How do you solve 9x + 2y = 5 and y - 2x + 3 = 0?

Dec 23, 2016

$x = \frac{11}{13}$ and $y = - \frac{17}{13}$

Explanation:

Step 1) Solve the second equation for $y$:

$y - 2 x + 3 + \textcolor{red}{2 x - 3} = 0 + \textcolor{red}{2 x - 3}$

$y - 2 x + \textcolor{red}{2 x} + 3 - \textcolor{red}{3} = 0 + 2 x - 3$

$y - 0 + 0 = 2 x - 3$

$y = 2 x - 3$

Step 2) Substitute $2 x - 3$ for $y$ in the first equation and solve for $x$:

$9 x + 2 \left(\textcolor{red}{2 x - 3}\right) = 5$

$9 x + 4 x - 6 = 5$

$\left(9 + 4\right) x - 6 = 5$

$13 x - 6 = 5$

$13 x - 6 + \textcolor{red}{6} = 5 + \textcolor{red}{6}$

$13 x - 0 = 11$

$13 x = 11$

$\frac{13 x}{\textcolor{red}{13}} = \frac{11}{\textcolor{red}{13}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{13}}} x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{13}}}} = \frac{11}{13}$

$x = \frac{11}{13}$

Step 3) Substitute $\frac{11}{13}$ for $x$ in the solution to the second equation in Step 1) and calculate $y$:

$y = 2 \left(\textcolor{red}{\frac{11}{13}}\right) - 3$

$y = \frac{22}{13} - 3$

$y = \frac{22}{13} - \left(3 \cdot \frac{13}{13}\right)$

$y = \frac{22}{13} - \frac{39}{13}$

$y = - \frac{17}{13}$