# How do you solve the following system?: 5x +y =2, 13x -5y = -2

Mar 29, 2017

See the entire solution process below:

#### Explanation:

Step 1) Solve the first equation for $y$:

$5 x + y = 2$

$- \textcolor{red}{5 x} + 5 x + y = - \textcolor{red}{5 x} + 2$

$0 + y = - 5 x + 2$

$y = - 5 x + 2$

Step 2) Substitute $- 5 x + 2$ for $y$ in the second equation and solve for $x$:

$13 x - 5 y = - 2$ becomes:

$13 x - 5 \left(- 5 x + 2\right) = - 2$

$13 x - \left(5 \times - 5 x\right) - \left(5 \times 2\right) = - 2$

$13 x + 25 x - 10 = - 2$

$38 x - 10 = - 2$

$38 x - 10 + \textcolor{red}{10} = - 2 + \textcolor{red}{10}$

$38 x - 0 = 8$

$38 x = 8$

$\frac{38 x}{\textcolor{red}{38}} = \frac{8}{\textcolor{red}{38}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{38}}} x}{\cancel{\textcolor{red}{38}}} = \frac{2 \times 4}{\textcolor{red}{2 \times 19}}$

$x = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \times 4}{\textcolor{red}{\cancel{2} \times 19}}$

$x = \frac{4}{19}$

Step 3) Substitute $\frac{4}{19}$ for $x$ in the solution to the first equation at the end of Step 1 and calculate $y$:

$y = - 5 x + 2$ becomes:

$y = \left(- 5 \times \frac{4}{19}\right) + 2$

$y = - \frac{20}{19} + 2$

$y = - \frac{20}{19} + \left(\frac{19}{19} \times 2\right)$

$y = - \frac{20}{19} + \frac{38}{19}$

$y = \frac{18}{19}$

The solution is: $x = \frac{4}{19}$ and $y = \frac{18}{19}$ or $\left(\frac{4}{19} , \frac{18}{19}\right)$