Question

To solve the system of equations below.Jose isolated the variable y in the first equation and then substituted it into the second equation. what was the resulting equation?

{3x=y=-2
{2x^2-y=10

a.2x^2-3x+2=10
b.2x^2-3x+2=10
c.2x^2+3x+2=10
d.2x^2-3x-2=10

Answer 1

Isolated y.

3x - y = -2

-y = -2 - 3x Multiply both sides by -1.

y = 2 + 3x

Now substituting into the second.

2x² - y = 10

2x² - (2 + 3x) = 10

2x² - 2 - 3x = 10

2x² - 3x - 2 = 10

So we can see it is option D.

Hope this helps.

Answer 2

The first step of solving this is to isolate the variable y in the first equation.

3x - y = -2 Subtract -3x from both sides
- y = - 3x - 2 Divide both sides by -1
y = 3x + 2

Now, substitute that y value into the second equation.

2x^{2} - y = 10 Substitute
2x^{2} - (3x + 2) = 10 Multiply all of the numbers in the parentheses by -1
2x^{2} - 3x - 2 = 10

So the correct answer choice is D) 2x^{2} -3x - 2 = 10.