# How do you solve the system of linear equations 5x - 4y = 9 and x - 2y = -3?

Jun 9, 2017

See a solution process below:

#### Explanation:

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

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

$z - 0 = - 3 + 2 y$

$z = - 3 + 2 y$

Step 2) Substitute $\left(- 3 + 2 y\right)$ for $x$ in the first equation and solve for $y$:

$5 x - 4 y = 9$ becomes:

$5 \left(- 3 + 2 y\right) - 4 y = 9$

$\left(5 \times - 3\right) + \left(5 \times 2 y\right) - 4 y = 9$

$- 15 + 10 y - 4 y = 9$

$- 15 + \left(10 - 4\right) y = 9$

$- 15 + 6 y = 9$

$\textcolor{red}{15} - 15 + 6 y = \textcolor{red}{15} + 9$

$0 + 6 y = 24$

$6 y = 24$

$\frac{6 y}{\textcolor{red}{6}} = \frac{24}{\textcolor{red}{6}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{6}}} y}{\cancel{\textcolor{red}{6}}} = 4$

$y = 4$

Step 3) Substitute $4$ for $y$ in the solution to the second equation at the end of Step 1 and calculate $x$:

$x = - 3 + 2 y$ becomes:

$x = - 3 + \left(2 \times 4\right)$

$x = - 3 + 8$

$x = 5$

The solution is: $x = 5$ and $y = 4$ or $\left(5 , 4\right)$