# How do you solve the system of equations 3y-4x=-33 and 5x-3y=40.5?

Jul 3, 2018

See a solution process below:

#### Explanation:

Step 1) Solve each equation for $3 y$:

• Equation 1:

$3 y - 4 x = - 33$

$3 y - 4 x + \textcolor{red}{4 x} = - 33 + \textcolor{red}{4 x}$

$3 y - 0 = - 33 + 4 x$

$3 y = - 33 + 4 x$

• Equation 2:

$5 x - 3 y = 40.5$

$5 x - \textcolor{red}{5 x} - 3 y = 40.5 - \textcolor{red}{5 x}$

$0 - 3 y = 40.5 - 5 x$

$- 3 y = 40.5 - 5 x$

$\textcolor{red}{- 1} \times - 3 y = \textcolor{red}{- 1} \left(40.5 - 5 x\right)$

$3 y = \left(\textcolor{red}{- 1} \times 40.5\right) - \left(\textcolor{red}{- 1} \times 5 x\right)$

$3 y = - 40.5 - \left(- 5 x\right)$

$3 y = - 40.5 + 5 x$

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

$- 33 + 4 x = - 40.5 + 5 x$

$- 33 + \textcolor{b l u e}{40.5} + 4 x - \textcolor{red}{4 x} = - 40.5 + \textcolor{b l u e}{40.5} + 5 x - \textcolor{red}{4 x}$

$7.5 + 0 = 0 + \left(5 - \textcolor{red}{4}\right) x$

$7.5 = 1 x$

$7.5 = x$

$x = 7.5$

Step 3) Substitute $7.5$ for $x$ into either of the solutions for the equations in Step 1 and calculate $y$:

$3 y = - 33 + 4 x$ becomes:

$3 y = - 33 + \left(4 \times 7.5\right)$

$3 y = - 33 + 30$

$3 y = - 3$

$\frac{3 y}{\textcolor{red}{3}} = - \frac{3}{\textcolor{red}{3}}$

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

$y = - 1$

The Solution Is:

$x = 7.5$ and $y = - 1$

Or

$\left(7.5 , - 1\right)$