# How do you solve 10x + 10y = 1 and x = y - 3  using substitution?

Feb 11, 2017

See the entire solution process below:

#### Explanation:

Step 1) Because the second equation is already solved for $x$ we can substitute $y - 3$ for $x$ in the first equation and solve for $y$:

$10 x + 10 y = 1$ becomes:

$10 \left(y - 3\right) + 10 y = 1$

$10 y - 30 + 10 y = 1$

$10 y + 10 y - 30 = 1$

$20 y - 30 = 1$

$20 y - 30 + \textcolor{red}{30} = 1 + \textcolor{red}{30}$

$20 y - 0 = 31$

$20 y = 31$

$\frac{20 y}{\textcolor{red}{20}} = \frac{31}{\textcolor{red}{20}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{20}}} y}{\cancel{\textcolor{red}{20}}} = \frac{31}{20}$

$y = \frac{31}{20}$

Step 2) Substitute $\frac{31}{20}$ for $y$ in the second equation and calculate $x$:

$x = y - 3$ becomes:

$x = \frac{31}{20} - 3$

$x = \frac{31}{20} - \left(\frac{20}{20} \times 3\right)$

$x = \frac{31}{20} - \frac{60}{20}$

$x = - \frac{29}{20}$

The solution is: $x = - \frac{29}{20}$ and $y = \frac{31}{20}$ or $\left(- \frac{29}{20} , \frac{31}{20}\right)$