# How do you solve 3x - 10y = -25 and 4x + 40y = 20?

May 31, 2018

See a solution process below:

#### Explanation:

Step 1) Solve each equation for a common term, in this problem it can be $40 y$:

• Equation 1:

$3 x - 10 y = - 25$

$\textcolor{red}{- 4} \left(3 x - 10 y\right) = \textcolor{red}{- 4} \times - 25$

$\left(\textcolor{red}{- 4} \times 3 x\right) + \left(\textcolor{red}{- 4} \times - 10 y\right) = 100$

$- 12 x + 40 y = 100$

$- 12 x + \textcolor{red}{12 x} + 40 y = 100 + \textcolor{red}{12 x}$

$0 + 40 y = 100 + 12 x$

$40 y = 100 + 12 x$

• Equation 2:

$4 x + 40 y = 20$

$4 x - \textcolor{red}{4 x} + 40 y = 20 - \textcolor{red}{4 x}$

$0 + 40 y = 20 - 4 x$

$40 y = 20 - 4 x$

Step 2) Because the left side of both equations are now equal we can equate the right side of the equations and solve for $x$:

$100 + 12 x = 20 - 4 x$

$100 - \textcolor{red}{100} + 12 x + \textcolor{b l u e}{4 x} = 20 - \textcolor{red}{100} - 4 x + \textcolor{b l u e}{4 x}$

$0 + \left(12 + \textcolor{b l u e}{4}\right) x = - 80 - 0$

$16 x = - 80$

$\frac{16 x}{\textcolor{red}{16}} = - \frac{80}{\textcolor{red}{16}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{16}}} x}{\cancel{\textcolor{red}{16}}} = - 5$

$x = - 5$

Step 3) Substitute $- 5$ for $x$ in the solution to either equation in Step 1 and solve for $y$:

$40 y = 100 + 12 x$ becomes:

$40 y = 100 + \left(12 \cdot - 5\right)$

$40 y = 100 + \left(- 60\right)$

$40 y = 100 - 60$

$40 y = 40$

$\frac{40 y}{\textcolor{red}{40}} = \frac{40}{\textcolor{red}{40}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{40}}} y}{\cancel{\textcolor{red}{40}}} = 1$

$y = 1$

The Solution Is:

$x = - 5$ and $y = 1$

Or

$\left(- 5 , 1\right)$