# How do you solve the following system: 6x+2y=-4, 6x+3y=-12?

May 8, 2018

The solution is $\left(2 , - 8\right)$.

#### Explanation:

Solve the system:

$\text{Equation 1} :$ $6 x + 2 y = - 4$

$\text{Equation 2} :$ $6 x + 3 y = - 12$

The solution to this system of linear equations is the point they have in common, which is the point at which they intersect. I'm going to solve this system by elimination.

Multiply Equation 1 by $- 1$. This will change the signs.

$- 1 \left(6 x + 2 y = - 4\right)$

$- 6 x - 2 y = 4$

$- 6 x - 2 y = \textcolor{w h i t e}{\ldots . .} 4$
$\textcolor{w h i t e}{. .} 6 x + 3 y = - 12$
$- - - - - - -$
$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots .} y = \textcolor{w h i t e}{.} - 8$

Substitute $- 8$ for $y$ into either equation and solve for $x$.

Equation 1

$6 x + 2 \left(- 8\right) = - 4$

$6 x - 16 = - 4$

Add $16$ to both sides of the equation.

$6 x = - 4 + 16$

$6 x = 12$

Divide both sides by $6$.

$x = \frac{12}{6}$

$x = 2$

The solution is $\left(2 , - 8\right)$.

graph{(-6x-2y-4)(6x+3y+12)=0 [-16.94, 15.09, -14.92, 1.09]}