SOLVING SYSTEM OF LINEAR EQUATIONS BY SUBSTITUTION

Solve the following system of linear equations using substitution method.

Example 1 :

x = 8 – 2y

2x + 3y = 13

Solution :

x = 8 – 2y ----(1)

2x + 3y = 13 ----(2)

(1) is solved for x.

Substitute x = 8 – 2y in (2),

2(8 – 2y) + 3y = 13

16 – 4y + 3y = 13

16 – y = 13

y = 3

Substitute y = 3 in (1).

x = 8 – 2(3)

x = 2

Hence, the solution is (x, y) = (2, 3).

Example 2 :

y  =  4 + x

5x - 3y  =  0

Solution :

y = 4 + x ----(1)

5x - 3y = 0 ----(2)

(1) is solved for y.

Substitute y = 4 + x in (2).

5x – 3(4 + x) = 0

5x – 12 – 3x = 0

2x – 12 = 0

x = 6

Substitute x = 6 in (1).

y = 4 + 6

y = 10

Hence, the solution is (x, y) = (6, 10).

Example 3 :

x = -10 – 2y

3y – 2x = -22

Solution :

x = -10 – 2y ----(1)

3y – 2x = -22 ----(2)

(1) is solved for x.

Substitute x = -10 – 2y in (2).

3y – 2(-10 – 2y) = -22

3y + 20 + 4y = -22

7y + 20 = -22

7y = -22 – 20

7y = -42

y = -6

Substitute y = -6 in (1).

x = -10 – 2(- 6)

x = -10 + 12

x = 2

Hence, the solution is (x, y) = (2, -6).

Example 4 :

x = -1 + 2y

x = 9 – 2y

Solution :

x = -1 + 2y ----(1)

x = 9 – 2y ----(2)

(1) and (2) are solved for x.

(1) = (2)

x = x

-1 + 2y = 9 – 2y

-1 + 4y = 9

4y = 10

y = 5/2

y = 2.5

Substitute y = 2.5 in (2).

x = 9 – 2(2.5)

x = 9 – 5

x = 4

Hence, the solution is (x, y) = (4, 2.5).

Example 5 :

3x – 2y = 8

x = 3y + 12

Solution :

3x – 2y = 8 ----(1)

x = 3y + 12 ----(2)

(2) is solved for x.

Substitute x = 3y + 12 in (1).

3(3y + 12) – 2y = 8

9y + 36 – 2y = 8

7y + 36 = 8

7y = -28

y = -4

Substitute y = -4 in (2).

x = 3(- 4) + 12

x = -12 + 12

x = 0

Hence, the solution is (x, y) = (0, -4).

Example 6 :

x + 2y = 8

y = 7 – 2x

Solution :

x + 2y = 8 ----(1)

y = 7 – 2x ----(2)

(2) is solved for y.

Substitute y = 7 – 2x in (1).

x + 2(7 – 2x) = 8

x + 14 – 4x = 8

-3x = 8 – 14

x = 2

Substitute x = 2 in (2).

y = 7 – 2(2)

y = 3

Hence, the solution is (x, y) = (2, 3).

Example 7 :

x = -1 – 2y

2x – 3y = 12

Solution :

x = -1 – 2y ----(1)

2x – 3y = 12 ----(2)

(1) is solved for x.

Substitute x = -1 – 2y in (2).

2(-1 – 2y) – 3y = 12

-2 - 4y – 3y = 12

-7y = 12 + 2

y = -2

Substitute y = -2 in (1).

x = -1 – 2y

x = -1 – 2(-2)

x = 3

Hence, the solution is (x, y) = (3, -2).

Example 8 :

y = 5x

7x – 2y = 3

Solution :

y = 5x ----(1)

7x – 2y = 3 ----(2)

(1) is solved for y.

Substitute y = 5x in (2).

7x – 2(5x) = 3

7x – 10x = 3

x = -1

Substitute x = -1 in equation (1).

y = -5

Hence, the solution is (x, y) = (-1, -5).

Example 9 :

An adult ticket to a museum costs $3 more than a children’s ticket. When 200 adult tickets and 100 children’s tickets are sold, the total revenue is $2100. What is the cost of a children’s ticket?

Solution :

Let x be the adult ticket cost and y be the children ticket cost.

y = x + 3 ------(1)

200x + 100y = 2100

Dividing by 100, we get

2x + y = 21 ------(2)

Applying (1) in (2), we get

2x + x + 3 = 21

3x = 21 - 3

3x = 18

x = 18/3

x = 6

Applying x = 6 in (1), we get

y = 6 + 3

y = 9

So, cost of each adult ticket is $6 and cost of children ticket is $9.

Example 10 :

A humane society has 73 dogs and cats to be adopted. The number of cats is 10 more than twice the number of dogs. Write a system of linear equations that represents this situation. How many of each animal is up for adoption?

Solution :

Let x be the number of dogs and y be the number of cats

y = 2x + 10 -------(1)

x + y = 73 -------(2)

Applying (1) in (2), we get

x + 2x + 10 = 73

3x = 73 - 10

3x = 63

x = 63/3

x = 21

Applying x = 21 in (2), we get

21 + y = 73

y = 73 - 21

y = 52

Number of dogs = 21 and number of cats = 52.

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