# How do you solve 12x-5y=8 and 3x=5/4y+2 using substitution?

May 23, 2017

See a solution process below:

#### Explanation:

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

$3 x = \frac{5}{4} y + 2$

$\textcolor{red}{\frac{1}{3}} \times 3 x = \textcolor{red}{\frac{1}{3}} \left(\frac{5}{4} y + 2\right)$

$\textcolor{red}{\frac{1}{\textcolor{b l a c k}{\cancel{\textcolor{red}{3}}}}} \times \textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} x = \left(\textcolor{red}{\frac{1}{3}} \times \frac{5}{4} y\right) + \left(\textcolor{red}{\frac{1}{3}} \times 2\right)$

$x = \frac{5}{12} y + \frac{2}{3}$

Step 2) Substitute $\frac{5}{12} y + \frac{2}{3}$ for $x$ in the first equation and solve for $y$:

$12 x - 5 y = 8$ becomes:

$12 \left(\frac{5}{12} y + \frac{2}{3}\right) - 5 y = 8$

$\left(12 \times \frac{5}{12} y\right) + \left(12 \times \frac{2}{3}\right) - 5 y = 8$

$\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{12}}} \times \frac{5}{\textcolor{red}{\cancel{\textcolor{b l a c k}{12}}}} y\right) + \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{12}}} 4 \times \frac{2}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}}\right) - 5 y = 8$

$5 y + 8 - 5 y = 8$

$5 y - 5 y + 8 = 8$

$0 + 8 = 8$

$8 = 8$

Because $8$ does in fact equal $8$ these two equations are parallel and the same. They have an infinite number of points the same.