91Ó°ÊÓ

We are given a system of equations.

$\left\{\begin{array}{l}3x+2y=-3\\ -x-2y=-19\end{array}\right.$

We need to solve the system using the method of elimination.

The two equations in the system are written in standard form and the coefficients of $y$are already opposites. So we add the two equations

$\begin{array}{rcl}3x+2y+(-x-2y)& =& -3-19\\ 3x+2y-x-2y& =& -22\\ 2x& =& -22\\ x& =& -\frac{22}{2}\\ x& =& -11\end{array}$

Substitute $-11$for $x$in the second equation $-x-2y=-19$.

$\begin{array}{rcl}-(-11)-2y& =& -19\\ 11-2y& =& -19\\ -2y& =& -19-11\\ -2y& =& -30\\ y& =& \frac{-30}{-2}\\ y& =& 15\end{array}$

So the solution for the given system is $(-11,15)$.

Substitute $-11$for $x$and $15$for $y$in the first equation.

$\begin{array}{rcl}3x+2y& =& -3\\ 3(-11)+2\left(15\right)& =& -3\\ -33+30& =& -3\\ -3& =& -3\end{array}$

As the statement is correct, so the solution satisfies the first equation.

Substitute $-11$for $x$and $15$for $y$in the second equation.

$\begin{array}{rcl}-x-2y& =& -19\\ -(-11)-2\left(15\right)& =& -19\\ 11-30& =& -19\\ -19& =& -19\end{array}$

As the statement is correct, so the solution satisfies the second equation.