# How do you solve the following system: x-4y=2, 5x+2y=20 ?

Jan 18, 2016

$x = \frac{42}{11} , y = \frac{5}{11}$

#### Explanation:

Take the first equation.

$x - 4 y = 2$

Add $4 y$ to both sides.

$x = 4 y + 2$

Since we know that $x = 4 y + 2$, we can place $x$ with $4 y + 2$ in the other equation.

$5 \textcolor{red}{x} + 2 y = 20$

$5 \textcolor{red}{\left(4 y + 2\right)} + 2 y = 20$

Distribute and solve for $y$.

$20 y + 10 + 2 y = 20$

$22 y = 10$

$y = \frac{10}{22} = \frac{5}{11}$

Now that we know the value for $y$, plug it into either equation to solve for $x$. I'm choosing the first equation since it's simpler, but the same answer would come from plugging the value into the second equation.

$x - 4 \textcolor{b l u e}{y} = 2$

$x - 4 \textcolor{b l u e}{\left(\frac{5}{11}\right)} = 2$

$x - \frac{20}{11} = 2$

$x = \frac{22}{11} + \frac{20}{11} = \frac{42}{11}$

Since $x = \frac{42}{11}$ and $y = \frac{5}{11}$, the two lines should intersect at the point $\left(\frac{42}{11} , \frac{5}{11}\right)$ or $\left(3.818 , 0.455\right)$.

We can check a graph:

graph{(x-4y-2)(5x+2y-20)=0 [-5.26, 10.54, -3.38, 4.516]}