How do you find an ordered pair for 4y + 1 = -3x and 5x + 6y = -3?

Feb 12, 2017

See the entire solution process below:

Explanation:

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

$4 y + 1 = - 3 x$

$4 y + 1 - \textcolor{red}{1} = - 3 x - \textcolor{red}{1}$

$4 y + 0 = - 3 x - 1$

$4 y = - 3 x - 1$

$\frac{4 y}{\textcolor{red}{4}} = \frac{- 3 x - 1}{\textcolor{red}{4}}$

$\frac{\textcolor{red}{\cancel{\textcolor{B l a c k}{4}}} y}{\cancel{\textcolor{red}{4}}} = \frac{- 3 x - 1}{4}$

$y = \frac{- 3 x - 1}{4}$

Step 2) Substitute $\frac{- 3 x - 1}{4}$ for $y$ in the second equation and solve for $x$:

$5 x + 6 y = - 3$ becomes:

$5 x + 6 \left(\frac{- 3 x - 1}{4}\right) = - 3$

$5 x - \frac{18 x}{4} - \frac{6}{4} = - 3$

$\textcolor{red}{4} \left(5 x - \frac{18 x}{4} - \frac{6}{4}\right) = \textcolor{red}{4} \times - 3$

$\left(\textcolor{red}{4} \times 5 x\right) - \left(\textcolor{red}{4} \times \frac{18 x}{4}\right) - \left(\textcolor{red}{4} \times \frac{6}{4}\right) = - 12$

$20 x - \left(\cancel{\textcolor{red}{4}} \times \frac{18 x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}}\right) - \left(\cancel{\textcolor{red}{4}} \times \frac{6}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}}\right) = - 12$

$20 x - 18 x - 6 = - 12$

$2 x - 6 = - 12$

$2 x - 6 + \textcolor{red}{6} = - 12 + \textcolor{red}{6}$

$2 x - 0 = - 6$

$2 x = - 6$

$\frac{2 x}{\textcolor{red}{2}} = - \frac{6}{\textcolor{red}{2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x}{\cancel{\textcolor{red}{2}}} = - 3$

$x = - 3$

Step 3) Substitute $- 3$ for $x$ in the solution to first equation at the end of Step 1 and calculate $y$:

$y = \frac{- 3 x - 1}{4}$ becomes:

$y = \frac{\left(- 3 \times - 3\right) - 1}{4}$

$y = \frac{9 - 1}{4}$

$y = \frac{8}{4}$

$y = 2$

The solution is: $x = - 3$ and $y = 2$ or $\left(- 3 , 2\right)$