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

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 KeyCopy and complete to simplify each expression.

Question 1.
0.2y + 0.6y

0.2y + 0.6y =
0.2y + 0.6y = 0.8y.

Explanation:

Question 2.
0.8p + 0.5p

0.8 p + 0.5p =
0.8 p + 0.5p = 1.3p.

Explanation:

Question 3.
1.3g + 0.9g
1.3g + 0.9g =
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 =

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 =
$$\frac{1}{2}$$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
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
$$\frac{1}{4}$$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.
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
0.5y + 0.2y = 0.7y.

Explanation:
0.5y + 0.2y = 0.7y.

Question 2.
0.2x + 0.8x
0.2x + 0.8x = 1.0x.

Explanation:
0.2x + 0.8x = 1.0x.

Question 3.
x + 0.6x
x + 0.6x = 1.6x.

Explanation:
x + 0.6x
= 1x + 0.6x
= 1.6x.

Question 4.
0.4a + 1.2a
0.4a + 1.2a = 1.6a.

Explanation:
0.4a + 1.2a = 1.6a.

Question 5.
0.7b + 0.9b
0.7b + 0.9b = 1.6b.

Explanation:
0.7b + 0.9b = 1.6b

Question 6.
0.6m + 0.8m
0.6m + 0.8m = 1.4m.

Explanation:
0.6m + 0.8m = 1.4m.

Question 7.
0.5k + 1.6k
0.5k + 1.6k = 2.1k.

Explanation:
0.5k + 1.6k = 2.1k.

Question 8.
0.8a + 1.8a
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
$$\frac{1}{5}$$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
$$\frac{3}{7}$$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
$$\frac{5}{8}$$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
$$\frac{4}{9}$$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
$$\frac{1}{6}$$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
$$\frac{1}{2}$$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
$$\frac{2}{5}$$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
$$\frac{7}{9}$$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
$$\frac{1}{4}$$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
$$\frac{1}{2}$$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
$$\frac{3}{5}$$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
$$\frac{4}{5}$$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.
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.
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.

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.

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?
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.

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?
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.
$$\frac{2}{3}$$x + $$\frac{1}{4}$$x = $$\frac{11}{12}$$x.
$$\frac{2}{3}$$x + $$\frac{1}{4}$$x
= 11x ÷ 12 or $$\frac{11}{12}$$x