# How do you solve the following system: 3x - 2y = 9, 2x + 3y = 12?

May 16, 2017

Arrange your equation and get the solution $x = \frac{51}{13}$ and $y = \frac{18}{13}$

#### Explanation:

$9 x - 6 y = 27$
$4 x - 6 y = 24$

when you expand the first equation (3) and the second (2)
Combine the above equations:

$9 x + 4 x = 51$
$x = \frac{51}{13}$

Put this value in any equation to get y

$2 \cdot \frac{51}{13} + 3 y = 12$

$3 y = \frac{156 - 102}{13}$

$y = \frac{54}{13 \cdot 3}$

$y = \frac{18}{13}$

Your $x = \frac{51}{13}$ and your $y = \frac{18}{13}$

May 16, 2017

See a solution process below:

#### Explanation:

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

$3 x - 2 y = 9$

$3 x - 2 y + \textcolor{red}{2 y} = 9 + \textcolor{red}{2 y}$

$3 x - 0 = 9 + 2 y$

$3 x = 9 + 2 y$

$\frac{3 x}{\textcolor{red}{3}} = \frac{9 + 2 y}{\textcolor{red}{3}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} x}{\cancel{\textcolor{red}{3}}} = \frac{9}{3} + \frac{2 y}{3}$

$x = 3 + \frac{2}{3} y$

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

$2 x + 3 y = 12$ becomes:

$2 \left(3 + \frac{2}{3} y\right) + 3 y = 12$

$\left(2 \cdot 3\right) + \left(2 \cdot \frac{2}{3} y\right) + 3 y = 12$

$6 + \frac{4}{3} y + 3 y = 12$

$6 + \frac{4}{3} y + \left(\frac{3}{3} \cdot 3 y\right) = 12$

$6 + \frac{4}{3} y + \frac{9}{3} y = 12$

$6 + \frac{13}{3} y = 12$

$- \textcolor{red}{6} + 6 + \frac{13}{3} y = - \textcolor{red}{6} + 12$

$0 + \frac{13}{3} y = 6$

$\frac{13}{3} y = 6$

$\frac{\textcolor{red}{3}}{\textcolor{b l u e}{13}} \times \frac{13}{3} y = \frac{\textcolor{red}{3}}{\textcolor{b l u e}{13}} \times 6$

$\frac{\cancel{\textcolor{red}{3}}}{\cancel{\textcolor{b l u e}{13}}} \times \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{13}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} y = \frac{18}{13}$

$y = \frac{18}{13}$

Step 3) Substitute $\frac{18}{13}$ for $y$ in the solution to the first equation at the end of Step 1 and calculate $x$:

$x = 3 + \frac{2}{3} y$ becomes:

$x = 3 + \left(\frac{2}{3} \times \frac{18}{13}\right)$

$x = 3 + \left(\frac{2}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}} \times \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{18}}} 6}{13}\right)$

$x = 3 + \frac{12}{13}$

$x = \left(\frac{13}{13} \times 3\right) + \frac{12}{13}$

$x = \frac{39}{13} + \frac{12}{13}$

$x = \frac{51}{13}$

The solution is: $x = \frac{51}{13}$ and $y = \frac{18}{13}$ or $\left(\frac{51}{13} , \frac{18}{131}\right)$