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How to solve systems of equation by substitution?

0 votes
How would I solve for these standard form equations by substitution?
#1. -6x+8y= 12
3x-5y= -3
#2. 7x+7y=7
-8x - 7y = -2
#3. 4x+3y=6
7x +4y= 18
asked Mar 7, 2013 in ALGEBRA 2 by payton Apprentice

1 Answer

0 votes

Substitution method :

1).

The system of equations are - 6x + 8y = 12 and 3x - 5y = - 3.

Divide equation 1 : - 6x + 8y = 12 by 2.

- 3x + 4y = 6

Subtract 4y from each side.

- 3x = 6 - 4y

Multiply each side by negative 1.

3x = 4y - 6.

Substitute 3x = 4y - 6 in equation 2 : 3x - 5y = - 3.

(4y - 6) - 5y = - 3

- y = - 3 + 6 = 3

⇒ y = - 3.

Substitute y = - 3 in equation 1 : - 6x + 8y = 12.

- 6x + 8(- 3) = 12

- 6x - 24 = 12

- 6x = 12 + 24 = 36

⇒ x = - 6.

The solution is x = - 6 and y = - 3.

2).

The system of equations are 7x + 7y = 7 and - 8x - 7y = - 2.

Simpify equation 1.

7x + 7y = 7

Subtract 7x from each side.

7y = 7 - 7x.

Substitute 7y = 7 - 7x in equation 2 : - 8x - 7y = - 2.

- 8x - (7 - 7x) = - 2

- 8x - 7 + 7x = - 2

- x = - 2 + 7 = 5

⇒ x = - 5.

Substitute x = - 5 in equation 2 : - 8x - 7y = - 2.

- 8(- 5) - 7y = - 2

40 - 7y = - 2

- 7y = - 2 - 40 = - 42

⇒ y = 6.

The solution is x = - 5 and y = 6.

3).

The system of equations are 4x + 3y = 6 and 7x + 4y = 18.

Simpify equation 1 : 4x + 3y = 6 for y.

4x + 3y = 6

Subtract 4x from each side.

3y = 6 - 4x

Divide each side by 3.

y = (6 - 4x)/3.

Substitute y = (6 - 4x)/3 in equation 2 : 7x + 4y = 18.

7x + 4((6 - 4x)/3) = 18

21x + 24 - 16x = 54

5x = 54 - 24 = 30

⇒ x = 6.

Substitute x = 6 in equation 1 : 4x + 3y = 6.

4(6) + 3y = 6

24 + 3y = 6

3y = 6 - 24 = - 18

⇒ y = - 6.

The solution is x = 6 and y = - 6.

answered Aug 4, 2014 by lilly Expert

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