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This handy Math in Focus Grade 8 Workbook Answer Key Chapter 9 Lesson 9.1 Understanding and Applying Congruent Figures detailed solutions for the textbook questions.

Math in Focus Grade 8 Course 3 B Chapter 9 Lesson 9.1 Answer Key Understanding and Applying Congruent Figures

Math in Focus Grade 8 Chapter 9 Lesson 9.1 Guided Practice Answer Key

Identify the figures that seem congruent. Explain why they seem so.

Question 1.

Answer: Figures A and F seems congruent because they have the same shape and same size.

Question 2.

Answer: C and D seems congruent because they have the same shape and same size.

Complete.

Question 3.
The quadrilaterals below are congruent. Write the statement of congruence.

So, the statement of congruence is PQRS ≅ ____________.
Answer:
The corresponding angles from the above figure is
∠P = ∠U
∠S = ∠T
∠R = ∠W
∠S = ∠V
So, the statement of congruence is PQRS ≅ UTWV.

Question 4.
To make two congruent quadrilaterals, Wendy cut two pieces of cardboard, one on top of the other. She measured the lengths of some sides of the quadrilaterals.
a) Find the values of the variables x, y, and z. All lengths are in inches.

Solve for x:
x = LM =
Solve for y.
2y – 5 = IJ =
2y – 5 + 5 = Add 5 to both sides.
2y = Simplify.
2y – y = Subtract y from both sides.
y = Simplify.

Solve for z:
x – 6 = JG =
– 6 = Substitute x = .
= z Simplify.
So, the value of x is , y is , and z is .
Answer:

Solve for x:
x = 12 LM = 12
Solve for y.
2y – 5 = y + 2 IJ = NK
2y – 5 + 5 = y + 2 + 5 Add 5 to both sides.
2y = y + 7 Simplify.
2y – y = 7 Subtract y from both sides.
y = 7 Simplify.

b) What is the length of each side of GHIJ? of NKLM?
GH = in., HI = in., IJ = in., JG = in.
NK = in., KL = in., LM = in., MN = in.
Answer:
GH = 12 in., HI = 10 in., IJ = 7 in.,
JG = in.
x – 6 = z
12 – 6 = z
z = 6
NK = 7 in., KL = 6 in., LM = 12 in.,
MN = 10 in.
2z – 2 = 2 (6) – 2 = 12 – 2 = 10

Technology Activity

Materials:
geometry software

OBSERVE THE CONGRUENCE IN TRIANGLES

STEP 1: Draw △ABC and △DEFso that m∠CAB m∠FDE, AB = DE, and AC = DF. Draw the two triangles such that point A is 10 units to the left of point D.

STEP 2: Translate △ABC 10 units to the right. Select the Translate function within the Transform menus. Do the two triangles overlap? Are the two triangles congruent?

STEP 3: Repeat STEP 1 and STEP 2 for triangles with different dimensions. Which pairs of corresponding parts did you specify to be congruent? Was this sufficient to make the triangles congruent?

STEP 4: Repeat STEP 1 to STEP 3 with △ABC and △DEF so that m∠CAB = m ∠FDE, AB = DE, and m∠CBA = m∠FED.

STEP 5: Repeat STEP 1 to STEP 3 with △ABC and △DEF so that BC = EF, AB = DE, and AC = DF.

STEP 6: Repeat STEP 1 to STEP 3 with △ABC and △DEF so that m∠CAB = m ∠FDE = 90°, BC = EF, and AC = DF.

Math Journal How are △ABC and △DEF related in each case? What are the four sets of the minimum conditions for two triangles to be congruent?
Answer:

Question 5.
Justify whether the triangles are congruent. If they are congruent, write the statement of congruence and state the test used.
a)

LN = , m∠ = m∠, and LM = .
By the SAS test, △LMN ≅ .
Caution
In the SAS test, the angle congruent to its corresponding angle must be the included angle, which is the angle between the congruent corresponding sides.
Answer:
LN = XZ, m∠ = NML = m∠YXZ, and LM = XY.
By the SAS test, △LMN ≅ △XYZ.

b)

m∠ABE = m∠ Corr. ∠s
m∠BAE = m∠ Corr.∠s
m∠AEB = m∠ ∠sum of triangle
Three pairs of congruent angles do not ensure that the triangles are .
Answer:
m∠ABE = m∠BCD Corr. ∠s
m∠BAE = m∠CBD Corr.∠s
m∠AEB = m∠BDC ∠sum of triangle
Three pairs of congruent angles do not ensure that the triangles are CONGRUENT.

Math in Focus Course 3B Practice 9.1 Answer Key

Name the figures that are congruent. Name the corresponding congruent line segments and angles.

Question 1.
ABCD is a parallelogram with diagonal \(\overline{B D}\).

Answer: Yes the above parallelogram is congruent (△BAD ≅ △DCB).
AB = CD and AD = BC because opposite sides of a parallelogram are equal.
∠ABD = ∠CDB And ∠BAD = ∠DCB

Question 2.

Answer:
ABCD ≅ PQRS
AB = PR = 5 cm
AD = RS = 9.22 cm
BC = PQ = 9 cm
DC = QS = 7 cm
∠ABC = ∠RPQ
∠DCB = ∠SQP

Solve. Show your work.

Question 3.
WXYZ is a rectangle with diagonal \(\overline{W Y}\). Explain, using the given test for congruent triangles, why △WXY is congruent to △YZW.

a) SSS
Answer: WX = ZY
WZ = XY
WY = WY

b) SAS
Answer:
WX = ZY
m∠ZWX= m∠ZYX
WZ = XY

c) ASA
Answer:
m∠ZWY= m∠XYW

d) HL
Answer:
ΔWXY = ΔWZY (right angled triangle)
WY = WY (Hypotenuse)

Question 4.
ABCDE is a regular pentagon with congruent diagonals \(\overline{B E}\) and \(\overline{B D}\).
a) Justify △ABE ≅ △CBD with a test for congruent triangles.
Answer:
We can test whether the given figures are congruent by testing SSS, ASA, SAS, and HL.
SSS test:
AB = BC
AE = CD
BE = BD

b) Name a pair of congruent quadrilaterals.

Answer: △ABE ≅ △CBD because the shape and size are equal.

Question 5.
△ABC is congruent to △PQR. Find the values of x, y, and z.

Answer:
Given,
△ABC is congruent to △PQR.
AB = PQ
BC = QR
AC = PR
AB = 8 cm
So, PQ = y = 8 cm
BC = x cm
We know that
BC = QR
So, BC = 3.4 cm
m∠BCA= m∠QPR
m∠BCA= 67°
So, m∠QPR = 67°

Question 6.

a) Name the figure that is congruent to trapezoid BADC.
Answer: The figure that is congruent to trapezoid BADC is AECD.

b) Find the length of each side of the trapezoid you named in a).
Answer:
From the figure, we can see that
BC = ED
AC = AD
BE = CD
Now we will find the unknown values of the figure.
BC = 6 in.
So, ED = 6 in.
AC = 10 in.
AC = AD = 10 in.
BE = CD
CD = 16 in
So, BE = CD = 16 in.

Question 7.
A piece of fabric has many printed shapes of congruent triangles and congruent quadrilaterals on it.
a) The pieces of fabric shown are congruent quadrilaterals. Find the value of x, y, and z.

Answer:
z + 3 = 6 cm
z = 6 – 3
z = 3 cm
9x° = 63°
x° = 63/9
x = 7
y° = 49.5

b) The pieces of fabric shown are congruent triangles. Find the value of p, q, and r.

Answer:
2p + q = 18 cm—- eq. 1
p – 2q = 16 cm—-eq. 2
2p + q = 18 cm—- eq. 1 × 2
4p + 2q = 36 cm
p – 2q = 16 cm
5p = 52 cm
p = 52/5
p = 10.4 cm
substitute the value of p in the eq.2
10.4 – 2q = 16 cm
10.4 – 16 = 2q
-5.6 = 2q
q = -5.6/2
q = -2.8 cm
The sum of all three angles = 180°
55° + 58° + r° = 180°
113° + r° = 180°
r° = 180° – 113°
r° = 67°

Question 8.
The origami cornflower is formed by making symmetrical folds in a paper square. In the diagram, AB = BC = DE = EF and OA = OC = OD = OF. Write two possible statements of congruence for ABCO.

Answer:
△ABO and △BCO seems congruent
AB = BC.

Question 9.
In making this suspension bridge, engineers connected two cables at the same point P on the concrete piers and at the same distance from the base of the concrete piers at R.
a) Which congruence test ensures that △PQR is congruent to △PSR? Explain.
Answer: SAS (Side Angle Side). Each triangle in the below figure is a right angled triangle at R and congruent with the same size and shape.

b) How many other pairs of congruent triangles are attached to each concrete pier? Explain.

Answer: There are 10 pairs of congruent triangles that are attached to each concrete pier.

Question 10.
This skyscraper has several symmetric triangles overlaid on a grid of horizontal and vertical lines. In each direction, these lines are parallel.

a) This face of the building is symmetric about \(\overline{G K}\), so AB = CD, AX = CX, and BX = DX. Give the statement of congruence for △ABX and tell which congruence test you used.
Answer: SSS test is used to tell whether the figure is congruent. △ABX ≅ △CDX

b) Is △EFG congruent to △HJK? Explain.
Answer: No because the shape of the triangles is same but the size of the triangles are different.

Practice the problems of Math in Focus Grade 1 Workbook Answer Key Chapter 13 Practice 4 Subtraction with Regrouping to score better marks in the exam.

Math in Focus Grade 1 Chapter 13 Practice 4 Answer Key Subtraction with Regrouping

Regroup.

Question 1.
27 = 1 ten ________ ones
Answer:

Question 2.
15 = 0 tens ________ ones
Answer:

Question 3.
30 = 2 tens ________ ones
Answer:

Subtract.

Question 4.

Answer:

Question 5.

Answer:

Question 6.

Answer:

Question 7.

Answer:

Question 8.

Answer:

Question 9.

Answer:

Fill in the missing numbers.

Question 10.
21 – 5 = ___

Answer:

Question 11.
36 – 7 = ___

Answer:

Question 12.
25 – 18 = ___

Answer:

Question 13.
31 – 18 = ___

Answer:

Question 14.
32 – 14 = ___

Answer:

Question 15.
30 – 17 = ___

Answer:

Subtract.

Then solve the riddle.

Question 16.
38 – 9 = 29
Answer:

Question 17.
30 – 18 = _____
Answer:

Question 18.
32 – 5 = ___
Answer:

Question 19.
35 – 8 = ___
Answer:

Question 20.
27 – 7 = _____
Answer:

Question 21.
34 – 19 = _____
Answer:

How do you cut the sea?

Match the letters to the answers below to find out.

Nafasha drops a ball into each number machine. Write the missing numbers in the blanks to show what happens to each ball.
Answer:

Question 22.

Answer:

Question 23.

Answer:

Practice the problems of Math in Focus Grade 1 Workbook Answer Key Chapter 13 Practice 5 Adding Three Numbers to score better marks in the exam.

Math in Focus Grade 1 Chapter 13 Practice 5 Answer Key Adding Three Numbers

Add.

Question 1.

4 + 5 + 6 = ____
Answer: 4+5+6=15
Explanation:
By adding 4 with 5 we get 9,
Then by adding 9 with 6 we get 15.

Question 2.

8 + 7 + 7 = ____
Answer:
Explanation:8+7+7=22
By adding 8 with 7 we get 15,
Then by adding 15 with 7 we get 22.

Question 3.

6 + 9 + 8 = ____
Answer:
Explanation:6+9+8=23
By adding 6 with 9 we get 15,
Then by adding 15 we get 23

Question 4.

5 + 4 + 8 = ____
Answer:
Explanation:5+4+8=17
By adding 5 with 4 we get 9,
Then by adding 9 with 8 we get 17.

Make ten. Then add.

Example

Question 5.
5 + 8 + 5 = ____
Answer:

Explanation:
By adding 5 with 5 we get 10,
Then by adding 8 with 10 we get 18.

Question 6.
8 + 9 + 2 =___
Answer:

Explanation:
By adding 8 with 2 we get 10,
Then by adding 9 with 10 we get 19.

Question 7.
9 + 7 + 2 = ___
Answer:

Explanation:
By adding 9 with 2 we get 11,
Then by adding 11 with 7 we get 18

Question 8.
9 + 4 + 4 = ___
Answer:

Explanation:
By adding 9 with 4 we get 13,
Then by adding 13 with 4 we get 17.

Question 9.
2 + 9 + 5 = ___
Answer:

Explanation:
By adding 2 with 9 we get 11,
Then by adding 11 with 5 we get 16.

This handy Math in Focus Grade 4 Workbook Answer Key Chapter 11 Practice 2 Properties of Squares and Rectangles detailed solutions for the textbook questions.

Math in Focus Grade 4 Chapter 11 Practice 2 Answer Key Properties of Squares and Rectangles

All the figures are rectangles. Find the measures of the unknown angles.

Example

Find the measure of ∠a.

Question 1.
Find the measure of ∠b.

Answer:
Measure of ∠b = 90⁰ – 27⁰
∠b= 63⁰
Explanation:
A Rectangle is a four sided-polygon, having all the internal angles equal to 90 degrees. The two sides at each corner or vertex, meet at right angles. Subtract 27⁰ from 90⁰the difference is equal to 63⁰.
Question 2.
Find the measure of ∠c.

Answer:
Measure of ∠c = 90⁰ – 45 ⁰
∠c= 45⁰
Explanation:
A Rectangle is a four sided-polygon, having all the internal angles equal to 90 degrees. The two sides at each corner or vertex, meet at right angles. Subtract 45⁰ from 90⁰the difference is equal to 45⁰.

All the figures are rectangles. Find the measures of the unknown angles.

Question 3.
Find the measure of ∠p.

Answer:
Measure of ∠p = 90⁰ – 36⁰– 18⁰
∠p= 36⁰
Explanation:
A Rectangle is a four sided-polygon, having all the internal angles equal to 90 degrees. The two sides at each corner or vertex, meet at right angles. Subtract 36⁰ and 18⁰ from 90⁰the difference is equal to 36⁰.

Question 4.
Find the measure of ∠m.

Answer:
Measure of ∠m = 90⁰ – 39⁰– 22⁰
∠m= 29⁰
Explanation:
A Rectangle is a four sided-polygon, having all the internal angles equal to 90 degrees. The two sides at each corner or vertex, meet at right angles. Subtract 39⁰ and 22⁰ from 90⁰the difference is equal to 29⁰.

The figure is a rectangle. Find the measure of the unknown angle.

Question 5.
Find the measure of ∠s.

Answer:
Measure of ∠s = 90⁰ – 25⁰– 12⁰
∠s= 53⁰
Explanation:
A Rectangle is a four sided-polygon, having all the internal angles equal to 90 degrees. The two sides at each corner or vertex, meet at right angles. Subtract 25⁰ and 12⁰ from 90⁰the difference is equal to 53⁰.

Find the lengths of the unknown sides.

Question 6.
The figure is made up of a rectangle and a square. Find BC and GE.

Answer:
In the above image we can observe rectangle and a square.
From the rectangle ABFG we know that
GA = BF = 12 cm
AB = FG = 15cm
From the square CDEF we know that
CD = DE = EF = FC = 8 cm
BC = GA – CF
BC = 12 – 8
BC = 4 cm
GE = AB + EF
GE = 15 + 8
GE = 23 cm

Find the lengths of the unknown sides.

Question 7.
The figure is made up of two rectangles. Find BD and FG.

Answer:
In the above image we can observe two rectangles.
From the rectangle ABCG we know that
AB = CG = 20 feet
BC = GA = 17 feet
From the rectangle CDEF we know that
CD = EF = 5 feet
DE = FC = 14 feet
BD = BC + CD
BD = 17 + 5
BD = 22 feet
FG = AB – FC
FG = 20 – 14
FG = 6 feet

Question 8.
The figure is made up of two rectangles. Find QR and RT.

Answer:
In the above image we can observe two rectangles.
From the rectangle PQRS we know that
PQ = RS = 5 yards
QR = SP
From the rectangle STUV we know that
ST = UV = 6 yards
TU = VS = 4 yards
QR = PV + VS
QR = 3 + 4
QR = 7 yards
RT = RS + ST
RT = 5 + 6
RT = 11 yards

Find the lengths of the unknown sides.

Question 9.
The figure is made up of two rectangles. Find FG.

Answer:
In the above image we can observe two rectangles.
From the rectangle ABGH we know that
BG = HA = 5 miles
AB = GH
From the rectangle CDEF we know that
DE = FC = 8 miles
CD = EF
CG = HA – BC
CG = 5 – 3
CG = 2 miles
FG = DE – CG
FG = 8 – 2
FG = 6 miles

Question 10.
The figure is made up of a square and a rectangle. Find BC.

Answer:
In the above image we can observe rectangle and a square.
From the rectangle ABGH we know that
BG = HA = 10 feet
AB = GH
From the square CDEF we know that
CD = DE = EF = FC = 4 feet
BC = HA – FC – FG
BC = 10 – 4 – 2
BC = 4 feet

Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 4 Lesson 4.4 Sketching Graphs of Linear Equations to finish your assignments.

Math in Focus Grade 8 Course 3 A Chapter 4 Lesson 4.4 Answer Key Sketching Graphs of Linear Equations

Math in Focus Grade 8 Chapter 4 Lesson 4.4 Guided Practice Answer Key

Use graph paper. Use 1 grid square to represent 1 unit for the x interval from -2 to 2, and the y interval from -2 to 4.

Question 1.
Graph the equation y = \(\frac{3}{2}\)x + 1.
Answer:

Use graph paper. Use 1 grid square to represent 1 unit on both axes for the x interval from 0 to 3, and the y Interval from -3 to 7.

Question 2.
Graph the equation y = 2x + 1.
Answer:

Question 3.
Graph the equation y = –\(\frac{1}{3}\)x – 2.
Answer:

Use graph paper. Use 1 grid square to represent 1 unit on both axes for the x interval from -2 to 2, and the y interval from 0 to 10.

Question 4.
Graph a line with slope -2 that passes through the point (2, 2).
Answer:

Question 5.
Graph a line with slope 2 that passes through the point (-2, 1).
Answer:

Math in Focus Course 3A Practice 4.4 Answer Key

For this practice, use 1 grid square to represent 1 unit on both axes for the interval from -6 to 6.

Graph each linear equation.

Question 1.
y = \(\frac{1}{3}\)x + 1
Answer:

Question 2.
y = \(\frac{1}{6}\)x + 3
Answer:

Question 3.
y = \(\frac{1}{2}\)x + 2
Answer:

Question 4.
y = \(\frac{2}{3}\)x – 1
Answer:

Question 5.
y = -x + 5
Answer:

Question 6.
y = 3 – \(\frac{1}{4}\)x
Answer:

Question 7.
y = 1 – \(\frac{1}{2}\)x
Answer:

Question 8.
y = –\(\frac{1}{5}\)x – 2
Answer:

Question 9.
Math Journal Graph the equation y = 2 – \(\frac{2}{3}\)x. Explain how to use the graph to find other solutions of the equation.
Answer:

First select a value of the x-coordinate on the x-axis. Then trace it along the vertical grid lines until the gridline intersects with the graph.
Then trace horizontally from the graph to the y-axis to obtain the corresponding y-coordinate. This pair of x-coordinate and y-coordinate is one solution to the equation.

Question 10.
Math Journal Martha says that the point (4, -2) lies on the graph of the equation y = –\(\frac{1}{4}\)x – 1. Explain how you can find out if she is right without actually graphing the equation.
Answer:

Graph each line with the given slope that passes through the given point.

Question 11.
Slope = \(\frac{2}{5}\); (5, 4)
Answer:
y = \(\frac{2}{5}\)x + 4

Question 12.
Slope = \(\frac{2}{3}\); (6, 1)
Answer:

Question 13.
Slope = -3; (1, 0)
Answer:

Question 14.
Slope = -2; (-1, -2)
Answer:

Question 15.
Math Journal Suppose that Emily shows you some of her homework:

Graph the equation y = -2x + \(\frac{1}{2}\).
Describe Emily’s mistake. Graph the equation correctly.
Answer:
Emily made three mistakes. She labeled the coordinates (6, 3) incorrectly, drew the slope of the line incorrectly, and mistook the value for the slope as the y-intercept.

Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 4 Lesson 4.3 Writing Linear Equations to finish your assignments.

Math in Focus Grade 8 Course 3 A Chapter 4 Lesson 4.3 Answer Key Writing Linear Equations

Math in Focus Grade 8 Chapter 4 Lesson 4.3 Guided Practice Answer Key

For each line, state its slope and its y-intercept.

Question 1.
5x + 4y = 8
First, write the equation in the slope-intercept form.
5x + 4y = 8
5x + 4y – = 8 –      Subtract 5x from both sides.
=     Simplify.
=     Divide both sides by .
y =       Write in slope-intercept form.
Comparing the equation y = with y = mx + b:
Slope: m =
y-intercept: b =
Answer:
First, write the equation in the slope-intercept form.
5x + 4y = 8
5x + 4y – 5x = 8 – 5x    Subtract 5x from both sides.
4y = 8 – 5x    Simplify.
y = 2 – 5/4 x    Divide both sides by 4.
y = -5/4 x + 2      Write in slope-intercept form.
Comparing the equation y = -5/4 x + 2  with y = mx + b:
Slope: m = -5/4
y-intercept: b = 2

Question 2.
2x – 3y = 7
First write the equation in the slope-intercept form.
2x – 3y = 7
2x – 3y + = 7 +    Add 3y to both sides.
=         Simplify.
=      Subtract from both sides.
=      Simplify.
=     Divide both sides by .
y =       Write in slope-intercept form.
Comparing the equation y = with y = mx + b:
Slope: m =
y-intercept: b =
Answer:
First write the equation in the slope-intercept form.
2x – 3y = 7
2x – 3y + 3y = 7 + 3y   Add 3y to both sides.
2x = 7 + 3y        Simplify.
2x – 2x = 7 + 3y – 2x     Subtract 2x from both sides.
0 = 7 + 3y – 2x     Simplify.
3y  = 2x – 7    Divide both sides by 3.
y = 2/3 x – 7/3      Write in slope-intercept form.
Comparing the equation y = 2/3 x – 7/3 with y = mx + b:
Slope: m = 2/3
y-intercept: b = -7/3

Question 3.
5y – x = 15
Answer:
First write the equation in the slope-intercept form.
Add x on both sides
5y – x + x = 15 + x
5y = 15 + x
Subtract 5y on both sides
5y – 5y = 15 + x – 5y
0 = 15 + x – 5y
5y = x + 15
Divide by 5 on both sides
y = 1/5 x + 3

Question 4.
2y – 3x = -4
Answer:
First write the equation in the slope-intercept form.
2y = 3x -4
Divide by 2 on both sides
y = 3/2 x – 2

Question 5.
6y + 5x = 24
Answer:
First write the equation in the slope-intercept form.
Subtract 5x on both sides
6y + 5x – 5x = 24 – 5x
6y = 24 – 5x
Divide by 6 on both sides
y = 4 – 5x
y = -5x + 4

Question 6.
3y + 4x = 3
Answer:
First write the equation in the slope-intercept form.
Subtract 4x on both sides
3y + 4x -4x = 3 – 4x
3y = -4x + 3
Divide by 3 on both sides
y = -4/3 x + 1

Use the given slope and y-intercept of a line to write an equation in slope-intercept form.

Question 7.
Slope, m = –\(\frac{2}{3}\)
y-intercept, b = 4
y = mx + b
y = Substitute the given values for and b.
Answer:
y = mx + b
y = –\(\frac{2}{3}\) Substitute the given values for 4 and b.

Question 8.
Slope, m = 4
y-intercept, b = -7
Answer:
y = mx + b
y = 4 Substitute the given values for -7 and b.

Solve.

Question 9.
A line has the equation 3y + 6 = 10x. Write an equation of a line parallel to this given line that has a y-intercept of 2.
First write the given equation in slope-intercept form.
3y + 6 = 10x
3y + 6 – = 10x –      Subtract both sides by 6.
=     Simplify.
y =     Divide both sides by .
y =      Write in slope-intercept form.
The line has slope m = and y-intercept b = .
Then write an equation for the parallel line with slope m = and y-intercept b = .
y = . Substitute the values of m and b.
So, an equation of the line parallel to 3y = 10x – 6 is .
Answer:
First write the given equation in slope-intercept form.
3y + 6 = 10x
3y + 6 – 6 = 10x – 6     Subtract both sides by 6.
3y = 10x – 6    Simplify.
y = 10/3 x – 2    Divide both sides by 3.
y = 10/3 x – 2      Write in slope-intercept form.
The line has slope m = 10/3 and y-intercept b = -2.
Then write an equation for the parallel line with slope m = 10/3 and y-intercept b = -2
y = 10/3 x + 2. Substitute the values of m and b.
So, an equation of the line parallel to 3y = 10x – 6 is 10/3 x + 2.

Question 10.
A line has slope -3 and passes through the point (-6, 8). Write an equation of the line.
Answer:
Given,
A line has slope -3 and passes through the point (-6, 8)
m = -3
y-intercept = 8
y = -3x + 8

Question 11.
A line has slope \(\frac{1}{3}\) and passes through the point (0, 1). Write an equation of the line.
Answer:
Given,
A line has slope \(\frac{1}{3}\) and passes through the point (0, 1).
m = \(\frac{1}{3}\)
y-intercept = 1
y = mx + b
y = \(\frac{1}{3}\)x + 1

Question 12.
A line has slope 2 and passes through the point (1, 5). Write an equation of the line.
Answer:
Given,
A line has slope 2 and passes through the point (1, 5).
m = 2
y-intercept = 5
y = mx + b
y = 2x + 5

Question 13.
Write an equation of the line that passes through the point (-2, 1) and is parallel to y = 5 – 3x.
First, write the equation in slope-intercept form.
y = 5 – 3x
y =     Write in slope-intercept form.
The line has slope m = .
So, the line parallel to y = 5 – 3x has slope m = .
Then use the slope m = . and the fact that (-2, 1) lies on the parallel line to find the y-intercept.
y = mx + b Write in slope-intercept form.
=      Substitute the values for m, x, and y.
=     Simplify.
=     Subtract from both sides.
=     Simplify.
The y-intercept is .
So, an equation of the line is .
Answer:
First, write the equation in slope-intercept form.
y = 5 – 3x
y = -3x + 5    Write in slope-intercept form.
The line has slope m = -2.
So, the line parallel to y = 5 – 3x has slope m = -3.
Then use the slope m = -2. and the fact that (-2, 1) lies on the parallel line to find the y-intercept.
y = mx + b Write in slope-intercept form.
y = -2x + 1 Substitute the values for m, x, and y.
So, an equation of the line is y = -2x + 1.

Question 14.
Write an equation of the line that passes through the pair of points (-2, -5) and (2, -1).
Answer:
The line passes through the points (-2, -5) and (2, -1).
Slope m = \(\frac{-1-(-5)}{2-(-2)}\)
= \(\frac{4}{4}\)
= 1
The line passes through the y-axis at the point (0, 0).
Thus m = 1 and y-intercept b is 0.
y = x

Math in Focus Course 3A Practice 4.3 Answer Key

Find the slope and the y-intercept of the graph of each equation.

Question 1.
y = -5x + 7
Answer:
Given,
y = -5x + 7
The equation of the line is y = mx + b
Slope, m = -4
y-intercept = 7

Question 2.
y = 2x + 3
Answer:
Given,
y = 2x + 3
The equation of the line is y = mx + b
Slope, m = 2
y-intercept = 3

Question 3.
5x + 2y = 6
Answer:
Given,
5x + 2y = 6
The equation of the line is y = mx + b
2y = -5x + 6
y = -5/2 x + 3
Slope, m = -5/2
y-intercept = 3

Question 4.
2x – 7y = 10
Answer:
Given,
The equation of the line is y = mx + b
2x – 7y = 10
-7y = -2x + 10
7y = 2x – 10
y = 2/7 x – 10/7
Slope, m = 2/7
y-intercept = 10/7

Use the given slope and y-intercept of a line to write an equation in slope-intercept form.

Question 5.
Slope, m =\(\frac{1}{2}\)
y-intercept, b = 3
Answer:
The equation of the line is y = mx + b
y = \(\frac{1}{2}\)x + 3

Question 6.
Slope, m = -2
y-intercept, b = 5
Answer:
The equation of the line is y = mx + b
y = -2x + 5

Solve. Show your work.

Question 7.
A line has the equation 4y = 3x – 8. Find an equation of a line parallel to this line that has a y-intercept of 2.
Answer:
A line has the equation 4y = 3x – 8.
y = 3/4 x – 2
A line parallel to this line that has a y-intercept of 2
y = 3/4 x + 2

Question 8.
A line has the equation 3y = 3 – 2x. Find an equation of a line parallel to this line that has a y-intercept of 5.
Answer:
A line has the equation 3y = 3 – 2x.
y = 1 – 2/3 x
y = -2/3 x + 1
y-intercept = 5
y = -2/3 x + 5

Question 9.
Math Journal Ira says that the graphs of the equations y = -3x + 7 and y = 3x – 7 are parallel lines. Do you agree? Explain.
Answer:
No. The slope of the equation y = -3x + 7 is -3 and the slope of the equation y = 3x – 7 is 3.
So, the graphs of the equations are not parallel lines.

Question 10.
Find an equation of the line that passes through the point (0, 4) and has a slope of –\(\frac{1}{3}\).
Answer:
Slope = –\(\frac{1}{3}\)
y-intercept = 4
The equation of the line is y = mx + b
y = –\(\frac{1}{3}\)x + 4

Question 11.
A line has slope –\(\frac{1}{2}\) and passes through the point (-4, -2). Write an equation of the line.
Answer:
The equation of the line is y = mx + b
slope = –\(\frac{1}{2}\)
y – intercept = -2
y = –\(\frac{1}{2}\)x – 2

Question 12.
Find an equation of the line that passes through the point (-5, 7) and is parallel to y = 4 – 3x.
Answer:
The equation of the line is y = mx + b
y = 4 – 3x
y = -3x + 4
y = -3x + 7

Question 13.
Find an equation of the line that passes through the point (0, 2) and is parallel to 6y = 5x – 24.
Answer:
6y = 5x – 24
The equation of the line is y = mx + b
y = 5/6 x – 4
The line that passes through the point (0, 2)
y = 5/6 x + 2

Question 14.
Find an equation of the line that passes through the pair of points (-5, -1) and (0, 4).
Answer:
The line passes through the points (-5, -1) and (0, 4).
Slope m = \(\frac{4-(-1)}{0-(-5)}\)
= \(\frac{5}{5}\)
= 1
The line passes through the y-axis at the point (0, 0).
Thus m = 1 and y-intercept b is 0.
y = x

Question 15.
Find an equation of the line that passes through the pair of points (-3, 2) and (-2, 5).
Answer:
The line passes through the points (-3, 2) and (-2, 5)
Slope m = \(\frac{5-2}{-2-(-3)}\)
= \(\frac{3}{1}\)
= 3
The line passes through the y-axis at the point (0, 0).
Thus m = 3 and y-intercept b is 0.
y = 3x

Question 16.
Math Journal Can you write a linear equation in the slope-intercept form using the points (3, 4) and (5, 8)? Explain.
Answer:
The line passes through the points (3, 4) and (5, 8)
Slope m = \(\frac{8-4}{5-3}\)
= \(\frac{4}{2}\)
= 2
The line passes through the y-axis at the point (0, 0).
Thus m = 2 and y-intercept b is 0.
y = 2x

Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 4 Lesson 4.2 Understanding Slope-Intercept Form to finish your assignments.

Math in Focus Grade 8 Course 3 A Chapter 4 Lesson 4.2 Answer Key Understanding Slope-Intercept Form

Technology Activity

Materials
graphing calculator

EXPLORE THE RELATIONSHIP BETWEEN y = mx AND y = mx + b

Work in pairs

Answer:

Math Journal The equation of another line is given by 2y = 5x – 4. It can also be written as y = 2.5x – 2. Predict the y-intercept. Use the graphing calculator to check your prediction. Is your prediction correct?
Answer:
Slope m = 2.5
y-intercept = -2

Math in Focus Grade 8 Chapter 4 Lesson 4.2 Guided Practice Answer Key

Write an equation for each line.

Question 1.
The line passes through the points (, ) and (, ).
Slope m = \(\frac{?-?}{?-?}\)
= \(\frac{?}{?}\)
=

The line intersects the y-axis at the point (, ).
So, the y-intercept is .
So, the equation of the line is .
Answer:
The line passes through the points (6, 3) and (-4, -2).
Slope m = \(\frac{6-(-4)}{3-(-2)}\)
= \(\frac{10}{5}\)
= 2
The line passes through the y-axis at the point (0, 0).
Thus m = 2 and y-intercept b is 0.
y = 2x

Question 2.
The line passes through the points (, ) and (, ).
Slope m = \(\frac{?-?}{?-?}\)
= \(\frac{?}{?}\)
=

The line intersects the y-axis at the point (, ).
So, the y-intercept is .
So, the equation of the line is .
Answer:
The line passes through the points (1, -6) and (-4, 9).
Slope m = \(\frac{9-(-6)}{-4-1}\)
= \(\frac{15}{-5}\)
= -3
The line intersects the y-axis at the point (0, -3).
So, the y-intercept is -3.
So, the equation of the line is y = -3x -3.

Question 3.

Answer:
The line passes through the points (-5, -2) and (1, -2).
Slope m = \(\frac{-2-(-2)}{1+5}\)
= \(\frac{0}{6}\)
= 0
The line intersects the y-axis at the point (0, -2).
So, the y-intercept is -2.
So, the equation of the line is y = 0x -2.

Question 4.

Answer:
The line passes through the points (0, 2) and (2, 4).
Slope m = \(\frac{4-2)}{2-0}\)
= \(\frac{2}{2}\)
= 1
The line intersects the y-axis at the point (0, 0).
So, the y-intercept is 0.
So, the equation of the line is y = -2x + 2

Math in Focus Course 3A Practice 4.2 Answer Key

Identify the y-Intercept. Then calculate the slope using the points indicated.

Question 1.

Answer:
The line passes through the points (-2, -2) and (3, 3).
Slope m = \(\frac{3-(-2)}{3-(-2)}\)
= \(\frac{5}{5}\)
= 1
The line intersects the y-axis at the point (0, 0).
So, the y-intercept is 0.
So, the equation of the line is y = x.

Question 2.

Answer:
The line passes through the points (0, -1) and (2, -3).
Slope m = \(\frac{-3-(-1)}{2-0}\)
= \(\frac{-2}{2}\)
= -1
The line intersects the y-axis at the point (0, -1).
So, the y-intercept is -1.
So, the equation of the line is y = -1x -1.

Question 3.

Answer:
The line passes through the points (0, 3) and (8, -6).
Slope m = \(\frac{-6-3}{8-0}\)
= \(\frac{-9}{8}\)
m = \(\frac{-9}{8}\)
The line intersects the y-axis at the point (0, 3).
So, the y-intercept is 3.

Question 4.

Answer:
The line passes through the points (0, 2) and (4, 6).
Slope m = \(\frac{6-2}{4-0}\)
= \(\frac{4}{4}\)
m = 1
The line intersects the y-axis at the point (0, -2).
So, the y-intercept is -2.

Write an equation in the form y = mx or y = mx + b for each line.

Question 5.

Answer:
The line passes through the points (4, 3) and (-4, -3).
Slope m = \(\frac{-3-3}{-4-4}\)
= \(\frac{-6}{-8}\)
= \(\frac{3}{4}\)
The line intersects the y-axis at the point (0, 0).
So, the y-intercept is 0.
So, the equation of the line is y = \(\frac{3}{4}\)x

Question 6.

Answer:
The line passes through the points (-2, 4) and (2, -2).
Slope m = \(\frac{-2-4}{2+2}\)
= \(\frac{-6}{4}\)
= \(\frac{-3}{2}\)
The line intersects the y-axis at the point (0, 1).
So, the y-intercept is 1.
So, the equation of the line is y = \(\frac{-3}{2}\)x +1.

Question 7.

Answer:
The line passes through the points (-8, 3) and (0, -2).
Slope m = \(\frac{-2-3}{0+8}\)
= \(\frac{-5}{8}\)
The line intersects the y-axis at the point (0, -2).
So, the y-intercept is -2.
So, the equation of the line is y = \(\frac{-5}{8}\)x -2.

Question 8.

Answer:
The line passes through the points (0, 4) and (-3, -2).
Slope m = \(\frac{-2-4}{-3-0}\)
= \(\frac{-6}{-3}\)
= 2
The line intersects the y-axis at the point (0, 4).
So, the y-intercept is 4.
So, the equation of the line is y = 2x + 4

Graph each line using 1 grid square to represent 1 unit on both axes for the interval from -4 to 4. Then write the equation for each line.

Question 9.
The line passes through the points (-4, 3) and (-4, -2).
Answer:

Question 10.
The line passes through the points (-3, 4) and (1, 4).
Answer:

Question 11.
Math Journal Line A passes through the origin and has a negative slope. Line B has a positive y-intercept and a positive slope. Line C has a negative slope and a negative y-intercept. Give a possible equation for each line. Justify your answer.
Answer:
Possible equations: Line A: y = -3x; Line B: y = 5x + 2; Line C: y = -2x – 7.
For line A, the value of m in the equation y = mx + b is negative and the value of b is 0.
For line B, both m and b have positive values. For line C, both m and b have negative values.

Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 4 Lesson 4.1 Finding and Interpreting Slopes of Lines to finish your assignments.

Math in Focus Grade 8 Course 3 A Chapter 4 Lesson 4.1 Answer Key Finding and Interpreting Slopes of Lines

Hands-On Activity

Materials:
graph paper

USE TRIANGLES TO FIND THE SLOPE OF A LINE

Work in pairs.

You can find the slope of any nonvertical line, not just. a line that represents a direct proportion, by finding the ratio of the rise to the run.

STEP 1: Graph the line below on graph paper. The line should pass through the points (0, 1) and (6, 4). Then draw and label three right triangles on the line as shown. The triangles should be the same shape but different sizes. Make sure that each right angle lies on the intersection of two grid lines.

STEP 2: Copy and complete the table.

STEP 3: What do you observe about the ratios in the last column of the table?

Math Journal As you have learned, the slope of a line is the ratio of the rise to the run. Which side of each triangle has a length equal to the rise? Which has a length equal to the run? For each line you drew, what did you notice about the three ratios you calculated?
Answer:

Hands-On Activity

Materials
graph paper

USE POINTS TO FIND THE SLOPE OF A LINE

STEP 1: Graph the line that passes through each pair of points, using 1 grid square to represent 1 unit on both axes for the interval from 0 to 7.
a) Line 1: (0, 1) and (4, 2)
b) Line 2: (0, 1) and (4, 5)
c) Line 3: (0, 1) and (4, 7)

STEP 2: For each line you drew, draw a right triangle. The segment connecting the two points should be the longest side. Find the length of the vertical side, the length of horizontal side, and the ratio \(\frac{\text { Length of vertical side }}{\text { Length of horizontal side }}\).

Math Journal Of the three lines drawn in STEP 1, which line has the greatest slope? Which line has the least slope? How can you tell by looking at the lines? How can you tell from the ratios you calculated?

Math in Focus Grade 8 Chapter 4 Lesson 4.1 Guided Practice Answer Key

Complete.

Question 1.
The graphs give information about the distance, d miles, traveled over time, t hours, by cars and trucks on a California highway. Which speed is slower?

Speed for cars:
Unit rate = \(\frac{\text { Rise }}{\text { Run }}\)
= \(\frac{?}{?}\)
= mi/h
Speed for trucks:
Unit rate = \(\frac{\text { Rise }}{\text { Run }}\)
= \(\frac{?}{?}\)
= mi/h
The slope for the car speed graph is , so the unit rate is miles per hour.
The slope for the truck speed graph is , so the unit rate is miles per hour.
The speed for is slower than the speed for .
Answer:
Speed for cars:
= \(\frac{100}{1.5}\)
= 66.6 mi/h
Speed for trucks:
= \(\frac{100}{2}\)
= 50 mi/h
The slope for the car speed graph is 66.6, so the unit rate is 67 miles per hour.
The slope for the truck speed graph is 50, so the unit rate is 50 miles per hour.
The speed for trucks is slower than the speed for car.

Question 2.
The graph passes through the points (, ) and (, )

Slope = \(\frac{\text { Rise }}{\text { Run }}\)
= \(\frac{?}{?}\)
= \(\frac{?}{?}\)
= \(\frac{?}{?}\)
The slope is .
Answer:
The graph passes through the points (-2, 4) and (2, -2)
Slope = \(\frac{\text { Rise }}{\text { Run }}\)
m = \(\frac{4+2}{-2-2}\)
= \(\frac{6}{-4}\)
= \(\frac{-3}{2}\)
The slope is \(\frac{-3}{2}\)

Question 3.
The graph passes through the points (, ) and (, )

Slope = \(\frac{\text { Rise }}{\text { Run }}\)
= \(\frac{?}{?}\)
= \(\frac{?}{?}\)
= \(\frac{?}{?}\)
The slope is .
Answer:
The graph passes through the points (4, 3) and (-2, -1)
Slope = \(\frac{\text { Rise }}{\text { Run }}\)
= \(\frac{-1-3}{-2-4}\)
= \(\frac{-4}{-6}\)
= \(\frac{2}{3}\)
The slope is 2/3.

Solve. Show your work.

Question 4.
The graphs represent the amount of water, w gallons, in Pool A over time, t hours, and the amount of water, w gallons, left in Pool B over time, t hours.

a) Find the slope of the line graph for Pool A. What does it represent?
Answer:
slope = 180/4
slope = 45

b) Find the slope of the line graph for Pool B. What does it represent?
Answer:
Slope = \(\frac{\text { Rise }}{\text { Run }}\)
= \(\frac{150}{4-0}\)
= 37.5

Complete.

Question 5.

Use the points (, ) and (, ):
Slope = \(\frac{\text { Rise }}{\text { Run }}\)
= \(\frac{?}{?}\)
= \(\frac{?}{?}\)
=
The slope is .
Answer:
Use the points (3, 4) and (3, 0):
Slope = \(\frac{\text { Rise }}{\text { Run }}\)
= \(\frac{0-4}{0}\)
= undefined
The slope is undefined.

Find the slope of the line passing through each pair of points.

Question 6.
M(-2, 0) and N(0, 4)
Let M (-2, 0) be (x₁, y₁) and N (0, 4) be (x₂, y₂).
Method 1
Slope = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\)
= \(\frac{?}{?}\)
= \(\frac{?}{?}\)
=

Method 2
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{?}{?}\)
= \(\frac{?}{?}\)
=
The slope is .
Answer:
Method 1
Slope = \(\frac{0-4}{-2-0}\)
= \(\frac{-4}{-2}\)
= 2
Slope = 2
Method 2
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{4-0}{0+2}\)
= \(\frac{4}{2}\)
= 2
The slope is 2.

Question 7.
S(-5, 8) and T(-2, 2)
Let S (-5, 8) be (x₁, y₁) and T(-2, 2) be (x₂, y₂).

Method 1
Slope = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\)
= \(\frac{?}{?}\)
= \(\frac{?}{?}\)
=

Method 2
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{?}{?}\)
= \(\frac{?}{?}\)
=
The slope is .
Answer:
Let S (-5, 8) be (x₁, y₁) and T(-2, 2) be (x₂, y₂).
Method 1
Slope = \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\)
= \(\frac{8-2}{-5+2}\)
= \(\frac{6}{-3}\)
= -2

Method 2
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{2-8}{-2+5}\)
= \(\frac{-6}{3}\)
= -2
The slope is -2.

Math in Focus Course 3A Practice 4.1 Answer Key

Find the slope of each line using the points indicated.

Question 1.

Answer:
Let the points be (-1,4) and (-4, -1)
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{-1-4}{-4+1}\)
= \(\frac{-5}{-3}\)
= \(\frac{5}{3}\)
The slope is \(\frac{5}{3}\).

Question 2.

Answer:
Let the points be (-5,10) and (-1, 0)
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{0-10}{-1+5}\)
= \(\frac{-10}{4}\)
= \(\frac{-5}{2}\)
The slope is \(\frac{-5}{2}\).

Question 3.

Answer:
Let the points be (-10,3) and (2, 3)
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{3-3}{2-10}\)
= 0
The slope is 0.

Question 4.

Answer:
Let the points be (4,2) and (4, 0)
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{0-2}{4-4}\)
= undefined
The slope is undefined.

Question 5.
Math Journal Jason says that the line in Graph B has a greater slope than the line in Graph A because it is steeper. What error is Jason making? Justify your answer.

Answer:
Jason compared the two lines visually. Because the two graphs have different scales for the vertical aces, the slopes cannot be accurately compared by their steepness.

Question 6.
Math Journal Andy graphs a vertical line through the points (5, 2) and (5, 5). He says the slope of the line is \(\frac{3}{0}\). What error is he making?
Answer:
Let the points be (5, 2) and (5, 5)
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{5-2}{5-5}\)
= undefined
The slope is undefined.

Find the slope of the line passing through each pair of points.

Question 7.
A (-10, 3), B (0, 3)
Answer:
Let the points be A (-10, 3), B (0, 3)
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{3-3}{0+10}\)
= 0
The slope is 0.

Question 8.
S(5, -2), T(2, -5)
Answer:
Let the points be S(5, -2), T(2, -5)
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{-5+2}{2-5}\)
= \(\frac{-3}{-3}\)
= 1
The slope is 1.

Question 9.
P(0, -7), Q(-3, 5)
Answer:
Let the points be P(0, -7), Q(-3, 5)
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{5+7}{-3-0}\)
= \(\frac{12}{-3}\)
= -4
The slope is -4.

Question 10.
X(4, 4), V(4, -2)
Answer:
Let the points be X(4, 4), V(4, -2)
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{-2-4}{4-4}\)
= undefined
The slope is undefined.

Question 11.
Math Journal Two points have the same x-coordinates but different y-coordinates. Make a prediction about the slope of a line drawn through the points. Justify your prediction.
Answer:
Two points have the same x-coordinates but different y-coordinates. A vertical line graph is drawn. Because slope = rise/run and the run is 0, the slope is undefined. Any number divided by zero is undefined.

Question 12.
MathJournal Two points have the same y-coordinates but different x-coordinates. Make a prediction about the slope of a line drawn through the points. Justify your prediction.
Answer:
It is a line parallel to the x-axis. No matter what the value of x is, y stays the same.

Solve. Show your work.

Question 13.
In the Fahrenheit system, water freezes at 32°F and boils at 212°F. In the Celsius system, water freezes at 0°C and boils at 100°C.
a) Translate the verbal description into a pair of points in the form (temperature in °C, temperature in °F).
Answer: (0, 32); (100, 212)

b) Find the slope of the line passing through the pair of points in a).
Answer: 9/5

c) Suppose the temperature in a room goes up by 5°C. By how much does the temperature go up in degrees Fahrenheit? Explain.
Answer: By 9° F; The slope of the line is 9/5,
so a horizontal change of 5 corresponds to a vertical change of 9.

Question 14.
The table shows how much a certain amount of gasoline costs at two gasoline stations on a particular day.

a) At which station is each additional gallon of gasoline more expensive? Explain.
Answer:
The cost at station A is more expensive.
(1, 3) a) We consider the points:
(3, 11)
Slope_B = \(\frac{11-3}{3-1}\) The cost of an additional gallon is the slope of the line representing the cost
= \(\frac{8}{2}\)
= 4
We determine the slope of the tine representing the cost at Station A:
= 4
(1,4) We consider the points:
(3, 10)
slope_B = \(\frac{10-4}{3-1}\)
= \(\frac{6}{2}\)
= 3
As Slope_A > Slope_B, each additional gallon is more expensive at Station A.

b) Graph the relationship between cost and gallons of gasoline purchased for each station. Use 1 unit on the horizontal axis to represent 1 gallon for the x interval from 0 to 5, and 1 unit on the vertical axis to represent $2 for the y interval from 0 to 20.
Answer:
b) We graph the cost of gasoline at Station A and Station B:

c) Which graph is steeper?
Answer:
We notice that Graph A is steeper than Graph B.

a) Station A
b) See graphs
C) Graph A

Go through the Math in Focus Grade 8 Workbook Answer Key Chapter 4 Lines and Linear Equations to finish your assignments.

Math in Focus Grade 8 Course 3 A Chapter 4 Answer Key Lines and Linear Equations

Math in Focus Grade 8 Chapter 4 Quick Check Answer Key

Tell whether each graph represents a direct proportion. If so, find the constant of proportionality. Then write a direct proportion equation.

Question 1.

Answer:
If y = kx then y is said to be directly proportional to x.
y = 10x
y/10 = x
Thus the above graph is said to be constant of proportionality.

Question 2.

Answer: y = 1/10 x
The above graph is not direct proportion.

Go through the Math in Focus Grade 8 Workbook Answer Key Cumulative Review Chapters 3-4 to finish your assignments.

Math in Focus Grade 8 Course 3 A Cumulative Review Chapters 3-4 Answer Key

Concepts and Skills

Solve each equation. Show your work. (Lesson 3.1)

Question 1.
0.2(x + 2) – 2 = 0.4
Answer:
0.2x + 0.4 – 2 = 0.4
0.2x = 2
x = 2/0.2
x = 10

Question 2.
2(x – 5) – 3(3 – x) = \(\frac{1}{2}\) (x – 2)
Answer:
2x – 10 – 9 + 3x = \(\frac{1}{2}\)x – 1
5x – 18 = \(\frac{1}{2}\)x
4 \(\frac{1}{2}\)x = 18
4x = 36
x = 36/4 = 9

Question 3.
\(\frac{x}{3}\) + \(\frac{3+x}{6}\) = 3
Answer:
\(\frac{x}{3}\) + \(\frac{3+x}{6}\) = 3
2x + 3 + x = 18
3x = 18 – 3
3x = 15
x = 15/3
x = 5

Question 4.
\(\frac{2(x+3)}{5}\) – \(\frac{x-1}{2}\) = 2
Answer:
\(\frac{2(x+3)}{5}\) – \(\frac{x-1}{2}\) = 2
\(\frac{2x+6}{5}\) – \(\frac{x-1}{2}\) = 2
4x + 12 – 5x + 5 = 10
-x + 17 = 10
-x = 10 – 17
-x = -7
x = 7

Express each decimal as a fraction, without the use of calculator. (Lesson 3.1)

Question 5.
\(0 . \overline{5}\)
Answer:
\(\frac{5}{9}\)

Question 6.
\(0 . \overline{8}\)
Answer: \(\frac{8}{9}\)

Question 7.
\(0.2 \overline{7}\)
Answer: \(\frac{5}{18}\)

Question 8.
\(0 . \overline{09}\)
Answer: \(\frac{90}{999}\)

Tell whether each equation has one solution, no solution, or an infinite number of solutions. Show your work. (Lesson 3.2)

Question 9.
3x – 2 = -3(\(\frac{2}{3}\) – x)
Answer:
3x – 2 = -3(\(\frac{2}{3}\) – x)
3x – = -2 + 3x
3x – 3x + 2 = 0
2
Infinite solutions

Question 10.
3x + 6 = -2(\(\frac{3}{2}\) – x)
Answer:
3x + 6 = -2(\(\frac{3}{2}\) – x)
3x + 6 = -3 + 2x
3x – 2x = -3 – 6
x = -9
One solution

Question 11.
5(6a – 6) + 40 = 3(10a – 7) + 31
Answer:
5(6a – 6) + 40 = 3(10a – 7) + 31
30a – 30 + 40 = 30a – 21 + 31
Infinte solution

Question 12.
3x + 7 = -8(\(\frac{3}{4}\) – x)
Answer:
3x + 7 = -8(\(\frac{3}{4}\) – x)
3x + 7 = -6 + 8x
3x – 8x = -6 – 7
-5x = -13
x = 13/5
One solution

Question 13.
\(\frac{1}{4}\)(2x – 1) = \(\frac{1}{2}\)x + \(\frac{3}{8}\)
Answer:
Given,
\(\frac{1}{4}\)(2x – 1) = \(\frac{1}{2}\)x + \(\frac{3}{8}\)
\(\frac{1}{4}\)x – \(\frac{1}{4}\) = \(\frac{1}{2}\)x + \(\frac{3}{8}\)
\(\frac{1}{4}\)x – \(\frac{1}{2}\)x = \(\frac{1}{4}\) + \(\frac{3}{8}\)
–\(\frac{1}{4}\)x = \(\frac{5}{8}\)
x = –\(\frac{5}{2}\)
No solution

Question 14.
\(\frac{1}{8}\)x + 6 = \(\frac{1}{16}\)(2x – 96)
Answer:
Given,
\(\frac{1}{8}\)x + 6 = \(\frac{1}{16}\)(2x – 96)
\(\frac{1}{8}\)x + 6 = \(\frac{1}{8}\)x – 6
No solution

Find the value of y when x = 4. (Lesson 3.3)

Question 15.
2x – 1 = \(\frac{1}{2}\) + y
Answer:
2x – 1 = \(\frac{1}{2}\) + y
x = 4
2(4) – 1 = \(\frac{1}{2}\) + y
7 = \(\frac{1}{2}\) + y
y = 7 – \(\frac{1}{2}\)
y = 6 \(\frac{1}{2}\) or 6.5

Question 16.
\(\frac{1}{4}\)(2y – 1) = 0.6 + \(\frac{5x}{8}\)
Answer:
Given,
\(\frac{1}{4}\)(2y – 1) = 0.6 + \(\frac{5x}{8}\)
X = 4
\(\frac{1}{4}\)(2y – 1) = 0.6 + \(\frac{5(4)}{8}\)
\(\frac{1}{4}\)(2y – 1) = 0.6 + \(\frac{20}{8}\)
0.25 (2y – 1) = 0.6 + 2.5
0.5y – 0.25 = 3.1
0.5y = 3.1 + 0.25
0.5y = 3.35
y = 6.7

Express y in terms of x. Find the value of y when x = 4. (Lesson 3.4)

Question 17.
6(3x + y) = 3
Answer:
Given equation
6(3x + y) = 3
18x + 6y = 3
Substitute the value of x in the given equation
x = 4
18(4) + 6y = 3
72 + 6y = 3
24 + 2y = 1
2y = -23
y = -23/2
y = -11 1/2

Question 18.
\(\frac{2 x-1}{4}\) = 3y
Answer:
Given,
\(\frac{2 x-1}{4}\) = 3y
Substitute the value of x in the given equation
3y = \(\frac{2 x-1}{4}\)
3y = (2(4) – 1)4
3y = 7/4
y = 7/12

Express x in terms of y. Find the value of x when y = -2. (Lesson 3.4)

Question 19.
(\(\frac{2x-y}{5}\)) = 9
Answer:
Given,
(\(\frac{2x-y}{5}\)) = 9
2x – y = 9 × 5
2x – y = 45
Substitute the value of y in the given equation
y = -2
2x – (-2) = 45
2x + 2 = 45
2x = 45 – 2
2x = 43
x = 43/2
x = 21.5

Question 20.
0.75(x + y) = 12
Answer:
Given,
0.75(x + y) = 12
0.75x + 0.75y = 12
Substitute the value of y in the given equation
y = -2
0.75x + 0.75 (-2) = 12
0.75x – 1.5 = 12
0.75x = 12 – 1.5
0.75x = 10.5
x = 14

Find the slope of the line passing through each pair of points. (Lesson 4.1)

Question 21.
A (1, 2), B (4, 8)
Answer:
The line passes through the points (1, 2) and (4, 8).
Slope m = \(\frac{8-2}{4-1}\)
= \(\frac{6}{3}\)
= 2
Thus the slope, m = 2

Question 22.
C(1, 4), D(2, 7)
Answer:
The line passes through the points (1, 4) and (2, 7).
Slope m = \(\frac{7-4}{2-1}\)
= \(\frac{3}{1}\)
= 3
Thus the slope, m = 3

Question 23.
E (0, 0), F (-7, 7)
Answer:
The line passes through the points (0, 0) and (-7, 7).
Slope m = \(\frac{7-0}{-7-0}\)
= \(\frac{7}{-7}\)
= -1
Thus the slope, m =-1

Question 24.
G (-3, 0), F (0, -6)
Answer:
The line passes through the points (-3, 0) and (0, -6).
Slope m = \(\frac{-6-0}{0+3}\)
= \(\frac{-6}{3}\)
= -2
Thus the slope, m = -2

Identify the y-intercept. Then calculate the slope using the points indicated. (Lessons 4.1, 4.2)

Question 25.

Answer:
The line passes through the points (0, 10) and (2.5, 30).
Slope m = \(\frac{30-10}{2.5-0}\)
= \(\frac{20}{2.5}\)
= 8
Thus the slope, m = 8
y-intercept = 10

Question 26.

Answer:
The line passes through the points (0, 3) and (0, 14).
m = 14/3
y-intercept = 14

For each equation, find the slope and the y-intercept of the graph of the equation. (Lesson 4.3)

Question 27.
y = 7x + 1
Answer:
Given,
y = 7x + 1
slope, m = 7
y-intercept, b = 1

Question 28.
y = -2x – 5
Answer:
Given,
y = -2x – 5
slope, m = -2
y-intercept, b = -5

Question 29.
2y = 4x + 6
Answer:
Given,
2y = 4x + 6
y = 2x + 3
slope, m = 2
y-intercept, b = 3

Question 30.
4y + 3x = 8
Answer:
Given,
4y + 3x = 8
4y = -3x + 8
y = 3/4 x + 2
slope, m = 3/4
y-intercept, b = 2

Use the given slope and y-intercept of a line to write an equation in slope-intercept form. (Lesson 4.3)

Question 31.
Slope, m = 3
y-intercept, b = 2
Answer:
Slope, m = 3
y-intercept, b = 2
The equation in slope-intercept is y = mx + b
y = 3x + 2

Question 32.
Slope, m = -1
y-intercept, b = 4
Answer:
Slope, m = -1
y-intercept, b = 4
The equation in slope-intercept is y = mx + b
y = -1x + 4

Question 33.
Slope, m = 5
y-intercept, b = -2
Answer:
Slope, m = 5
y-intercept, b = -2
The equation in slope-intercept is y = mx + b
y = 5x – 2

Question 34.
Slope, m = –\(\frac{3}{2}\)
y-intercept, b = -5
Answer:
Slope, m = –\(\frac{3}{2}\)
y-intercept, b = -5
The equation in slope-intercept is y = mx + b
y = –\(\frac{3}{2}\)x – 5

Solve. Show your work. (Lesson 4.3)

Question 35.
Write an equation of the line parallel to 2y = 4x + 3 that has a y-intercept of 4.
Answer: y = 2x + 4

Question 36.
A line has slope -4 and passes through the point (\(\frac{3}{4}\), 3). Write an equation of the line.
Answer:
y = -4x + 6

Question 37.
Write an equation of the line that passes through the point (2, 3) and is parallel to 3y + 2x = 7.
Answer:
y = –\(\frac{2}{3}\)x + \(\frac{13}{3}\)

Use graph paper. Graph each linear equation. Use 1 grid scale to represent 1 unit on both axes for the interval -5 to 5. (Lesson 4.4)

Question 38.
y = -2x + 8
Answer:

Question 39.
y = -2 – 3x
Answer:

Question 40.
y = \(\frac{1}{2}\)x – 3
Answer:

Solve. Show your work. (Lesson 4.5)

Question 41.
Bobby and Chloe each have a bank account. The balance, y dollars, in each account for x weeks, is shown in the graph.

a) Who saved money and who withdrew money during the 10 weeks?
Answer: Bobby’s savings increased, so he saved money. Chloe’s savings decreased, so she withdrew money.

b) Whose balance changed more over 10 weeks?
Answer:
Bobby changed by $100, Chloe’s changed by -$50; Bobby’s changed more.

c) Explain what information the coordinates of P give about the situation.
Answer:
After 5 weeks, Bobby and Chloe have the same amount of money, which is $75.

Problem Solving

Solve. Show your work.

Question 42.
The diagram shows a sheet of metal of width y inches. It is bent into a U-shaped gutter that is used to channel rain from a roof. The horizontal section of the gutter shown on the right is 10 inches wide and the heights are in the ratio of 2 : 3. (Chapter 3).

a) Let x represents the longer height of the gutter, in inches. Write a linear equation for the width of the sheet of metal, y inches-, in terms of the longer height of the gutter, x inches.
Answer: y = 5/3 x + 10

b) The width of the sheet of metal is 30 inches. Calculate the longer height of the gutter.
Answer: 12 in

Question 43.
In a grocery store, each apple costs $0.50, each orange costs $0.40, and each pear costs $0.30. Mrs. Fortney bought y apples, three times as many oranges as apples, and 7 fewer pears than apples. $he spent a total of $19.90 on the fruits. (Chapter 3)
a) Write a linear equation to find the amount spent on each fruit.
Answer: 50y + 120y + 30y – 210 = 1990
200y = 2200

b) Find the total cost spent on apples and pears.
Answer: $6.70

Question 44.
Jack traveled from his home to Denver at an average speed of x miles per hour. He arrived in \(\frac{3}{4}\) hour and took a 15-minute break. From Denver, he traveled at an average speed of (x + 2) miles per hour and reached his grandmother’s place in 1.5 hours. (Chapter 3)
a) Write a linear equation for the total distance traveled, D miles, in terms of average speed, x miles per hour.
Answer:
D = \(\frac{3}{4}\)x + \(\frac{3}{2}\)
D = \(\frac{3}{4}\)x + 1.5
D = \(\frac{9}{4}\)x + 3

b) The total distance traveled for the whole journey was 120 miles. Find the average speed for both parts of the journey.
Answer:
First part: 52 mi/h
Second part: 54 mi/h

Use graph paper. Solve.

Question 45.
Xavier walks into an elevator in the basement of a building. Its control panel displays “0” for the floor number. As Xavier goes up, the numbers increase one by one on the display. The table shows the floor numbers and the distance from ground level. (Chapter 4)

a) Graph the relationship between the distance of the elevator from ground level at different floor numbers. Use 1 grid square to represent 1 unit on the horizontal axis for the x interval 0 to 4, and 1 grid square for 10 units on the vertical axis for the y interval -10 to 30.
Answer:

b) Find the vertical intercept of the graph and explain what information it gives about the situation.
Answer: -10
It is the distance of the elevator from ground level, 10 ft below ground level when it is at the basement level.

c) Find the slope of the graph and explain what information it gives about the situation.
Answer: 10
The slope represents the increase in distance, in ft, of the elevator from ground level for every increase of the 1-floor number

d) Write an equation relating the distance of the elevator from ground level and the floor number on the display.
Answer: y = 10x – 10

e) What is the distance of the elevator from the ground level of the highest floor that is less than 165 feet? Is there a floor number with a distance from the ground level of exactly 165 feet?
Answer: 160 ft
There is no floor number with distance from ground level for exactly 165 ft.