Q. 28 - Precalculus Enhanced with Graphing Utilities

Question

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

$\left\{\begin{array}{c}3x+3y=-1\\ 4x+y=\frac{8}{3}\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the given system of equation is $x=1, y=-\frac{4}{3}; \left(1,-\frac{4}{3}\right)$.

Step 1. Given information.

The given equations are:

$\begin{array}{rcl}3x+3y& =& -1\\ 4x+y& =& \frac{8}{3}\end{array}$

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

The second equation can be written as:

$\begin{array}{rcl}4x+y& =& \frac{8}{3}\\ y& =& \frac{8}{3}-4x\\ y& =& \frac{8-12x}{3}\end{array}$

Step 3. Determine the value of one variable.

Substitute the above value in the first equation.

$\begin{array}{rcl}3x+3\left(\frac{8-12x}{3}\right)& =& -1\\ 3x+8-12x& =& -1\\ -9x& =& -1-8\\ x& =& \frac{-9}{-9}\\ x& =& 1\end{array}$

Step 4. Determine the value of another variable.

Substitute $x=1$ in $y=\frac{8-12x}{3}$.

$$\begin{gathered} y=\frac{8-12\left(1\right)}{3} \\ y=\frac{8-12}{3} \\ y=-\frac{4}{3} \end{gathered}$$

The solution of the given system of equation is $x=1, y=-\frac{4}{3}; \left(1,-\frac{4}{3}\right)$.