# How do solve the following linear system?:  -4x-14y=28 , x-2y=8 ?

Mar 12, 2017

See the entire solution process below:

#### Explanation:

Step 1) Solve the second equation for $x$:

$x - 2 y = 8$

$x - 2 y + \textcolor{red}{2 y} = 8 + \textcolor{red}{2 y}$

$x - 0 = 8 + 2 y$

$x = 8 + 2 y$

Step 2) Substitute $8 + 2 y$ for $x$ in the first equation and solve for $y$:

$- 4 x - 14 y = 28$ becomes:

$- 4 \left(8 + 2 y\right) - 14 y = 28$

$\left(- 4 \times 8\right) + \left(- 4 \times 2 y\right) - 14 y = 28$

$- 32 - 8 y - 14 y = 28$

$- 32 - 22 y = 28$

$\textcolor{red}{32} - 32 - 22 y = \textcolor{red}{32} + 28$

$0 - 22 y = 60$

$- 22 y = 60$

$\frac{- 22 y}{\textcolor{red}{- 22}} = \frac{60}{\textcolor{red}{- 22}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 22}}} y}{\cancel{\textcolor{red}{- 22}}} = \frac{2 \times 30}{\textcolor{red}{2 \times - 11}}$

$y = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} \times 30}{\textcolor{red}{\cancel{2} \times - 11}}$

$y = - \frac{30}{11}$

Step 3) Substitute $- \frac{30}{11}$ for $y$ in the solution to the second equation at the end of Step 1 and calculate $x$:

$x = 8 + 2 y$ becomes:

$x = 8 + \left(2 \times - \frac{30}{11}\right)$

$x = \left(\frac{11}{11} \times 8\right) - \frac{60}{11}$

$x = \frac{88}{11} - \frac{60}{11}$

$x = \frac{88 - 60}{11}$

$x = \frac{28}{11}$

The solution is: $x = \frac{28}{11}$ and $y = - \frac{30}{11}$ or $\left(\frac{28}{11} , - \frac{30}{11}\right)$