SOLVING SYSTEMS OF EQUATIONS BY SUBSTITUTION

Solve each system by substitution.

Problem 1 :

y = -3x + 4

y = 4x - 10

Solution :

y = -3x + 4 ---> (1)

y = 4x - 10 ---> (2)

Substitute y = -3x + 4 in equation (2)

-3x + 4 = 4x - 10

-3x - 4x = - 10 - 4

-7x = - 14

x = 2

By applying x = 2 in equation (1), we get

y = - 3(2) + 4

y = - 6 + 4

y = - 2

Therefore, the solution is x = 2 and y = -2.

Problem 2 :

y = -4x + 2

y = 6x - 8

Solution :

y = -4x + 2 ---> (1)

y = 6x - 8 ---> (2)

Substitute y = -4x + 2 in equation (2)

-4x + 2 = 6x - 8

-4x - 6x = - 8 - 2

-10x = -10

x = 1

By applying x = 1 in equation (1), we get

y = -4(1) + 2

y = -4 + 2

y = - 2

Therefore, the solution is x = 1 and y = -2.

Problem 3 :

x = - 2y

x - y = 9

Solution :

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

x - y = 9 ---> (2)

Substitute x = -2y in equation (2)

-2y - y = 9

-3y = 9

y = 9/-3

y = -3

By applying y = -3 in equation (1), we get

x = -2(-3)

x = 6

Therefore, the solution is x = 6 and y = -3.

Problem 4 :

y = 2x

-6x + 3y = 16

Solution :

y = 2x ---> (1)

-6x + 3y = 16 ---> (2)

Substitute y = 2x in equation (2)

-6x + 3(2x) = 16

-6x + 6x = 16

0 = 16

So, there is no solution.

Problem 5 :

y = -3x

4x - 2y = -20

Solution :

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

4x - 2y = -20 ---> (2)

Substitute y = -3x in equation (2), we get

4x - 2(-3x) = - 20

4x + 6x = - 20

10x = - 20

x = -2

By applying x = -2 in equation (1), we get

y = -3(-2)

y = 6

Therefore, the solution is x = -2 and y = 6.

Problem 6 :

y = 3x - 4

4x + 3y = 1

Solution :

y = 3x - 4 ---> (1)

4x + 3y = 1---> (2)

Substitute y = 3x - 4 in equation (2), we get

4x + 3(3x - 4) = 1

4x + 9x - 12 = 1

13x = 1 + 12

13x = 13

x = 1

By applying x = 1 in equation (1), we get

y = 3(1) - 4

y = 3 - 4

y = -1

Therefore, the solution is x = 1 and y = -1.

Problem 7 :

y = x - 4

-4x - 6y = -16

Solution:

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

-4x - 6y = -16 ---> (2)

Substitute y = x - 4 in equation (2)

-4x - 6(x - 4) = -16

-4x - 6x + 24 = -16

-10x + 24 = -16

-10x = -16 - 24

-10x = - 40

x = 4

By applying x = 4 in equation (1), we get

y = 4 - 4

y = 0

Therefore, the solution is x = 4 and y = 0.

Problem 8 :

x = 3y + 1

2x + 4y = 12

Solution :

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

2x + 4y = 12 ---> (2)

Substitute x = 3y + 1 in equation (2), we get

2(3y + 1) + 4y = 12

6y + 2 + 4y = 12

10y + 2 = 12

10y = 12 - 2

10y = 10

y = 1

By applying y = 1 in equation (1), we get

x = 3(1) + 1

x = 3 + 1

x = 4

Therefore solution is x = 4 and y = 1.

Problem 9 :

x = y - 4

-2x + 3y = 6

Solution :

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

-2x + 3y = 6 ---> (2)

Substitute x = y - 4 in equation (2), we get

-2(y - 4) + 3y = 6

-2y + 8 + 3y = 6

y = 6 - 8

y = -2

By applying y = -2 in equation (1), we get

x = - 2 - 4

x = - 6

Therefore, the  solution is x = -6 and y = -2.

Problem 10 :

Next week your math teacher is giving a chapter test. The test will consist of 35 questions. Some problems are worth 2 points and some problems are worth 4 points. There are 20 questions worth 2 points. How many problems of 4 points are on the test?

Solution:

Let x represents the number of problems worth 2 points

And y represents the number of problems worth 4 points.

x + y = 35 ---> (1)

There are 20 questions worth 2 points,

Hence x = 20.

20 + y = 35

y = 35 - 20

y = 15

Therefore 20 questions are worth 2 points and 15 questions are worth 4 points.