# How do you solve the following linear system: 3x + y = 4 , x = 5y + 2 ?

Jun 27, 2018

See a solution process below:

#### Explanation:

Step 1) Because the second equation is already solved for $x$ we can substitute $\left(5 y + 2\right)$ for $x$ in the first equation and solve for $y$:

$3 x + y = 4$ becomes:

$3 \left(5 y + 2\right) + y = 4$

$\left(3 \times 5 y\right) + \left(3 \times 2\right) + y = 4$

$15 y + 6 + y = 4$

$15 y + 6 - \textcolor{red}{6} + y = 4 - \textcolor{red}{6}$

$15 y + 0 + y = - 2$

$15 y + y = - 2$

$15 y + 1 y = - 2$

$\left(15 + 1\right) y = - 2$

$16 y = - 2$

$\frac{16 y}{\textcolor{red}{16}} = - \frac{2}{\textcolor{red}{16}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{16}}} y}{\cancel{\textcolor{red}{16}}} = - \frac{2}{16}$

$y = - \frac{1}{8}$

Step 2) Substitute $- \frac{1}{8}$ for $y$ in the second equation and calculate $x$:

$x = 5 y + 2$ becomes:

$x = \left(5 \times - \frac{1}{8}\right) + 2$

$x = - \frac{5}{8} + \left(\frac{8}{8} \times 2\right)$

$x = - \frac{5}{8} + \frac{16}{8}$

$x = \frac{- 5 + 16}{8}$

$x = \frac{11}{8}$

The Solution Is:

$x = \frac{11}{8}$ and $y = - \frac{1}{8}$

Or

$\left(\frac{11}{8} , - \frac{1}{8}\right)$