# How do you solve the following system: 11y + 3x = 4, -3y + x = -3 ?

Mar 24, 2018

$\textcolor{b l u e}{y = \frac{13}{20}}$

$\textcolor{b l u e}{x = - \frac{21}{20}}$

#### Explanation:

$11 y + 3 x = 4 \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left[1\right]$

$- 3 y + x = - 3 \setminus \setminus \setminus \setminus \setminus \left[2\right]$

Multiply equation $\left[2\right]$ by 3 and subtract it from equation $\left[1\right]$

$11 y + 3 x = 4 \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left[1\right]$

$- 9 y + 3 x = - 9 \setminus \setminus \setminus \setminus \setminus \left[2\right]$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus 20 y = 13$

Divide by 20:

$\textcolor{b l u e}{y = \frac{13}{20}}$

We now substitute this into one of the original equations, it dosen't matter which one we choose.

In $\left[2\right]$

$- 3 \left(\frac{13}{20}\right) + x = - 3$

Add $- 3 \left(\frac{13}{20}\right)$ to both sides:

$x = - 3 + 3 \left(\frac{13}{20}\right)$

$\textcolor{b l u e}{x = - \frac{21}{20}}$