# How do you solve the following system: -2x + 5y = 20, x +4y = 16?

Mar 21, 2018

Let's use substitution:

$- 2 x + 5 y = 20$

$x + 4 y = 16$

We need to solve for $x$ in the second equation

$x = 16 - 4 y$

Now we substitute $\left(16 - 4 y\right)$ for $x$ in the first equation

$- 2 \left(16 - 4 y\right) + 5 y = 20$

distribute the $- 2$

$- 32 + 8 y + 5 y = 20$

Solve for $y$. Add $32$ to both sides

$13 y = 52$

Divide by $13$ on both sides

$y = 4$

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We have $y$, let's find $x$:

$x = 16 - 4 y$

$x = 16 - 4 \left(4\right)$

$x = 16 - 16$

$x = 0$

$\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot$

To check our work, let's plug our values for $x$ and $y$ into the first equation and see if it equals $20$:

$- 2 \left(0\right) + 5 \left(4\right)$

$0 + 20$

$20$ EQUALS $20$! We were right