# How do you solve the following system: x - 3y = -1 , 7x+15y=32 ?

May 8, 2017

See a solution process below:

#### Explanation:

Step 1) Solve the first equation for $x$:

$x - 3 y = - 1$

$x - 3 y + \textcolor{red}{3 y} = - 1 + \textcolor{red}{3 y}$

$x - 0 = - 1 + 3 y$

$x = - 1 + 3 y$

Step 2) Substitute $- 1 + 3 y$ for $x$ in the second equation and solve for $y$:

$7 x + 15 y = 32$ becomes:

$7 \left(- 1 + 3 y\right) + 15 y = 32$

$\left(7 \cdot - 1\right) + \left(7 \cdot 3 y\right) + 15 y = 32$

$- 7 + 21 y + 15 y = 32$

$- 7 + \left(21 + 15\right) y = 32$

$- 7 + 36 y = 32$

$\textcolor{red}{7} - 7 + 36 y = \textcolor{red}{7} + 32$

$0 + 36 y = 39$

$36 y = 39$

$\frac{36 y}{\textcolor{red}{36}} = \frac{39}{\textcolor{red}{36}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{36}}} y}{\cancel{\textcolor{red}{36}}} = \frac{13 \times 3}{\textcolor{red}{12 \times 3}}$

$y = \frac{13 \times \textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}}{\textcolor{red}{12 \times \textcolor{b l a c k}{\cancel{\textcolor{red}{3}}}}}$

$y = \frac{13}{12}$

Step 3) Substitute $\frac{13}{12}$ for $y$ in the solution to the first equation at the end of Step 1 and calculate $x$:

$x = - 1 + 3 y$ becomes:

$x = - 1 + \left(3 \cdot \frac{13}{12}\right)$

$x = - 1 + \frac{39}{12}$

$x = \left(\frac{12}{12} \cdot - 1\right) + \frac{39}{12}$

$x = - \frac{12}{12} + \frac{39}{12}$

$x = \frac{27}{12}$

$x = \frac{3 \times 9}{3 \times 4}$

$x = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \times 9}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \times 4}$

$x = \frac{9}{4}$

The solution is: $x = \frac{9}{4}$ and $y = \frac{13}{12}$ or $\left(\frac{9}{4} , \frac{13}{12}\right)$