# How do you solve this system of equations: 2t + 4z = 68 and 4t + 3z = 76?

Jun 30, 2018

See a solution process below:

#### Explanation:

Step 1) Solve both equations for $4 t$:

• Equation 1:

$2 t + 4 z = 68$

$\textcolor{red}{2} \left(2 t + 4 z\right) = \textcolor{red}{2} \times 68$

$\left(\textcolor{red}{2} \times 2 t\right) + \left(\textcolor{red}{2} \times 4 z\right) = 136$

$4 t + 8 z = 136$

$4 t + 8 z - \textcolor{red}{8 z} = 136 - \textcolor{red}{8 z}$

$4 t + 0 = 136 - 8 z$

$4 t = 136 - 8 z$

• Equation 2:

$4 t + 3 z = 76$

$4 t + 3 z - \textcolor{red}{3 z} = 76 - \textcolor{red}{3 z}$

$4 t + 0 = 76 - 3 z$

$4 t = 76 - 3 z$

Step 2) Because the left side of both equations are now equal we can equate the right side of both equations and solve for $z$:

$136 - 8 z = 76 - 3 z$

$136 - \textcolor{b l u e}{76} - 8 z + \textcolor{red}{8 z} = 76 - \textcolor{b l u e}{76} - 3 z + \textcolor{red}{8 z}$

$60 - 0 = 0 + \left(- 3 + \textcolor{red}{8}\right) z$

$60 = 5 z$

$\frac{60}{\textcolor{red}{5}} = \frac{5 z}{\textcolor{red}{5}}$

$12 = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} z}{\cancel{\textcolor{red}{5}}}$

$12 = z$

$z = 12$

Step 3) Substitute $12$ for $z$ in the solution to either equation in Step 1 and solve for $t$:

$4 t = 136 - 8 z$ becomes:

$4 t = 136 - \left(8 \times 12\right)$

$4 t = 136 - 96$

$4 t = 40$

$\frac{4 t}{\textcolor{red}{4}} = \frac{40}{\textcolor{red}{4}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} t}{\cancel{\textcolor{red}{4}}} = 10$

$t = 10$

The Solution Is:

$t = 10$ and $z = 12$