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).
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 )
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.
To check if your solution is correct, we can substitute the values of x and y back into the original equations and see if they equal the given values.
Let's start with the first equation:
2x - 3y = 6
Substituting x = -4 and y = 2/3:
2(-4) - 3(2/3) = -8 - 2 = -10
Since the left side does not equal the right side (6), the solution you provided does not satisfy this equation. Therefore, (-4, 2/3) is not the correct solution.
Now let's check the second equation:
3x + 6y = 16
Substituting x = -4 and y = 2/3:
3(-4) + 6(2/3) = -12 + 4 = -8
Again, the left side does not equal the right side (16), so (-4, 2/3) is not a solution to this equation either.
Therefore, your solution (-4, 2/3) doesn't satisfy the system of equations. You may want to go through the steps again to find the correct solution.