How do you solve the following linear system: {(4x + 5y = -3) ,( y = 1/2x - 3/2):} ?

Mar 13, 2018

See explanation.

Explanation:

You can solve the equation by replacing $y$ in the first equation with $\frac{1}{2} x - \frac{3}{2}$ from the second equation:

$\left\{\begin{matrix}4 x + 5 \left(\frac{1}{2} x - \frac{3}{2}\right) = - 3 \\ y = \frac{1}{2} x - \frac{3}{2}\end{matrix}\right.$

$\left\{\begin{matrix}4 x + \frac{5}{2} x - \frac{15}{2} = - 3 \\ y = \frac{1}{2} x - \frac{3}{2}\end{matrix}\right.$

Now we can calculate $x$ from the first equation:

$\left\{\begin{matrix}\frac{13}{2} x = - 3 + \frac{15}{2} \\ y = \frac{1}{2} x - \frac{3}{2}\end{matrix}\right.$

$\left\{\begin{matrix}\frac{13}{2} x = \frac{9}{2} \\ y = \frac{1}{2} x - \frac{3}{2}\end{matrix}\right.$

$\left\{\begin{matrix}13 x = 9 \\ y = \frac{1}{2} x - \frac{3}{2}\end{matrix}\right.$

$\left\{\begin{matrix}x = \frac{9}{13} \\ y = \frac{9}{26} - \frac{3}{2}\end{matrix}\right.$

Last step is to put the calculated value of $x$ into the second equation to calculate $y$:

$\left\{\begin{matrix}x = \frac{9}{13} \\ y = \frac{9}{26} - \frac{39}{26}\end{matrix}\right.$

$\left\{\begin{matrix}x = \frac{9}{13} \\ y = - 1 \frac{2}{13}\end{matrix}\right.$