How do you solve the following linear system:  –3x + 2y = 12 , 5x + 2y = -4 ?

Mar 31, 2018

$x = - 2$
$y = 3$

Explanation:

1. The first step is to find similar coefficients between the two equations. Notice that both equations have a $2 y$ .

$- 3 x + 2 y = 12$
$5 x + 2 y = - 4$

2. Because the coefficients are equal and not opposite, you have to times one of the equations by -1. For simplicity, lets use $- 3 x + 2 y = 12$.

$- 1 \cdot \left(- 3 x + 2 y = 12\right)$
$5 x + 2 y = - 4$

$3 x - 2 y = - 12$
$5 x + 2 y = - 4$

3. Now that the coefficients in front of $y$ are opposite, we can add the two equations and the $y$ values would cancel out.

$3 x - 2 y = - 12$
$+$
$5 x + 2 y = - 4$

$8 x = - 16$

4. Divide both side by $8$ to get the value of $x$.

$x = - 2$

5. Substitute the value of $x$ into one of the equations. For simplicity, lets use $5 x + 2 y = - 4$ .

$5 \left(- 2\right) + 2 y = - 4$

6. Simplify

$- 10 + 2 y = - 4$

7. Add 10 to both sides so that $2 y$ is alone.

$2 y = 6$

8. Divide both side by $2$ to get the value of $y$.

$y = 3$

You now have found both the value of $x$ and the value of $y$.

$x = - 2$
$y = 3$