### Home > CC3 > Chapter 6 > Lesson 6.1.4 > Problem6-38

6-38.

Solve the system of equations below using the Equal Values Method.

$a=12b+3$
$a=-2b-4$

Set the two equations equal to each other.
$12b+3=-2b-4$

Solve for $b$.
$14b=-7$

$b=-\frac{1}{2}$

Substitute $b$ back into one of the original equations to find $a$.
$a=12\left(-\frac{1}{2}\right)+3$
$a=-6+3$

Remember to check by substituting $a$ and $b$ back into both of the original equations.
$a=-3$