# How do you solve this system of equations by substitution: x+ 4y = - 14 and y = 2x - 8?

Mar 26, 2018

$\left(x , y\right) = \left(2 , - 4\right)$

#### Explanation:

We're given $y = 2 x - 8.$ This means that we can replace all instances of $y$ in the first equation with $2 x - 8.$ The goal when given a system of equations in two variables is two get an equation with one variable and solve for that variable, and subsequently back-substitute your solution in to get the other variable.

$x + 4 \left(2 x - 8\right) = - 14$

$x + 8 x - 32 = - 14$

Solve for $x :$

$9 x - 32 = - 14$

$9 x = 18$

$x = 2$

To solve for $y ,$ simply plug in $x = 2$ into $y = 2 x - 8.$

$y = 2 \left(2\right) - 8 = - 4$

So, the solution is

$\left(x , y\right) = \left(2 , - 4\right)$

Mar 26, 2018

See a solution process below:

#### Explanation:

Step 1) Because the second equation is already solve for $y$ we can substitute $\left(2 x - 8\right)$ for $y$ in the first equation and solve for $x$:

$x + 4 y = - 14$ becomes:

$x + 4 \left(2 x - 8\right) = - 14$

$x + \left(4 \times 2 x\right) - \left(4 \times 8\right) = - 14$

$x + 8 x - 32 = - 14$

$1 x + 8 x - 32 = - 14$

$\left(1 + 8\right) x - 32 = - 14$

$9 x - 32 = - 14$

$9 x - 32 + \textcolor{red}{32} = - 14 + \textcolor{red}{32}$

$9 x - 0 = 18$

$9 x = 18$

$\frac{9 x}{\textcolor{red}{9}} = \frac{18}{\textcolor{red}{9}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{9}}} x}{\cancel{\textcolor{red}{9}}} = 2$

$x = 2$

Step 2) Substitute $2$ for $x$ in the second equation and calculate $y$:

$y = 2 x - 8$ becomes:

$y = \left(2 \times 2\right) - 8$

$y = 4 - 8$

$y = - 4$

The Solution Is:

$x = 2$ and $y = - 4$

Or

$\left(2 , - 4\right)$