# How do you solve the system of equations 3x + y = 136 and x - 5y = - 40?

Jan 19, 2017

$3 x + y = 136$
$x - 5 y = - 40$

$\implies x = 40$ and $y = 16$

#### Explanation:

This system

$3 x + y = 136$
$x - 5 y = - 40$

$\to$ can also be written as

$\left(\begin{matrix}3 & 1 \\ 1 & - 5\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}136 \\ - 40\end{matrix}\right)$ a matrix equation

and if a matrix is invertible it is true that

$A \vec{x} = \vec{c} b \implies \vec{x} = {A}^{- 1} \vec{b}$

To find the inverse ${A}^{- 1}$

$A = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right) \implies {A}^{- 1} = \frac{1}{\det \left(A\right)} \left(\begin{matrix}d & - b \\ - c & a\end{matrix}\right)$

In this case $a = 3 , b = 1 , c = 1 , d = - 5$. Therefore

$\det \left(A\right) = a d - b c \implies \det \left(A\right) = - 15 - 1 = - 16$

following the formula for ${A}^{- 1}$ we get

$\implies {A}^{- 1} = \left(\begin{matrix}\frac{5}{16} & \frac{1}{16} \\ \frac{1}{16} & - \frac{3}{16}\end{matrix}\right)$

$\implies \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}\frac{5}{16} & \frac{1}{16} \\ \frac{1}{16} & - \frac{3}{16}\end{matrix}\right) \left(\begin{matrix}136 \\ - 40\end{matrix}\right)$

So we plug in and get

$\implies x = \frac{\left(5\right) \left(136\right) - 40}{16} = \frac{640}{16} = 40$

and

$y = \frac{136 + 120}{16} = \frac{256}{16} = 16$