Q. 35 - Precalculus Enhanced with Graphing Utilities

Question

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

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

Step-by-Step Solution

Answer

The solution of the given system of equation is $x=4, y=3; \left(4,3\right)$.

Step 1. Given information.

The given equations are:

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

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

The first equation can be written as:

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

Step 3. Determine the value of one variable.

Substitute the above value in the second equation.

$\begin{array}{rcl}\frac{1}{4}\left(\frac{2(9-y)}{3}\right)-\frac{2}{3}y& =& -1\\ \frac{9-y}{6}-\frac{2}{3}y& =& -1\\ \frac{9-y-4y}{6}& =& -1\\ 9-5y& =& -6\end{array}$

On further simplification, we get

$\begin{array}{rcl}-5y& =& -6-9\\ -5y& =& -15\\ y& =& \frac{-15}{-5}\\ y& =& 3\end{array}$

Step 4. Determine the value of another variable.

Substitute $y=3$ in $x=\frac{2(9-y)}{3}$.

$\begin{array}{rcl}x& =& \frac{2(9-3)}{3}\\ x& =& \frac{2\left(6\right)}{3}\\ x& =& 2\left(2\right)\\ x& =& 4\end{array}$

The solution of the given system of equation is $x=4, y=3; \left(4,3\right)$.