answer:
x + y = 5
x – y = 2
We are given the system of equations:

For the equation x + y = 5 we build the table of values:

For the equation x – y = 2 we build the table of values:

We graph the two lines:

b) Graph x + y = 5 and x — y = 2 on the same coordinate plane. Find the point of intersection.
Answer:
From the graph, the slope of x + y = 5 is -1 and the y-intercept is 5
The slope of x – y = 2 is 1 and the y-intercept is -2.
The point of intersection is (3.5, 1.5)

c) Use the graph in b) to solve the system of linear equations.
Answer:
Therefore the solution oi me system or equations is x = 3.5 and y = 1.5.
x = 3.5; y = 1.5

Question 3.
x + 2y = 5
2x – 2y = 1
a) Graph x + 2y = 5 and 2x — 2y = 1 on the same coordinate plane. Find the point of intersection of the graphs.
Answer:
x + 2y = 5
2x – 2y = 1
We are given the system of equations:

For the equation x + 2y = 5 we build the table of values:

For the equation 2x — 2y = 1 we build the table of values:

We graph the two lines:

b) Use the graph in a) to solve the system of linear equations.
Answer:
From the graph, the slope of x + 2y = 5 is -0.5 and the y-intercept is 25.
The slope of 2x – 2y = 1 is 1 and the y-intercept is -0.5.
The point of intersection is (2, 1.5). Therefore the solution of the system of equations is x = 2 and y = 1.5.
x = 2; y = 1.5

Question 4.
2x + 3y= -1
x – 2y = 3

a) Graph 2x + 3y = -1 and x – 2y = 3 on the same coordinate plane. Find the point of intersection of the graphs.
Answer:
2x + 3y = -1
x – 2y = 3
We are given the system of equations:

For the equation 2x + 3y = -1 we build the table of values:

For the equation x — 2y = 3 we build the table of values:

We graph the two lines:

b) Use the graph in a) to solve the system of linear equations.
Answer:
From the graph, the slope of 2x + y = 5 is –\(\frac{2}{3}\) and the y-intercept is –\(\frac{1}{3}\)
The slope of x – 2y = 3 is 0.5 and the y-intercept is -1.5.
The point of intersection is (1, -1). Therefore the solution of the system of equations is x = 1 and y = -1.
x = 1: y = -1

Solve each system of equations using the graphical method.

Question 5.
x = 2y
y = x + 2
Answer:
x = 2y Equation 1
y = x + 2 Equation 2
We are given the system of equations:
y = 0.5x We rewrite Equation 1 in slope-intercept form:
y = x + 2 Equation 2 is written in slope-intercept form:

We graph the two equations: the first has slope 0.5 and y-intercept 0, while the second has slope 1 and y-intercept 2:
The point of intersection is (-4, -2). Therefore the soLution of the system of equations is x = -4 and y = -2.

Question 6.
y = 3
y = 2x + 1
Answer:
y = 3 Equation 1
y = 2x + 1 Equation 2
We are given the system of equations:
y = 3 Equation 1 is written in slope-intercept form:
y = 2x + 1 Equation 2 is written in slope-intercept form:

We graph the two equations: the first has slope 0 and y-intercept 3 (horizontal line), while the second has slope 2 and y-intercept 1:
The point of intersection is (1, 3). Therefore the solution of the system of equations is x = 1 and y = 3.
x = 1; y = 3

Question 7.
x = 2
y = 2x – 8
Answer:
x = 2 Equation 1
y = 2x — 8 Equation 2
We are given the system of equations:
x = 2 Equation 1 is the equation of a vertical line:
y = 2x – 8 Equation 2 is written in slope-intercept form:

We graph the two equations: the first is vertical passing through (2, 0), while the second slope 2 and y-intercept-8:
The point of intersection is (2, -4). Therefore the solution of the system of equations is x = 2 and y = -4.
x = 2 ; y = -4

Question 8.
3x – 2y = 19
3y = 2x – 21
Answer:
3x – 2y = 19 Equation 1
3y = 2x — 21 Equation 2
We are given the system of equations:
3x – 2y – 3x = 19 – 3x We rewrite Equation 1 in slope-intercept form:
-2y = -3x + 19
y = \(\frac{3}{2}\)x – \(\frac{19}{2}\)
y = \(\frac{3}{2}\)x – \(\frac{19}{2}\)
y = \(\frac{2}{3}\)x – 7
We rewrite Equation 2 in slope-intercept form:

We graph the two equations: the first has slope \(\frac{3}{2}\) and y-intercept –\(\frac{19}{2}\), while the second has slope \(\frac{1}{2}\) and y-intercept -7:
The point of intersection is (3, -5) Therefore the solution of the system of equations is x = 3 and y = -5.
x = 3; y = -5

Question 9.
2x + y = 11
x + 3y = 18
Answer:
2x + y = 11 Equation 1
x + 3y = 18 Equation 2
We are given the system of equations:
2x + y — 2x = 11 – 2x We rewrite Equation 1 in slope-intercept form:
y = 11 — 2x
x + 3y — x = 18 – x We rewrite Equation 2 in slope-intercept form:
3y = 18 – x
y = –\(\frac{1}{3}\)x + 6

We graph the two equations: the first has slope -2 and y-intercept 11, while the second has slope –\(\frac{1}{3}\) and y-intercept 6:
The point of intersection is (3, 5). Therefore the solution of the system of equations is x = 3 and y = 5
x = 3, y = 5

Question 10.
x + 2y = 1
4y – x = 17
Answer:
x + 2y = 1 Equation 1
4y — x = 17 Equation 2
We are given the system of equations:
x + 2y – x = 1 – x We rewrite Equation 1 in slope-intercept form:
2y = -x + 1
y = –\(\frac{1}{2}\)x + \(\frac{1}{2}\)
4y — x + x = 17 + x We rewrite Equation 2 in slope-intercept form:
4y = x + 17
y = \(\frac{1}{4}\)x + \(\frac{17}{4}\)

We graph the two equations: the first has slope –\(\frac{1}{2}\) and y-intercept \(\frac{1}{2}\), while the second has slope \(\frac{1}{4}\) and y-intercept \(\frac{17}{4}\)
The point of intersection is (-5, 3) Therefore the solution of the system of equations is x = -5 and y = 3.
x = -5; y = -3

Solve. Show your work.

Question 11.
Marianne jogged from point P to point Q while Walter jogged from point Q to point P. Point P and point Q are 7.5 kilometers apart. Marianne’s motion is represented by d = 8t and Walter’s motion is represented by 2d + 14t = 15, where t hours is the time and d kilometers is the distance from point P.

a) Solve the system of linear equations using the graphical method.
Answer:
We are given the system of equations:
d = 8t Equation 1
2d + 14t = 15 Equation 2
d = 8t a) Equation 1 is written in slope-intercept form:
2d + 14t — 14t = 15 — 14t We rewrite Equation 2 in slope-intercept form:
2d = -14t + 15
d = -7t + 7.5
img 36
We graph the two equations the first has slope 8 and y-intercept 0, while the second has sope -7 and y-
intercept 7.5:
The point of intersection is (0.5, 4). Therefore the solution of the system of equations is t = 0.5 and
d = 4.

b) When did Marianne and Walter meet? How far from point Q did they meet?
Answer:
t = 0.5 hours b) The t-coordinate of the intersection point s the time after which they met
d = 4 km The d-coordinate of the intersection point is the distance from point P when they met:
7.5 — 4 = 3.5 km The distance from point Q is
a) t = 0.5; d = 4
b) 0.5 hours; 3.5 km

Question 12.
Math Journal Explain when it is convenient to use each method of solving a system of linear equations: Elimination, substitution, and graphical. Give an example for each method.
Answer:
3x – 4y = 25
2x + 5y = 77
The elimination method is used when none of the variables has coefficient 1.
For example:
y = 2x + 3
3x – 2y = 7 The substitution method ¡s used when one of the variables has coefficient 1
For example:
y = 2x + 1
y = -x + 2
The graphical method is used when one of the variables has coefficient 1 in both equations.
For example:
The choice depends on the coefficients of the variables

Question 13.
Two cyclists are traveling along a track in the same direction. Their motions are described by the linear equations d = 10t and d – 8t = 2, where t hours is the time and d miles is the distance from point A on the-Track.
Answer:
d = 10t Equation 1
d – 8t = 2 Equation 2
We are given the system of equations:

a) Solve the system of linear equations using a graphing calculator.
Answer:
d = 8t a) Equation 1 is written in sLope-intercept form:
d — 8t + 8t = 2 + 8t We rewrite Equation 2 in slope-intercept form:
d = 8t + 2

We graph the two equations using a graphing calculator:
3d?
The point of intersection is (1,10). Therefore the solution of the system of equations is t = 1 and d = 10.

b) When will the cyclists meet?
Answer:
t = 1 hour b) The t-coordinate of the intersection point is the time after which they met
a) t = 1; d = 10
b) 1 hour

Question 14.
Dr. Murray is heating a beaker containing Liquid A and a beaker containing Liquid B. The temperature of Liquid A is represented by T = 2t + 140 and the temperature of Liquid B is represented by T = t + 160, where T°F is the temperature of the liquid after t seconds.
Answer:
T = 2t + 140 Equation 1
T = t + 160 Equation 2
We are given the system of equations:

a) Solve the system of linear equations using a graphing calculator.
Answer:
Equation 1 and Equation 2 are written in slope-intercept

We graph the two equations using a graphing calculator:
The point of intersection is (20, 180). Therefore the solution of the system of equations is t = 20 and T = 180.

b) When will the temperatures of the liquids be the same?
Answer:
t = 20 seconds b) The t-coordinate of the intersection point is the time after which the temperatures are
the same:
a) t = 20;T = 180;
b) 20 seconds

Question 15.
A carpenter is hammering an iron nail and another carpenter is hammering a copper nail. The temperatures of the nails increase during the hammering. The temperature of the iron nail is represented by T = 2t + 70 and the temperature of the copper nail is represented by T = 3t + 65, where T°F is the temperature of the nail after t seconds.
Answer:
T = 2t + 70 Equation 1
T = 3t + 65 Equation 2
We are given the system of equations:

a) Solve the system of linear equations using a graphing calculator.
Answer:
Equation 1 and Equation 2 are written in slope-intercept form.

We graph the two equations using a graphing calculator
The point of intersection is (5, 80). Therefore the solution of the system of equations is t = 5 and T = 80.

b) When will the temperatures of the nails be the same?
Answer:
t = 5 seconds b) The t-coordinate of the intersection point is the time after which the temperatures are
the same:

a) t = 5; T = 80;
b) 5 seconds