﻿ Solving Linear Systems (using Substitution Method) Worksheet | Problems & Solutions

# Solving Linear Systems (using Substitution Method) Worksheet

Solving Linear Systems (using Substitution Method) Worksheet
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1.
Which of the following ordered pairs satisfies the linear system?
$x$ - $y$ = 6     [Equation 1]
2$x$ + 5$y$ = -9 [Equation 2]
 a. (-3, -3) b. (3, 3) c. (-3, 3) d. (3, -3)

#### Solution:

y = x - 6
[Rearrange Equation 1.]

2x + 5(x - 6) = -9
[Substitute y = (x - 6) in equation 2.]

7x - 30 = -9
[Combine like terms.]

7x = 21

x = 3
[Divide each side by 7.]

y = x - 6 = 3 - 6
[Substitute x = 3 in the revised equation 1.]

y = -3
[Simplify.]

The solution for the linear system is (3, -3).

2.
Which of the following ordered pairs satisfies the linear system?
3$x$ - $y$ = 0      [Equation 1]
$x$ - 3$y$ = 0      [Equation 2]
 a. (3, -3) b. (-3, 3) c. (0, 0) d. (3, 3)

#### Solution:

y = 3x
[Rearrange Equation 1.]

x - 3(3x) = 0
[Substitute y = 3x in Equation 2.]

-8x = 0
[Combine like terms.]

x = 0
[Divide each side by -8.]

y = 3x = 3(0)
[Substitute x = 0 in revised Equation 1.]

y = 0
[Simplify.]

The solution for the linear system is (0, 0).

3.
A shopkeeper sold 60 softballs and basketballs last week for a total sum of $516. The price of a softball was$8 and that of a basketball was $20. How many softballs and basketballs did the shopkeeper sell?  a. 56 softballs, 4 basketballs b. 4 softballs, 57 basketballs c. 57 softballs, 3 basketballs d. None of the above #### Solution: Let x be the number of softballs and y be the number of basket balls. x + y = 60 [Equation 1.] 8x + 20y = 516 [Equation 2.] x = -y + 60 [Revise Equation 1.] 8(-y + 60) + 20y = 516 [Substitute x = -y + 60 in Equation 2.] 12y + 480 = 516 [Combine like terms.] 12y = 36 [Subtract 480 from each side.] y = 3 [Divide each side by 12.] x = -(3) + 60 = 57 [Substitute y = 3 in revised equation 1.] The shopkeeper sold 57 softballs and 3 basketballs. Correct answer : (3) 4. The price of each marble was$4 and the price of each dice was $2. A total of 90 marbles and dice were purchased for$262. Find the number of marbles and the number of dice purchased.
 a. 41 marbles, 49 dice b. 37 marbles, 53 dice c. 39 marbles, 51 dice d. 43 marbles, 47 dice

#### Solution:

Let x be the number of marbles and y be the number of dice purchased.

x + y = 90
[Equation 1.]

4x + 2y = 262
[Equation 2.]

x = 90 - y
[Rearrange Equation 1.]

4(90 - y) + 2y = 262
[Substitute x = 90 - y in Equation 2.]

-2y + 360 = 262
[Combine like terms.]

-2y = -98
[Subtract 360 from each side.]

y = 49
[Divide each side by -2.]

x = 90 - 49 = 41
[Substitute y = 49 in the revised equation 1.]

41 marbles and 49 dice are purchased.

5.
Which of the following linear systems has the solution (3, 0)?
 a. x + 3y = -3 and x + 4y = -3 b. x - 3y = -3 and x + 4y = 3 c. -x + 3y = 3 and x - 4y = 3 d. x + 3y = 3 and x - 4y = 3

#### Solution:

Substitute (3, 0) in all the linear systems and verify.

3 + 3(0) = 3 + 0 = 3
For the linear system x + 3y = 3, x - 4y = 3:
[Substitute (3, 0) in x + 3y = 3.]

3 - 4(0) = 3 - 0 = 3
[Substitute (3, 0) in x - 4y = 3. ]

So, the linear system in the choice D has the solution (3, 0).

6.
Which of the following equations would you use to isolate the variable?
6$x$ + 9$y$ =1   [Equation 1]
$x$ - 5$y$ = -4    [Equation 2]
 a. Equation 1 b. Equation 2

#### Solution:

The coefficient of x in Equation 2 is 1, so it is easier to isolate the variable.

So, Equation 2 can be used for isolating the variable.

7.
Which of the following equations would you use to isolate the variable?
-5$x$ + $y$ = 14   [Equation 1]
3$x$ + 2$y$ = 8    [Equation 2]
 a. Equation 1 b. Equation 2

#### Solution:

Since the coefficient of y in Equation 1 is 1, it is easier to isolate the variable.

So, Equation 1 can be used for isolating the variable.

8.
Which of the following ordered pairs satisfies the linear system?
2$x$ - 2$y$ = 2   [Equation 1.]
-$x$ + 5$y$ = -37 [Equation 2.]
 a. (-8, 9) b. (-9, 9) c. (-8, -9) d. (-9, -9)

#### Solution:

-x + 5y = -37
[Original equation 2.]

x = 37 + 5y
[Revise equation 2.]

2(37 + 5y) - 2y = 2
[Substitute 37 + 5y for x in equation 1.]

8y + 74 = 2
[Combine like terms.]

8y = -72
[Subtract 74 from each side.]

y = -9
[Divide each side by 8.]

x = 37 + 5y = 37 + 5(-9)
[Substitute -9 for y in the revised equation 2.]

x = -8
[Simplify.]

The solution for the linear system is (-8, -9).

9.
Lindsay purchased a total of 42 books and toys for the Ashton play school. Each book cost $11 and each toy cost$6. How many books and toys did she buy for $367?  a. 19 books and 23 toys b. 22 books and 20 toys c. 23 books and 19 toys d. 20 books and 22 toys #### Solution: Let x be the number of books and y be the number of toys, Lindsay purchased. x + y = 42 --- (1) [Linear equation for the total books and toys.] 11x + 6y = 367 --- (2) [Equation for the total cost of the books and toys.] 11x + 6(-x + 42) = 367 [From equation 1, y = -x + 42. Substitute it in equation 2.] 5x + 252 = 367 [Combine like terms.] 5x = 115 [Subtract 252 from each side.] x = 23 [Divide each side by 5.] y = -(23) + 42 = 19 [Substitute x = 23 in equation 1.] Lindsay bought 23 books and 19 toys. Correct answer : (3) 10. Tony purchased 24 pens and markers for$128. The price of a marker was $9 and the price of a pen was$5. How many pens and markers did Tony buy?
 a. 22 markers, 2 pens b. 21 markers, 3 pens c. 2 markers, 22 pens d. 1 markers, 23 pens

#### Solution:

Let x be the number of markers and y be the number of pens.

x + y = 24
[Equation 1.]

9x + 5y = 128
[Equation 2.]

y = 24 - x
[Revise Equation 1.]

9x + 5(24 - x) = 128
[Substitute y = 24 - x in Equation 2.]

4x + 120 = 128
[Combine like terms.]

4x = 8
[Subtract 120 from each side.]

x = 2
[Divide each side by 4.]

y = 24 - 2 = 22
[Substitute x = 2 in the revised equation 1.]

Tony bought 2 markers and 22 pens.