# How do solve the following linear system?:  x+2y=1 , -2x -8y = 4 ?

Feb 1, 2017

$x = 4$ and $y = - \frac{3}{2}$ or $- 1 \frac{1}{2}$

#### Explanation:

$x + 2 y = 1$
$- 2 x - 8 y = 4$

Reduce the second equation by dividing all terms by $2$.

$- 2 x - 8 y = 4$

$- x - 4 y = 2$

Multiply all terms by $- 1$.

$x + 4 y = - 2$

Separate $4 y$ into two.

$x + 2 y + 2 y = - 2$

In this equation, substitute $\left(x + 2 y\right)$ with $\textcolor{red}{1}$, the value from the first equation.

$x + 2 y + 2 y = - 2$

$\textcolor{red}{1} + 2 y = - 2$

Subtract $1$ from both sides.

$2 y = - 3$

Divide both sides by $2$.

$y = - \frac{3}{2} = - 1 \frac{1}{2}$

In the first equation, substitute $y$ with $\textcolor{b l u e}{- \frac{3}{2}}$.

$x + 2 y = 1$

$x + 2 \textcolor{b l u e}{\left(- \frac{3}{2}\right)} = 1$

Open the brackets and simplify.

$x - 3 = 1$

Add $3$ to both sides.

$x = 4$

Feb 1, 2017

$y = - 1 \frac{1}{2} , x = 4$

#### Explanation:

$x + 2 y = 1$-----------(1)

$- 2 x - 8 y = 4$---------(2)

$2 x + 4 y = 2$ $\left(1\right) \times 2$ ---------(3)

$- 4 y = 6$--------(2)+(3)

$4 y = - 6$

$y = - \frac{6}{4}$

$y = - 1 \frac{1}{2}$

Substitute $y = - \frac{6}{4}$ in ----------1

$x + 2 \left(- \frac{6}{4}\right) = 1$

$x - \frac{12}{4} = 1$

$x - 3 = 1$

$x = 1 + 3$

$x = 4$
Substitute #y= -6/4,x=4 in------1

$4 + 2 \left(- \frac{6}{4}\right) = 1$

$4 - \frac{12}{4} = 1$

multiply both sides by 4
$16 - 12 = 4$

$4 = 4$