# How do you solve the following system: 3x-y=12, 3x+4y=-10 ?

May 29, 2017

See a solution process below:

#### Explanation:

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

$3 x - y = 12$

$3 x - y + \textcolor{red}{y} - \textcolor{b l u e}{12} = \textcolor{red}{y} + 12 - \textcolor{b l u e}{12}$

$3 x \cancel{- y + \textcolor{red}{y}} - 12 = y + \cancel{12 - \textcolor{b l u e}{12}}$

$3 x - 12 = y$

$y = 3 x - 12$

Step 2) Substitute $\left(3 x - 12\right)$ for $y$ in the second equation and solve for $x$:

$3 x + 4 y = - 10$ becomes:

$3 x + 4 \left(3 x - 12\right) = - 10$

$3 x + \left(4 \times 3 x\right) - \left(4 \times 12\right) = - 10$

$3 x + 12 x - 48 = - 10$

$\left(3 + 12\right) x - 48 = - 10$

$15 x - 48 = - 10$

$15 x - 48 + \textcolor{red}{48} = - 10 + \textcolor{red}{48}$

$15 x \cancel{- 48 + \textcolor{red}{48}} = 38$

$15 x = 38$

$15 \frac{x}{\textcolor{red}{15}} = \frac{38}{\textcolor{red}{15}}$

$\textcolor{red}{\cancel{\textcolor{b l a c k}{15}}} \frac{x}{\cancel{\textcolor{red}{15}}} = \frac{38}{15}$

$x = \frac{38}{15}$

Step 3) Substitute $\frac{38}{15}$ for $x$ in the solution to the first equation at the end of Step 1 and calculate $y$:

$y = 3 x - 12$ becomes:

$y = \left(3 \times \frac{38}{15}\right) - 12$

$y = \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} \times \frac{38}{\textcolor{red}{\cancel{\textcolor{b l a c k}{15}}} 5}\right) - 12$

$y = \frac{38}{5} - 12$

$y = \frac{38}{5} - \left(\frac{5}{5} \times 12\right)$

$y = \frac{38}{5} - \frac{60}{5}$

$y = - \frac{22}{5}$

The solution is: $x = \frac{38}{15}$ and $y = - \frac{22}{5}$ or $\left(\frac{38}{15} , - \frac{22}{5}\right)$