# How do you solve the following system?: 3x - 2y = 1 , 6x+5y=18

Mar 30, 2018

See a solution process below: $\left(\frac{41}{27} , \frac{16}{9}\right)$

#### Explanation:

Step 1) Solve each equation for $6 x$:

• Equation 1:

$3 x - 2 y = 1$

$3 x - 2 y + \textcolor{red}{2 y} = 1 + \textcolor{red}{2 y}$

$3 x - 0 = 1 + 2 y$

$3 x = 1 + 2 y$

$\textcolor{red}{2} \times 3 x = \textcolor{red}{2} \left(1 + 2 y\right)$

$6 x = \left(\textcolor{red}{2} \times 1\right) + \left(\textcolor{red}{2} \times 2 y\right)$

$6 x = 2 + 4 y$

• Equation 2:

$6 x + 5 y = 18$

$6 x + 5 y - \textcolor{red}{5 y} = 18 - \textcolor{red}{5 y}$

$6 x + 0 = 18 - 5 y$

$6 x = 18 - 5 y$

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

$2 + 4 y = 18 - 5 y$

$2 - \textcolor{red}{2} + 4 y + \textcolor{b l u e}{5 y} = 18 - \textcolor{red}{2} - 5 y + \textcolor{b l u e}{5 y}$

$0 + \left(4 + \textcolor{b l u e}{5}\right) y = 16 - 0$

$9 y = 16$

$\frac{9 y}{\textcolor{red}{9}} = \frac{16}{\textcolor{red}{9}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{9}}} y}{\cancel{\textcolor{red}{9}}} = \frac{16}{9}$

$y = \frac{16}{9}$

Step 3) Substitute $\frac{16}{9}$ for $y$ in either of the equations in Step 1 and solve for $x$:

$6 x = 2 + 4 y$ becomes:

$6 x = 2 + 4 \left(\frac{16}{9}\right)$

$6 x = \left(\frac{9}{9} \times 2\right) + \frac{64}{9}$

$6 x = \frac{18}{9} + \frac{64}{9}$

$6 x = \frac{82}{9}$

$6 x = \frac{82}{9}$

$\frac{1}{6} \times 6 x = \frac{1}{6} \times \frac{82}{9}$

$\frac{6}{6} x = \frac{82}{54}$

$x = \frac{41}{27}$

The Solution Is:

$x = \frac{41}{27}$ and $y = \frac{16}{9}$

Or

$\left(\frac{41}{27} , \frac{16}{9}\right)$