Solve a System of Equations by Elimination -

$$\begin{gathered} 6x+4y=-4 \\ -6x-5y=8 \end{gathered}$$

Equations - $$\begin{gathered} 6x+4y=-4 \\ -6x-5y=8 \end{gathered}$$

Solve a System of Equations by Elimination and find the value of x and y.

Equation -

$$\begin{gathered} 6x+4y=-4 \\ -6x-5y=8 \end{gathered}$$

We can eliminate y's by multiplying the first equation by 5 and second equation by 4 -

$$\begin{gathered} (6x+4y)\times 5=-4\times 5 \\ (-6x-5y)\times 4=8\times 4 \end{gathered}$$

$$\begin{gathered} 30x+20y=-20 \\ -24x-20y=32 \end{gathered}$$

the equation become -

$$\begin{gathered} 30x+20y=-20 \\ -24x-20y=32 \end{gathered}$$

We Add the x's and y's and constants -

localid="1647454876610" $\begin{array}{l}30x+20y=-20\\ \frac{-24x-20y=32}{6x=12}\end{array}$

Solve for x -

$x=2$

Substitute localid="1647455102226" $\mathit{x}\mathbf{=}\mathbf{2}$into the second equation localid="1647455109274" $\mathbf{-}\mathbf{6}\mathit{x}\mathbf{-}\mathbf{5}\mathit{y}\mathbf{=}\mathbf{8}$Then solve for y -

$$\begin{gathered} -6x-5y=8 \\ -6\times 2-5y=8 \\ -12-5y=8 \\ -5y=8+12 \\ -5y=20 \\ y=-4 \end{gathered}$$

Written as $(x,y)$will be $(2,-4)$.

Check that the ordered pair is a solution to both original equations -

Put $(2,-4)$in the equation -

$$\begin{gathered} 6x+4y=-4 \\ 6\times 2+4\times (-4)=-4 \\ 12+(-16)=-4 \\ -4=-4 \end{gathered}$$

$$\begin{gathered} -6x-5y=8 \\ -6\times 2-5\times (-4)=8 \\ -12+20=8 \\ 8=8 \end{gathered}$$

The Solution is -$(2,-4)$