Q. 36 - Precalculus Enhanced with Graphing Utilities

Question

Solve each system of equations. If the system has no solution, say that it is inconsistent.

$\left\{\begin{array}{c}\frac{1}{3}x-\frac{3}{2}y=-5\\ \frac{3}{4}x+\frac{1}{3}y=11\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the given system of equation is $x=12, y=6; \left(12,6\right)$.

Step 1. Given information.

The given equations are:

$\begin{array}{rcl}\frac{1}{3}x-\frac{3}{2}y& =& -5\\ \frac{3}{4}x+\frac{1}{3}y& =& 11\end{array}$

Step 2. Determine one variable in terms of another variable.

The first equation can be written as:

$\begin{array}{rcl}\frac{1}{3}x-\frac{3}{2}y& =& -5\\ \frac{1}{3}x& =& \frac{3}{2}y-5\\ \frac{1}{3}x& =& \frac{3y-10}{2}\\ x& =& \frac{3(3y-10)}{2}\\ x& =& \frac{9y-30}{2}\end{array}$

Step 3. Determine the value of one variable.

Substitute the above value in the second equation.

$\begin{array}{rcl}\frac{3}{4}\left(\frac{9y-30}{2}\right)+\frac{1}{3}y& =& 11\\ \frac{27y-90}{8}+\frac{1}{3}y& =& 11\\ \frac{81y-270+8y}{24}& =& 11\\ \frac{89y-270}{24}& =& 11\end{array}$

On further simplification, we get

$\begin{array}{rcl}89y-270& =& 264\\ 89y& =& 264+270\\ y& =& \frac{534}{89}\\ y& =& 6\end{array}$

Step 4. Determine the value of another variable.

Substitute $y=6$ in $x=\frac{9y-30}{2}$.

$\begin{array}{rcl}x& =& \frac{9\left(6\right)-30}{2}\\ x& =& \frac{54-30}{2}\\ x& =& \frac{24}{2}\\ x& =& 12\end{array}$

The solution of the given system of equation is $x=12, y=6; \left(12,6\right)$.