# How do you solve 1 / 5x + 1/3y = - 1 / 15 and 1 / 5x + 2y = 8 / 5 using substitution?

We can solve each equation for $\frac{1}{5} x$ and set them equal to each other:

$\frac{1}{5} x + \frac{1}{3} y = - \frac{1}{15} \implies \frac{1}{5} x = - \frac{1}{15} - \frac{1}{3} y$

$\frac{1}{5} x + 2 y = \frac{8}{5} \implies \frac{1}{5} x = \frac{8}{5} - 2 y$

$\therefore - \frac{1}{15} - \frac{1}{3} y = \frac{8}{5} - 2 y$

We can now solve for $y$. Let's first multiply through by 15:

$15 \left(- \frac{1}{15} - \frac{1}{3} y\right) = 15 \left(\frac{8}{5} - 2 y\right)$

$- 1 - 5 y = 24 - 30 y$

$25 y = 25$

$y = 1$

And now we can substitute back into one of the original equations to solve for $x$:

$\frac{1}{5} x + 2 \left(1\right) = \frac{8}{5}$

$\frac{1}{5} x = \frac{8}{5} - 2$

$\frac{1}{5} x = \frac{8}{5} - \frac{10}{5} = - \frac{2}{5}$

$x = - 2$

Which gives us the point $\left(- 2 , 1\right)$ which we can see on the graph:

graph{(1/5x+1/3y+1/15)(1/5x+2y-8/5)=0}

Apr 16, 2017

$x = - 2 \mathmr{and} y = 1$

#### Explanation:

I would usually simplify the equations to get rid of the fractions first, but in this case we have identical terms in each equation, so I will solve for those first.

$\frac{1}{5} x = - \frac{1}{15} - \frac{1}{3} y \text{ and } \frac{1}{5} x = \frac{8}{5} - 2 y$

$\frac{1}{5} x = \frac{1}{5} x$

We can substitute the other sides of each equation for $\frac{1}{5} x$

$- \frac{1}{15} - \frac{1}{3} y = \frac{8}{5} - 2 y \text{ } \times 15$

$- 1 - 5 y = 24 - 30 y$

$30 y - 5 y = 24 + 1$

$25 y = 25$

$y = 1$

Substitute into $\frac{1}{5} x = \frac{8}{5} - 2 y$

$\frac{1}{5} x = \frac{8}{5} - 2 \left(1\right) \text{ } \times 5$

$x = 8 - 10$

$x = - 2$

Check:
$\frac{1}{5} x + \frac{1}{3} y = - \frac{1}{15} \text{ } \times 5$

$x + \frac{5}{3} y = - \frac{1}{3}$

$- 2 + 1 \frac{2}{3} = - \frac{1}{3}$

$- \frac{1}{3} = - \frac{1}{3}$