# How do you find the solution of the system of equations 4x - 5y = 40  and 2x + 10y=20?

Jun 16, 2018

See a solution process below:

#### Explanation:

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

$2 x + 10 y = 20$

$2 x + 10 y - \textcolor{red}{10 y} = 20 - \textcolor{red}{10 y}$

$2 x + 0 = 20 - 10 y$

$2 x = 20 - 10 y$

$\frac{2 x}{\textcolor{red}{2}} = \frac{20 - 10 y}{\textcolor{red}{2}}$

$x = \frac{20}{\textcolor{red}{2}} - \frac{10 y}{\textcolor{red}{2}}$

$x = 10 - 5 y$

Step 3) Substitute $\left(10 - 5 y\right)$ for $x$ in the first equation and solve for $y$:

$4 x - 5 y = 40$ becomes:

$4 \left(10 - 5 y\right) - 5 y = 40$

$\left(4 \times 10\right) - \left(4 \times 5 y\right) - 5 y = 40$

$40 - 20 y - 5 y = 40$

$40 + \left(- 20 - 5\right) y = 40$

$40 + \left(- 25\right) y = 40$

$40 - 25 y = 40$

$40 - \textcolor{red}{40} - 25 y = 40 - \textcolor{red}{40}$

$0 - 25 y = 0$

$- 25 y = 0$

$\frac{- 25 y}{\textcolor{red}{- 25}} = \frac{0}{\textcolor{red}{- 25}}$

$y = 0$

Step 3) Substitute $0$ for $y$ in the solution to the second equation at the end of Step 1 and calculate $x$:

$x = 10 - 5 y$ becomes:

$x = 10 - \left(5 \times 0\right)$

$x = 10 - 0$

$x = 10$

The Solution Is:

$x = 10$ and $y = 0$

Or

$\left(10 , 0\right)$

Here's the solution graphically:

#graph{(4x-5y-40)(2x+10y-20)=0[-5,15,-5,5]}