Q. 24 - Precalculus Enhanced with Graphing Utilities

Question

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

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

Step-by-Step Solution

Answer

The solution of the given system of equation is $x=-\frac{5}{3}, y=1; \left(-\frac{5}{3},1\right)$.

Step 1. Given information.

The given equations are:

$\begin{array}{rcl}2x+4y& =& \frac{2}{3}\\ 3x-5y& =& -10\end{array}$

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

The second equation can be written as:

$\begin{array}{rcl}3x-5y& =& -10\\ 3x& =& -10+5y\\ x& =& \frac{5y-10}{3}\end{array}$

Step 3. Determine the value of one variable.

Substitute the above value in the first equation.

$\begin{array}{rcl}2\left(\frac{5y-10}{3}\right)+4y& =& \frac{2}{3}\\ \frac{10y-20}{3}+4y& =& \frac{2}{3}\\ \frac{10y-20+12y}{3}& =& \frac{2}{3}\\ 22y-20& =& 2\\ 22y& =& 2+20\\ y& =& \frac{22}{22}\\ y& =& 1\end{array}$

Step 4. Determine the value of another variable.

Substitute $y=1$ in $x=\frac{5y-10}{3}$.

$\begin{array}{rcl}x& =& \frac{5\left(1\right)-10}{3}\\ x& =& \frac{5-10}{3}\\ x& =& -\frac{5}{3}\end{array}$

The solution of the given system of equation is $x=-\frac{5}{3}, y=1; \left(-\frac{5}{3},1\right)$.