# How do you solve 0.5x+0.25y=36 and y+18=16x using substitution?

Mar 5, 2016

$x = 5$
$y = 62$

#### Explanation:

In either equation, express one of the variables in terms of the other terms.

$\left[1\right] 0.5 x + 0.25 y = 36$
$\left[2\right] y + 18 = 16 x$

If we solve for $y$ in $\left[2\right]$, we have

$\left[3\right] \implies y = 16 x - 18$

Now, substitute this relation into $\left[1\right]$

$\left[1\right] 0.5 x + 0.25 y = 36$

$\implies 0.5 x + 0.25 \left(16 x - 18\right) = 18$

Multiplying the entire equation by 4, we have

$\implies 4 \left(0.5 x + 0.25 \left(16 x - 18\right) = 18\right)$

$\implies 2 x + 1 \left(16 x - 18\right) = 72$
$\implies 2 x + 16 x - 18 = 72$

$\implies 18 x = 90$
$\implies x = 5$

Using $\left[3\right]$ (or $\left[1\right]$ or $\left[2\right]$), solve for $y$

$y = 16 x - 18$
$\implies y = 16 \left(5\right) - 18$

$\implies y = 80 - 18$

$\implies y = 62$