# How do you solve -3w+ z=4 and -9w+ 5z= -1?

Jun 20, 2018

See a solution process below:

#### Explanation:

Step 1)Solve both equations for $- 9 w$:

• Equation 1:

$- 3 w + z = 4$

$\textcolor{red}{3} \left(- 3 w + z\right) = \textcolor{red}{3} \times 4$

$\left(\textcolor{red}{3} \times - 3 w\right) + \left(\textcolor{red}{3} \times z\right) = 12$

$- 9 w + 3 z = 12$

$- 9 w + 3 z - \textcolor{red}{3 z} = 12 - \textcolor{red}{3 z}$

$- 9 w + 0 = 12 - 3 z$

$- 9 w = 12 - 3 z$

• Equation 2:

$- 9 w + 5 z = - 1$

$- 9 w + 5 z - \textcolor{red}{5 z} = - 1 - \textcolor{red}{5 z}$

$- 9 w + 0 = - 1 - 5 z$

$- 9 w = - 1 - 5 z$

Step 2)* Now that the left side of each equation is equal we can equate the right side of each equation and solve for $z$

$12 - 3 z = - 1 - 5 z$

$12 - \textcolor{red}{12} - 3 z + \textcolor{b l u e}{5 z} = - 1 - \textcolor{red}{12} - 5 z + \textcolor{b l u e}{5 z}$

$0 + \left(- 3 + \textcolor{b l u e}{5}\right) z = - 13 - 0$

$2 z = - 13$

$\frac{2 z}{\textcolor{red}{2}} = - \frac{13}{\textcolor{red}{2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} z}{\cancel{\textcolor{red}{2}}} = - \frac{13}{2}$

$z = - \frac{13}{2}$

Step 3) Substitute $- \frac{13}{2}$ for $z$ in the solution to either equation in Step 1 and solve for $w$:

$- 9 w = 12 - 3 z$ becomes:

$- 9 w = 12 - \left(3 \times - \frac{13}{2}\right)$

$- 9 w = 12 - \left(- \frac{39}{2}\right)$

$- 9 w = 12 + \frac{39}{2}$

$- 9 w = \left(\frac{2}{2} \times 12\right) + \frac{39}{2}$

$- 9 w = \frac{24}{2} + \frac{39}{2}$

$- 9 w = \frac{24 + 39}{2}$

$- 9 w = \frac{63}{2}$

$\frac{1}{\textcolor{red}{- 9}} \times - 9 w = \frac{1}{\textcolor{red}{- 9}} \times \frac{63}{2}$

$\frac{1}{\cancel{\textcolor{red}{- 9}}} \times \textcolor{red}{\cancel{\textcolor{b l a c k}{- 9}}} w = - \frac{63}{18}$

$w = - \frac{7}{2}$

The Solution Is:

$w = - \frac{7}{2}$ and $z = - \frac{13}{2}$