# How do you solve 3x + y = 1 and y + 4 = 5x  using substitution?

Aug 27, 2017

See a solution process below: $\left(\frac{5}{8} , - \frac{7}{8}\right)$

#### Explanation:

Step 1)* Solve the second equation for $y$:

$y + 4 = 5 x$

$y + 4 - \textcolor{red}{4} = 5 x - \textcolor{red}{4}$

$y + 0 = 5 x - 4$

$y = 5 x - 4$

Step 2) Substitute $\left(5 x - 4\right)$ for $y$ in the first equation and solve for $x$:

$3 x + y = 1$ becomes:

$3 x + \left(5 x - 4\right) = 1$

$3 x + 5 x - 4 = 1$

$\left(3 + 5\right) x - 4 = 1$

$8 x - 4 = 1$

$8 x - 4 + \textcolor{red}{4} = 1 + \textcolor{red}{4}$

$8 x - 0 = 5$

$8 x = 5$

$\frac{8 x}{\textcolor{red}{8}} = \frac{5}{\textcolor{red}{8}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}} x}{\cancel{\textcolor{red}{8}}} = \frac{5}{8}$

$x = \frac{5}{8}$

**Step 3)$S u b s t i t u t e$5/8$f \mathmr{and}$x$\in t h e s o l u t i o n \to t h e \sec o n d e q u a t i o n a t t h e e n d o f S t e p 1 \mathmr{and} c a l c \underline{a} t e$y#:

$y = 5 x - 4$ becomes:

$y = \left(5 \times \frac{5}{8}\right) - 4$

$y = \frac{25}{8} - 4$

$y = \frac{25}{8} - \left(\frac{8}{8} \times 4\right)$

$y = \frac{25}{8} - \frac{32}{8}$

$y = \frac{25 - 32}{8}$

$y = - \frac{7}{8}$

The Solution Is: $x = \frac{5}{8}$ and $y = - \frac{7}{8}$ or $\left(\frac{5}{8} , - \frac{7}{8}\right)$