# How do you solve the following system: 4x+3y= -7, 2x-5y=-19 ?

Nov 2, 2017

See a solution process below:

#### Explanation:

Step 1) Solve each equation for $4 x$:

Equation 1:

$4 x + 3 y - \textcolor{red}{3 y} = - 7 - \textcolor{red}{3 y}$

$4 x + 0 = - 7 - 3 y$

$4 x = - 7 - 3 y$

Equation 2:

$2 x - 5 y + \textcolor{red}{5 y} = - 19 + \textcolor{red}{5 y}$

$2 x - 0 = - 19 + 5 y$

$2 x = - 19 + 5 y$

$\textcolor{red}{2} \times 2 x = \textcolor{red}{2} \left(- 19 + 5 y\right)$

$4 x = \left(\textcolor{red}{2} \times - 19\right) + \left(\textcolor{red}{2} \times 5 y\right)$

$4 x = - 38 + 10 y$

Step 2) Because the left sides of each equation is equal, we can equate the right sides of each equation and solve for $y$:

$- 7 - 3 y = - 38 + 10 y$

$\textcolor{b l u e}{38} - 7 - 3 y + \textcolor{red}{3 y} = \textcolor{b l u e}{38} - 38 + 10 y + \textcolor{red}{3 y}$

$31 - 0 = 0 + \left(10 + \textcolor{red}{3}\right) y$

$31 = 13 y$

$\frac{31}{\textcolor{red}{13}} = \frac{13 y}{\textcolor{red}{13}}$

$\frac{31}{13} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{13}}} y}{\cancel{\textcolor{red}{13}}}$

$\frac{31}{13} = y$

$y = \frac{31}{13}$

Step 3) Substitute $\frac{31}{13}$ for $y$ in the solution to either equation in Step 1 and calculate $x$:

$4 x = - 7 - 3 y$ becomes:

$4 x = - 7 - \left(3 \times \frac{31}{13}\right)$

$4 x = - 7 - \frac{93}{13}$

$4 x = - \left(\frac{13}{13} \times 7\right) - \frac{93}{13}$

$4 x = - \frac{91}{13} - \frac{93}{13}$

$4 x = - \frac{184}{13}$

$\textcolor{red}{\frac{1}{4}} \times 4 x = \textcolor{red}{\frac{1}{4}} \times - \frac{184}{13}$

$\frac{4}{\textcolor{red}{4}} x = \textcolor{red}{\frac{1}{\textcolor{b l a c k}{\cancel{\textcolor{red}{4}}}}} \times - \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{184}}} 46}{13}$

$x = - \frac{46}{13}$

The Solution Is: $x = - \frac{46}{13}$ and $y = \frac{31}{13}$ or $\left(- \frac{46}{13} , \frac{31}{13}\right)$