Question

What is the solution of the system?

Answer 1

We have $\begin{bmatrix}2x+y=20\\ 6x-5y=12\end{bmatrix}$

$\mathrm{Multiply\:}2x+y=20\mathrm{\:by\:}3:\ 6x+3y=60$

$\begin{bmatrix}6x+3y=60\\ 6x-5y=12\end{bmatrix}$

6x - 5y = 12
-
6x + 3y = 60
/
-8y = -48

$-8y=-48 \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}-8 \ \textgreater \ \frac{-8y}{-8}=\frac{-48}{-8} \ \textgreater \ y=6$

$\mathrm{For\:}6x+3y=60\mathrm{\:plug\:in\:}\ \:y=6$

$6x+3\cdot \:6=60 \ \textgreater \ \mathrm{Multiply\:the\:numbers:}\:3\cdot \:6=18 \ \textgreater \ 6x+18=60$

$\mathrm{Subtract\:}18\mathrm{\:from\:both\:sides} \ \textgreater \ 6x+18-18=60-18 \ \textgreater \ 6x=42$

$\mathrm{Divide\:both\:sides\:by\:}6 \ \textgreater \ \frac{6x}{6}=\frac{42}{6} \ \textgreater \ x = 7$

Therefore....
$The\:solutions\:to\:the\:system\:of\:equationts\:are \ \textgreater \ y=6,\:x=7$