# How do you solve the system of equations 1/2x + 3y = 5/6 and 1/3x - 5y = 16/9?

##### 1 Answer
Jul 12, 2018

See a solution process below:

#### Explanation:

Step 1) Solve both equations for $x$:

• Equation 1:

$\frac{1}{2} x + 3 y = \frac{5}{6}$

$\textcolor{red}{2} \left(\frac{1}{2} x + 3 y\right) = \textcolor{red}{2} \times \frac{5}{6}$

$\left(\textcolor{red}{2} \times \frac{1}{2} x\right) + \left(\textcolor{red}{2} \times 3 y\right) = \cancel{\textcolor{red}{2}} \times \frac{5}{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} 3}$

$\frac{\textcolor{red}{2}}{2} x + 6 y = \frac{5}{3}$

$1 x + 6 y = \frac{5}{3}$

$x + 6 y = \frac{5}{3}$

$x + 6 y - \textcolor{red}{6 y} = \frac{5}{3} - \textcolor{red}{6 y}$

$x + 0 = \frac{5}{3} - 6 y$

$x = \frac{5}{3} - 6 y$

• Equation 2:

$\frac{1}{3} x - 5 y = \frac{16}{9}$

$\textcolor{red}{3} \left(\frac{1}{3} x - 5 y\right) = \textcolor{red}{3} \times \frac{16}{9}$

$\left(\textcolor{red}{3} \times \frac{1}{3} x\right) - \left(\textcolor{red}{3} \times 5 y\right) = \cancel{\textcolor{red}{3}} \times \frac{16}{\textcolor{red}{\cancel{\textcolor{b l a c k}{9}}} 3}$

$\frac{\textcolor{red}{3}}{3} x - 15 y = \frac{16}{3}$

$1 x - 15 y = \frac{16}{3}$

$x - 15 y = \frac{16}{3}$

$x - 15 y + \textcolor{red}{15 y} = \frac{16}{3} + \textcolor{red}{15 y}$

$x - 0 = \frac{16}{3} + 15 y$

$x = \frac{16}{3} + 15 y$

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

$\frac{5}{3} - 6 y = \frac{16}{3} + 15 y$

$\frac{5}{3} - \textcolor{b l u e}{\frac{16}{3}} - 6 y + \textcolor{red}{6 y} = \frac{16}{3} - \textcolor{b l u e}{\frac{16}{3}} + 15 y + \textcolor{red}{6 y}$

$\frac{5 - \textcolor{b l u e}{16}}{3} - 0 = 0 + \left(15 + \textcolor{red}{6}\right) y$

$- \frac{11}{3} = 21 y$

$\frac{1}{\textcolor{red}{21}} \times - \frac{11}{3} = \frac{21 y}{\textcolor{red}{21}}$

$- \frac{1 \times 11}{\textcolor{red}{21} \times 3} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{21}}} y}{\cancel{\textcolor{red}{21}}}$

$- \frac{11}{63} = y$

$y = - \frac{11}{63}$

Step 3) Substitute $- \frac{11}{63}$ for $y$ in the solution to either equation in Step 1 and calculate $x$:

$x = \frac{16}{3} + 15 y$ becomes:

$x = \frac{16}{3} + \left(15 \times - \frac{11}{63}\right)$

$x = \left(\frac{21}{21} \times \frac{16}{3}\right) + \frac{15 \times - 11}{63}$

$x = \frac{336}{63} + \frac{- 165}{63}$

$x = \frac{336 - 165}{63}$

$x = \frac{171}{63}$

The Solution Is:

$x = \frac{171}{63}$ and $y = - \frac{11}{63}$

Or

$\left(\frac{171}{63} , - \frac{11}{63}\right)$