# How do you solve the following system?: -2x+10y=-20 , 2x+3y=11

Jan 21, 2016

$\left\{\begin{matrix}x = \frac{85}{13} \\ y = - \frac{9}{13}\end{matrix}\right.$

#### Explanation:

$\left\{\begin{matrix}- 2 x + 10 y = - 20 \text{ S1" \\ 2x+3y=11" S2}\end{matrix}\right.$

You can use Linear Systems with Addition or Subtraction doing $S 1 + S 2$ to remove $x$ (in the equations $x$ has same coefficent and opposite sign).

You also could divide the first equation $S 1$ by $2$

$\left\{\begin{matrix}- \cancel{2} x + {\cancel{10}}^{5} y = - {\cancel{20}}^{10} \text{ S1/2" \\ 0x+13y=-9 " S1+S2}\end{matrix}\right.$

$\implies \left\{\begin{matrix}- x + 5 y = - 10 \\ y = - \frac{9}{13}\end{matrix}\right.$

$\left\{\begin{matrix}- x = - 10 - 5 \cdot \left(- \frac{9}{13}\right) \\ y = - \frac{9}{13}\end{matrix}\right.$

$\left\{\begin{matrix}x = 10 - \frac{45}{13} \\ y = - \frac{9}{13}\end{matrix}\right.$

$\left\{\begin{matrix}x = \frac{130 - 45}{13} \\ y = - \frac{9}{13}\end{matrix}\right. \implies \left\{\begin{matrix}x = \frac{85}{13} \\ y = - \frac{9}{13}\end{matrix}\right.$

graph{(10y-2x+20)(3y+2x-11)=0 [3.734, 8.059, -1.165, 0.998]}