How do you solve the following system of equations?: 2x + 15y = -4 , -x + 3y=12?

Mar 7, 2017

See the entire solution process below:

Explanation:

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

$- x + 3 y = 12$

$- x + 3 y - \textcolor{red}{3 y} = 12 - \textcolor{red}{3 y}$

$- x + 0 = 12 - 3 y$

$- x = 12 - 3 y$

$- 1 \times - x = - 1 \left(12 - 3 y\right)$

$x = \left(- 1 \times 12\right) - \left(- 1 \times 3 y\right)$

$x = - 12 + 3 y$

Step 2) Substitute $- 12 + 3 y$ for $x$ in the first equation and solve for $y$:

$2 x + 15 y = - 4$ becomes:

$2 \left(- 12 + 3 y\right) + 15 y = - 4$

$\left(2 \times - 12\right) + \left(2 \times 3 y\right) + 15 y = - 4$

$- 24 + 6 y + 15 y = - 4$

$- 24 + 21 y = - 4$

$\textcolor{red}{24} - 24 + 21 y = \textcolor{red}{24} - 4$

$0 + 21 y = 20$

$21 y = 20$

$\frac{21 y}{\textcolor{red}{21}} = \frac{20}{\textcolor{red}{21}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{21}}} y}{\cancel{\textcolor{red}{21}}} = \frac{20}{21}$

$y = \frac{20}{21}$

Step 3) Substitute $\frac{20}{21}$ for $y$ in the solution to the second equation at the end of Step 1 and calculate $x$:

$x = - 12 + 3 y$ becomes:

$x = - 12 + \left(3 \times \frac{20}{21}\right)$

$x = - 12 + \frac{20}{7}$

$x = \left(\frac{7}{7} \times - 12\right) + \frac{20}{7}$

$x = - \frac{84}{7} + \frac{20}{7}$

$x = - \frac{64}{7}$

The solution is: $x = - \frac{64}{7}$ and $y = \frac{20}{21}$ or $\left(- \frac{64}{7} , \frac{20}{21}\right)$