Go through the Math in Focus Grade 7 Workbook Answer Key Chapter 3 Lesson 3.1 Adding Algebraic Terms to finish your assignments.

## Math in Focus Grade 7 Course 2 A Chapter 3 Lesson 3.1 Answer Key Adding Algebraic Terms

### Math in Focus Grade 7 Chapter 3 Lesson 3.1 Guided Practice Answer Key

**Copy and complete to simplify each expression.**

Question 1.

0.2y + 0.6y

0.2y + 0.6y =

Answer:

0.2y + 0.6y = 0.8y.

Explanation:

Question 2.

0.8p + 0.5p

0.8 p + 0.5p =

Answer:

0.8 p + 0.5p = 1.3p.

Explanation:

Question 3.

1.3g + 0.9g

1.3g + 0.9g =

Answer:

1.3g + 0.9g = 2.2g.

Explanation:

1.3g + 0.9g = 2.2g.

**Copy and complete to simplify each expression.
**Question 4.

x + \(\frac{3}{4}\)x

x + \(\frac{3}{4}\)x =

Answer:

Explanation:

x + \(\frac{3}{4}\)x = (4x + 3x) ÷ 4

= 7x ÷ 4 or \(\frac{7}{4}\)x

Question 5.

\(\frac{1}{2}\)p + \(\frac{2}{5}\)p

The LCD of \(\frac{1}{2}\) and \(\frac{2}{5}\) is .

So, p is divided into p sections.

\(\frac{1}{2}\)p + \(\frac{2}{5}\)p =

Answer:

\(\frac{1}{2}\)p + \(\frac{2}{5}\)p = \(\frac{15}{10}\)p

Explanation:

The LCD of \(\frac{1}{2}\) and \(\frac{2}{5}\) is 10.

So, p is divided into 10 p sections.

\(\frac{1}{2}\)p + \(\frac{2}{5}\)p

= [(1 ×5)p + (2 × 5)p] ÷ 10

= (5p + 10p) ÷ 10

= 15p ÷ 10 or \(\frac{15}{10}\)p

**Simplify each expression.
**Question 6.

m + \(\frac{5}{6}\)m

Answer:

m + \(\frac{5}{6}\)m = \(\frac{11}{6}\)m

Explanation:

m + \(\frac{5}{6}\)m

= (6m + 5m) ÷ 6

= 11m ÷ 6 or \(\frac{11}{6}\)m.

Question 7.

\(\frac{1}{4}\)x + \(\frac{2}{3}\)x

Answer:

\(\frac{1}{4}\)x + \(\frac{2}{3}\)x = \(\frac{7}{12}\)x

Explanation:

\(\frac{1}{4}\)x + \(\frac{2}{3}\)x

= LCM of 4 n 3 = 12.

\(\frac{1}{4}\)x + \(\frac{2}{3}\)x

= [(1 × 3) + (2 × 4)]x ÷ 12

= (3 + 4)x ÷ 12

= 7x ÷ 12 or \(\frac{7}{12}\)x

Question 8.

Math Journal In questions 5 and 7 you can multiply the two denominators to get the LCD. Does this always work? Explain.

Answer:

You can multiply the two denominators to get the LCD because there is no common factors between the denominators, so you can multiply the numbers to simplify the problem.

Explanation:

Question5:

\(\frac{1}{2}\)p + \(\frac{2}{5}\)p

LCD of 2 n 5 = 2 × 5 = 10.

Question7:

\(\frac{1}{4}\)x + \(\frac{2}{3}\)x

LCD of 4 n 3 = 4 × 3 = 12.

### Math in Focus Course 2A Practice 3.1 Answer Key

**Simplify each expression with decimal coefficients.**

Question 1.

0.5y + 0.2y

Answer:

0.5y + 0.2y = 0.7y.

Explanation:

0.5y + 0.2y = 0.7y.

Question 2.

0.2x + 0.8x

Answer:

0.2x + 0.8x = 1.0x.

Explanation:

0.2x + 0.8x = 1.0x.

Question 3.

x + 0.6x

Answer:

x + 0.6x = 1.6x.

Explanation:

x + 0.6x

= 1x + 0.6x

= 1.6x.

Question 4.

0.4a + 1.2a

Answer:

0.4a + 1.2a = 1.6a.

Explanation:

0.4a + 1.2a = 1.6a.

Question 5.

0.7b + 0.9b

Answer:

0.7b + 0.9b = 1.6b.

Explanation:

0.7b + 0.9b = 1.6b

Question 6.

0.6m + 0.8m

Answer:

0.6m + 0.8m = 1.4m.

Explanation:

0.6m + 0.8m = 1.4m.

Question 7.

0.5k + 1.6k

Answer:

0.5k + 1.6k = 2.1k.

Explanation:

0.5k + 1.6k = 2.1k.

Question 8.

0.8a + 1.8a

Answer:

0.8a + 1.8a = 2.6a.

Explanation:

0.8a + 1.8a = 2.6a.

**Simplify each expression with fractional coefficients.
**Question 9.

\(\frac{1}{5}\)x + \(\frac{2}{5}\)x

Answer:

\(\frac{1}{5}\)x + \(\frac{2}{5}\)x = \(\frac{3}{5}\)x

Explanation:

\(\frac{1}{5}\)x + \(\frac{2}{5}\)x

= [(1 + 2)x] ÷ 5

= 3x ÷ 5

= \(\frac{3}{5}\)x

Question 10.

\(\frac{3}{7}\)p + \(\frac{2}{7}\)p

Answer:

\(\frac{3}{7}\)p + \(\frac{2}{7}\)p = \(\frac{5}{7}\)p

Explanation:

\(\frac{3}{7}\)p + \(\frac{2}{7}\)p

= [(3 + 2)p] ÷ 7

= 5p ÷ 7

= \(\frac{5}{7}\)p

Question 11.

\(\frac{5}{8}\)m + \(\frac{7}{8}\)m

Answer:

\(\frac{5}{8}\)m + \(\frac{7}{8}\)m = \(\frac{3}{2}\)m

Explanation:

\(\frac{5}{8}\)m + \(\frac{7}{8}\)m

= (5 + 7)m ÷ 8

= 12m ÷ 8

= 6m ÷ 4

= 3m ÷ 2

= \(\frac{3}{2}\)m

Question 12.

\(\frac{4}{9}\)n + \(\frac{7}{9}\)n

Answer:

\(\frac{4}{9}\)n + \(\frac{7}{9}\)n = \(\frac{13}{9}\)n.

Explanation:

\(\frac{4}{9}\)n + \(\frac{7}{9}\)n

= (4 + 9)n ÷ 9

= 13n ÷ 9

= \(\frac{13}{9}\)n

**Simplify each expression with fractional coefficients by rewriting the fractions.
**Question 13.

\(\frac{1}{6}\)a + \(\frac{1}{3}\)a

Answer:

\(\frac{1}{6}\)a + \(\frac{1}{3}\)a = \(\frac{1}{2}\)a

Explanation:

\(\frac{1}{6}\)a + \(\frac{1}{3}\)a

= LCD of 6 n 3 = 3.

= (1 + 2)a ÷ 6

= 3a ÷ 6

= a ÷ 2 or \(\frac{1}{2}\)a

Question 14.

\(\frac{1}{2}\)k + \(\frac{3}{8}\)k

Answer:

\(\frac{1}{2}\)k + \(\frac{3}{8}\)k =

Explanation:

\(\frac{1}{2}\)k + \(\frac{3}{8}\)k

= LCD of 2 n 8 = 8.

= (4 + 3)k ÷ 8

= 7k ÷ 8 or \(\frac{7}{8}\)k

Question 15.

\(\frac{2}{5}\)p + \(\frac{7}{10}\)p

Answer:

\(\frac{2}{5}\)p + \(\frac{7}{10}\)p = \(\frac{11}{10}\)p

Explanation:

\(\frac{2}{5}\)p + \(\frac{7}{10}\)p

= LCD of 5 n 10 = 10.

= [(2 ×2) + 7]p ÷ 10

= (4 + 7)p ÷ 10

= 11p ÷ 10 or \(\frac{11}{10}\)p

Question 16.

\(\frac{7}{9}\)r + \(\frac{2}{3}\)r

Answer:

\(\frac{7}{9}\)r + \(\frac{2}{3}\)r = \(\frac{13}{9}\)r

Explanation:

\(\frac{7}{9}\)r + \(\frac{2}{3}\)r

= LCD of 9 n 3 = 9.

= [7 + (2 × 3)]r ÷ 9

= (7 + 6)r ÷ 9

= 13r ÷ 9 or \(\frac{13}{9}\)r

Question 17.

\(\frac{1}{4}\)a + \(\frac{1}{3}\)a

Answer:

\(\frac{1}{4}\)a + \(\frac{1}{3}\)a = \(\frac{7}{12}\)a

Explanation:

\(\frac{1}{4}\)a + \(\frac{1}{3}\)a

= LCD of 4 n 3 = 12.

= [(1 × 3) + (1 × 4)]a ÷12

= (3 + 4)a ÷ 12

= 7a ÷ 12 or \(\frac{7}{12}\)a

Question 18.

\(\frac{1}{2}\)x + \(\frac{2}{5}\)x

Answer:

\(\frac{1}{2}\)x + \(\frac{2}{5}\)x = \(\frac{9}{10}\)x

Explanation:

\(\frac{1}{2}\)x + \(\frac{2}{5}\)x

= LCD of 2 n 5 = 10.

= [(1 × 5) + (2 × 2)]x ÷ 10

= (5 + 4)x ÷ 10

= 9x ÷ 10 or \(\frac{9}{10}\)x

Question 19.

\(\frac{3}{5}\)p + \(\frac{3}{4}\)p

Answer:

\(\frac{3}{5}\)p + \(\frac{3}{4}\)p = \(\frac{27}{20}\)p

Explanation:

\(\frac{3}{5}\)p + \(\frac{3}{4}\)p

= LCD of 5 n 4 = 20.

= [(3 × 4) + (3 × 5)]p ÷ 20

= (12 + 15)p ÷ 20

= 27p ÷ 20 or \(\frac{27}{20}\)p

Question 20.

\(\frac{4}{5}\)y + \(\frac{1}{3}\)y

Answer:

\(\frac{4}{5}\)y + \(\frac{1}{3}\)y = \(\frac{17}{15}\)y

Explanation:

\(\frac{4}{5}\)y + \(\frac{1}{3}\)y

= LCD of 5 n 3 = 15.

= [(4 × 3) + (1 × 5)]y ÷ 15

= (12 + 5)y ÷ 15

= 17y ÷ 15 or \(\frac{17}{15}\)y

**Solve. Show your work.
**Question 21.

The figures show rectangle A and rectangle B. Write and simplify an algebraic expression for each of the following.

a) The perimeter of rectangle A.

Answer:

Perimeter of rectangle A = \(\frac{28}{3}\)m

Explanation:

Perimeter of rectangle A = 2(Length + Width)

= 2(\(\frac{2}{3}\)m + 4m)

= 2[(2 + 12)m ÷ 3]

= 2(14m ÷ 3)

= 28m ÷ 3 or \(\frac{28}{3}\)m

b) The perimeter of rectangle B.

Answer:

Perimeter of rectangle B = \(\frac{35}{9}\)m

Explanation:

Perimeter of rectangle B = 2(Length + Width)

= 2(1.5m + \(\frac{4}{9}\)m)

= 2[(1.5 × 9) + 4)m ÷ 9]

= 2[(13.5 + 4) ÷ 9]m

= 2(17.5m ÷ 9)

= 35m ÷ 9 or \(\frac{35}{9}\)m

c) The sum of the perimeters of the two rectangles.

Answer:

Sum of the perimeters of the two rectangles = \(\frac{119}{9}\)m.

Explanation:

Perimeter of rectangle A = \(\frac{28}{3}\)m

Perimeter of rectangle B = \(\frac{35}{9}\)m

Sum of Perimeter of rectangle A n Perimeter of rectangle B

= \(\frac{28}{3}\)m + \(\frac{35}{9}\)m

LCD of 3 n 9 = 9.

= [(28 × 3) + (35 × 1)]m ÷ 9

= (84 + 35)m ÷ 9

= 119m ÷ 9 or \(\frac{119}{9}\)m

Question 22.

The length and width of two rectangular gardens are shown. Find the sum of the areas of the two gardens in simplest form.

Answer:

Sum of the areas of the two gardens = 13.1y ft.

Explanation:

Area of rectangular garden 1 = Length × Width

= 2y ft × 3.3 ft

= 6.6y square ft.

Area of rectangular garden 2 = Length × Width

= 5y ft × 1.3y ft

= 6.5y square ft.

Sum of Area of rectangular garden 1 and Area of rectangular garden 2

= 6.6y square ft + 6.5y square ft

= 13.1y square ft.

Question 23.

Math Journal James and Evan simplified the same algebraic expression. Their work is shown.

Describe a method each person might have used to get his answer. Which method do you prefer? Why?

Answer:

Both the method are easy to solve. I would like to prefer to simplified method.

Explanation:

James algebraic expression: Simplified method

\(\frac{1}{5}\)x + 0.3x = \(\frac{1}{2}\)x.

Evan algebraic expression: Fractional method

\(\frac{1}{5}\)x + 0.3x = 0.5x.

Question 24.

Which of the following expressions has a greater value if y is a positive number? Explain your reasoning.

Answer:

1.4y + \(\frac{2}{5}\)y has a greater value if y is a positive number.

Explanation:

Expression 1:

1.4y + \(\frac{2}{5}\)y

If y = 1.

= (1.4 × 1) + \(\frac{2}{5}\) × 1

= 1.4 + \(\frac{2}{5}\)

= [(1.4 × 5) + 2] ÷ 5

= (7 + 2) ÷ 5

= 9 ÷ 5 or \(\frac{9}{5}\)

= 1.8.

Expression 2:

\(\frac{1}{3}\)y + \(\frac{3}{4}\)y

If y = 1.

= [\(\frac{1}{3}\) × 1] + \(\frac{3}{4}\) × 1

= \(\frac{1}{3}\) + \(\frac{3}{4}\)

LCD of 3 n 4 = 12.

= [(1 × 4) + (3 × 3)] ÷ 12

= (4 + 9) ÷ 12

= 13 ÷ 12 or \(\frac{13}{12}\)

= 1.08.

Question 25.

A restaurant serves x meals of chicken quarters daily and makes soup each day using \(\frac{1}{2}\) of a chicken. The chef expresses the number of chickens she uses each 1 1 day as \(\frac{1}{4}\)x + \(\frac{1}{2}\). How many chickens does she use in three days?

Answer:

Amount of chicken used in 3 days = 9x ÷ 4. or \(\frac{9}{4}\)x

Explanation:

A restaurant serves x meals of chicken quarters daily.

makes soup each day using \(\frac{1}{2}\) of a chicken.

=> Amount of soup each day used = \(\frac{1}{2}\) × x = \(\frac{1}{2}\).

Amount of chicken used daily = \(\frac{1}{4}\)x + \(\frac{1}{2}\).

Number of days = 3.

Amount of chicken used in 3 days = Amount of chicken used daily × Number of days

= [\(\frac{1}{4}\)x + \(\frac{1}{2}\)] × 3

= LCD of 4 n 2 = 4.

= [[(1 × 1)x + (1 × 2)] ÷ 4 ] × 3

= [(1x + 2x) ÷ 4]× 3

= (3x ÷ 4) × 3

= 9x ÷ 4.

Question 26.

Mary simplified the algebraic expression \(\frac{2}{3}\)x + \(\frac{1}{4}\)x as shown below.

Describe and correct the error Mary made.

Answer:

Mary did not take the LCD and solved the problem, that is her error.

\(\frac{2}{3}\)x + \(\frac{1}{4}\)x = \(\frac{11}{12}\)x.

Explanation:

\(\frac{2}{3}\)x + \(\frac{1}{4}\)x

= LCD of 3 n 4 = 12.

= [(2 ×4)x + (1 × 3)x] ÷ 12

= (8x + 3x) ÷ 12

= 11x ÷ 12 or \(\frac{11}{12}\)x