# How do you solve 15x-20y=1 and 10y=1+5x using substitution?

Aug 15, 2016

$x = \frac{3}{5}$ and $y = \frac{2}{5}$

#### Explanation:

We have two equations $15 x - 20 y = 1$ and $10 y = 1 + 5 x$. Here we can easily substitute the value of $10 y$ from the second equation into the first equation and doing so we get

$15 x - 20 y = 1$

$15 x - 2 \times 10 y = 1$

15x-2×(1+5x)=1 or

$15 x - 2 - 10 x = 1$ or

$15 x - 10 x = 1 + 2$ or

$5 x = 3$ and $x = \frac{3}{5}$.

Hence $10 y = 1 + 5 x$

= 1+5×3/5=1+3=4 or

$y = \frac{4}{10} = \frac{2}{5}$