Solve the following system of equations using elimination:

$$\{\begin{array}{l}7x+8y=4\\ 3x-5y=27\end{array}$$

Step 1 - Write both equations in standard form. If any coefficients are fractions, clear them.

$7x+8y=4$

$3x-5y=27$

Step 2 - Check to see if the coefficients of one variable are opposites or equivalent. To get opposite coefficients of $y$, we will need to multiply the first equation by 5 and the second equation by 8.

$5(7x+8y)=5(4)$

$8(3x-5y)=8(27)$

Step 3 - Simplify.

$35x+40y=20$

$24x-40y=216$

Step 4 - Add the equations to eliminate one variable.

$\begin{array}{c}\hfill 35x+40y=20\\ \hfill 24x-40y=216\\ \hfill 59x=236\end{array}$

Step 5 - Solve for the remaining variable.

$59x=236$

$x=4$

Step 6 - Substitute the solution into one of the original equations. Then solve for the other variable.

$59x=236$

$x=4$

$\begin{array}{ccc}\hfill 7x+8y& \hfill =\hfill & 4\hfill \\ \hfill 7(4)+8y& \hfill =\hfill & 4\hfill \\ \hfill 28+8y& \hfill =\hfill & 4\hfill \\ \hfill 8y& \hfill =\hfill & -24\hfill \\ \hfill y& \hfill =\hfill & -3\hfill \end{array}$

Step 7 - Check that the ordered pair is a solution to both original equations.

$\begin{array}{cc}\hfill 7x+8y=4& \hfill 3x-5y=277\\ \hfill 7(4)+8(-3)\stackrel{?}{=}4& \hfill 3(4)-5(-3)\stackrel{?}{=}277\\ \hfill 20-24\stackrel{?}{=}4& \hfill 3(4)-5(-3)\stackrel{?}{=}277\\ \hfill 4=4✓& \hfill 27=27✓\end{array}$

The solution is $(0,-3)$.