# Solving Linear Systems by Substitution Worksheet

Solving Linear Systems by Substitution Worksheet
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1.
A shopkeeper sold 32 softballs and basketballs last week for a total sum of $224. The price of a softball is$6 and that of a basketball is $14. How many softballs and basketballs did the shopkeeper sell?  a. 5 softballs, 28 basketballs b. 27 softballs, 5 basketballs c. 28 softballs, 4 basketballs d. 30 softballs, 30 basketballs #### Solution: Let x be the number of softballs and y be the number of basket balls. x + y = 32 [Equation 1.] 6x + 14y = 224 [Equation 2.] x = - y + 32 [Rearrange equation 1.] 6(- y + 32) + 14y = 224 [Substitute the values.] 8y + 192 = 224 [Group the like terms.] 8y = 32 [Subtract 192 from the two sides of the equation.] y = 4 [Divide throughout by 8.] x = - (4) + 32 = 28 [Substitute the values.] The shopkeeper sold 28 softballs and 4 basketballs. Correct answer : (3) 2. Danielle purchased a total of 48 books and toys for the Boone play school. Each book costs$23 and each toy costs $6. How many books and toys did she buy for$730?
 a. 25 books and 23 toys b. 23 books and 25 toys c. 22 books and 26 toys d. 26 books and 22 toys

#### Solution:

Let x be the number of books and y be the number of toys, Danielle purchased.

x + y = 48 --- (1)
[Express as a linear equation.]

23x + 6y = 730 --- (2)
[Equation for the total cost of the books and toys.]

23x + 6(- x + 48) = 730
[From equation 1, y = - x + 48. Substitute it in equation 2.]

17x + 288 = 730
[Group the like terms.]

17x = 442
[Subtract 288 from the two sides of the equation.]

x = 26
[Divide throughout by 17.]

y = - (26) + 48 = 22
[Substitute the values.]

Danielle bought 26 books and 22 toys.

3.
A circus company sold 201 tickets on a particular day. The entry fee is $22 for an adult and$7 for a child. The total amount collected was $2,802. How many adults and children went to the circus on that day?  a. 108 adults and 93 children b. 109 adults and 92 children c. 92 adults and 109 children d. 93 adults and 108 children #### Solution: Let x be the number of adults. Let y be the number of children. Total number of tickets = 201 So, Number of adults + Number of children = 201 x + y = 201 [Equation 1.] Total amount collected =$2802

So, (Cost of an adult ticket) × (Number of adults) + (Cost of a child ticket) × (Number of children) = $2802 22x + 7y = 2802 [Substitute the values.] 22(- y + 201) + 7y = 2802 [Substitute the values.] -15y + 4422 = 2802 [Group the like terms.] -15y = -1620 [Subtract 4422 from the two sides of the equation.] y = 108 [Divide throughout by -15.] x = - (108) + 201 = 93 [Substitute the values.] 93 adults and 108 children went to the circus on that day. Correct answer : (4) 4. A mechanical plant hires 1,199 labors on a daily wage scheme paying$9,691. Men are paid $9 and women are paid$7. Find the number of men and women hired.
 a. 649 men, 550 women b. 550 men, 649 women c. 551 men, 648 women d. 648 men, 501 women

#### Solution:

Let number of men be x.

Let number of women be y.

x + y = 1199 --- (1)
[Express as a linear equation.]

9x + 7y = 9691 --- (2)
[Equation for the daily wages paid.]

y = 1199 - x
[Rearrange equation 1.]

9x + 7(1199 - x) = 9691
[Substitute the values.]

2x + 8393 = 9691
[Group the like terms.]

2x = 1298
[Subtract 8393 from the two sides of the equation.]

x = 649
[Divide throughout by 2.]

y = 1199 - 649 = 550
[Substitute the values.]

550 women and 649 men are hired.

5.
Find the value of $m$ such that the lines $y$ = 5$x$ - 1, $x$ = 4 and $y$ = $m$$x$ + 4 are concurrent.
 a. $\frac{15}{4}$ b. $\frac{1}{19}$ c. $\frac{19}{4}$ d. $\frac{4}{15}$

#### Solution:

The equations of three lines are
y = 5x - 1 -------- (1)
x = 4 -------- (2)
y = mx + 4 -------- (3)
[Given.]

The given equations are concurrent . So, they pass through the same point.
[Given.]

From equation (2), x = 4

Consider equation (1), y = 5x - 1

y = 5(4) - 1 = 19
[Substitute x = 4, from step 3.]

Consider equation (3), y = mx + 4

19 = m(4) + 4
[Substitute x = 4 and y = 19, from steps 3 and 5.]

19 = 4m + 4
[Multiply.]

15 = 4m
[Subtract 4 from both sides.]

m = 15 / 4
[Divide both sides by 4.]

Therefore, the value of m = 15 / 4.

6.
Kelsey purchased a total of 31 books and toys for the Lake City play school. Each book costs $35 and each toy costs$9. How many books and toys did she buy for \$747?
 a. 18 books and 13 toys b. 28 books and 59 toys c. 31 books and 1 toy d. 21 books and 83 toys

#### Solution:

Let x be the number of books and y be the number of toys Kelsey purchased.

x + y = 31 --- (1)
[Linear equation for the total books and toys.]

35x + 9y = 747 --- (2)
[Equation for the total cost of the books and toys.]

35x + 9(- x + 31) = 747
[From equation 1, y = - x + 31. Substitute it in equation 2.]

26x + 279 = 747
[Combine like terms.]

26x = 468
[Subtract 279 from each side.]

x = 18
[Divide each side by 26.]

y = - (18) + 31 = 13
[Substitute x = 18 in equation 1.]

Kelsey bought 18 books and 13 toys.

7.
Solve the system by substitution.
4$x$ + $y$ = 5
- 16$x$ - 4$y$ = - 20
 a. (20.3, -76.2) b. no solution c. infinite solutions d. (1, 1)

#### Solution:

4x + y = 5
[First equation.]

y = 5 - 4x
[Solve for y.]

- 16x - 4y = - 20
[Second equation.]

- 16x - 4(5 - 4x) = - 20
[Substitute the values.]

- 16x - 20 + 16x = - 20

0 = 0
[Always true]

So, the number of solutions is infinite.

The solution is {(x, y): 4x + y = 5}.

8.
Solve the system by substitution:
10$x$ + 10$y$ = 100
12$x$ = - 12$y$ + 108
 a. (100, 0) b. No solution c. (100, 10) d. {($x$, $y$): 10$x$ + 10$y$ = 100}

#### Solution:

12x = - 12y + 108
[Second equation.]

x = - y + 9
[Divide throughout by 12 .]

10x + 10y = 100
[First equation.]

10( - y + 9) + 10y = 100
[Substitute the values.]

- 10y + 90 + 10y = 100

90 = 100

So, there is no solution and the system is inconsistent.

9.
Find the value of $a$ for the given dependent system.
$y$ = 5$x$ + $a$
30$x$ - 6$y$ = 54
 a. - 9 b. 55 c. 54 d. 6

#### Solution:

y = 5x + a
[First equation.]

30x - 6y = 54
[Second equation.]

30x - 6(5x + a) = 54
30x - 30x - 6a = 54
[Substitute the values.]

- 6a = 54

a = - 9
[Divide throughout by - 6.]

10.
Find the value of $a$ for the given dependent system.
6$y$ = 7$x$
30$y$ - 41$a$ - 35$x$ = 0
 a. 13 b. 8 c. 9

#### Solution:

6y = 7x
[First equation.]

y = 7 / 6x
[Simplify.]

30y - 41a - 35x = 0
[Second equation.]

30(7 / 6x) - 41a - 35x = 0
[Substitute the values.]

35x - 41a - 35x = 0

41a = 0 a = 0