Question

Solve each system by elimination by multiplying both equations. eliminate y first. show all work.

Answer 1

First $\mathrm{Multiply\:}10x-3y=18\mathrm{\:by\:}3:\quad 30x-9y=54$
Then $\mathrm{Multiply\:}6x-10y=-22\mathrm{\:by\:}5:\quad 30x-50y=-110$
To get $\begin{bmatrix}30x-9y=54\\ 30x-50y=-110\end{bmatrix}$
So...
$30x-50y=-110 \ \textgreater \ \begin{bmatrix}30x-9y=54\\ -41y=-164\end{bmatrix}$
$-41y=-164 \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}-41 \ \textgreater \ \frac{-41y}{-41}=\frac{-164}{-41}$
To get y = 4.
$\mathrm{For\:}30x-9y=54\mathrm{\:plug\:in\:} \:y=4 \ \textgreater \ 30x-9\cdot \:4=54$
$\mathrm{Multiply\:the\:numbers:}\:9\cdot \:4=36 \ \textgreater \ 30x-36=54 \ \textgreater$
Next $\mathrm{Add\:}36\mathrm{\:to\:both\:sides} \ \textgreater \ 30x-36+36=54+36 \ \textgreater \ \mathrm{Simplify} \ \textgreater \ 30x=90$
Finally $\mathrm{Divide\:both\:sides\:by\:}30 \ \textgreater \ \frac{30x}{30}=\frac{90}{30} \ \textgreater \ x = 3$
Therefore our solutions are y = 4, x = 3