Question

Only need my answer to be check. Problem:
2x-3y=6
3x+6y=16

I rewrote both as 1) y=2/3x-2
2)1/2x+8/3

set them both equal to each other and got -4 for x. I plugged it in for one of the original equations and got 2/3 as y for the point (-4,2/3).

Answer 1

1 step is correct.

2 x - 3 y = 6

Subtract 2 x to both sides

- 3 y = 6 - 2x

- 3 y = - 2x + 6

Divide both sides by - 3

y = - 2 x / - 3 + 6 / - 3

y = ( 2 / 3 ) x - 2

2 step:

3 x + 6 y = 16

3 x + 6 ∙ [ ( 2 / 3 ) x - 2 ] = 16

3 x + 12 x / 3 - 12 = 16

3 x + 4 x - 12 = 16

7 x - 12 = 16

Add 12 to both sides

7 x = 16 + 12

7 x = 28

Divide both sides by 7

x = 28 / 7

x = 4


y = ( 2 / 3 ) x - 2

y = 2 ∙ 4 / 3 - 2

y = 8 / 3 - 2

y = 8 / 3 - 6 / 3

y = 2 / 3


OR

2 x - 3 y = 6 Multiply both sides by 2

4 x - 6 y = 12


4 x - 6 y = 12
+
3 x + 6 y = 16
___________

7 x = 28

Divide both sides by 7

x = 28 / 7

x = 4

Replace this value in equation:

2 x - 3 y = 6

2 ∙ 4 - 3 y = 6

8 - 3 y = 6

Subtract 8 to both sides

- 3 y = 6 - 8

- 3 y = - 2

Divide both sides by - 3

y = - 2 / - 3

y = 2 / 3

Intersection point ( 4 , 2 / 3 )

Answer 2

Multiply Eq1 by 2 and add Eq1 and Eq2:
4x - 6y = 12
3x + 6y = 16
Sum: 7x = 28,
X = 4.

In Eq2, replace X with 4 and solve for Y:
3*4 + 6y = 16,
Y = 2/3.