# How do you solve x-2y=4 and 8y+24x=5 using substitution?

Mar 10, 2016

$\left\{\begin{matrix}x = \frac{3}{4} \\ y = - \frac{13}{8}\end{matrix}\right.$

#### Explanation:

Given:

$\left\{\left(x - 2 y = 4 , \text{S1"),(8y+24x=5,"S2"):} <=> {(x-2y=4, "S1"),(24x+8y=5,"S2}\right)\right.$

The Substitution solving method tells us that it's possible to find the value of one of the two variables from one of the two equations, then substitute it in the other one.

In your excercise it's easy to find $x$ value from $S 1$ as follows:

$\left\{\begin{matrix}x = 4 + 2 y & \text{S1" \\ 24x+8y=5 & "S2}\end{matrix}\right.$

Now we can substituite it in $S 2$

$\left\{\begin{matrix}x = 4 + 2 y & \text{S1" \\ 24(4+2y)+8y=5 & "S2}\end{matrix}\right.$

$\left\{\begin{matrix}x = 4 + 2 y \\ 96 + 48 y + 8 y = 5\end{matrix}\right. \iff \left\{\begin{matrix}x = 4 + 2 y \\ 56 y = 5 - 96\end{matrix}\right. \iff \left\{\begin{matrix}x = 4 + 2 y \\ 56 y = - 91\end{matrix}\right.$

$\left\{\begin{matrix}x = 4 + 2 y \\ y = - \left(\frac{91}{56}\right) = - \frac{13}{8}\end{matrix}\right.$

Now we can substitute the $y$ value in $S 1$ to find $x$ value:

$\left\{\begin{matrix}x = 4 + \cancel{2} \left(- \frac{13}{\cancel{8}} ^ 4\right) \\ y = - \frac{13}{8}\end{matrix}\right.$

$\left\{\begin{matrix}x = 4 - \frac{13}{4} = \frac{16 - 13}{4} = \frac{3}{4} \\ y = - \frac{13}{8}\end{matrix}\right.$

$\therefore \left\{\begin{matrix}x = \frac{3}{4} \\ y = - \frac{13}{8}\end{matrix}\right.$

In a graph the system solution is the point where $S 1$ intercepts $S 2$

graph{(x-2y-4)(24x+8y-5)=0 [-3.347, 6.52, -4.167, 0.766]}