Q 15 - Intermediate Algebra

Question

In the following exercises, solve the following system of equations by graphing.

$\left\{\begin{array}{l}x+3y=-6\\ y=-\frac{4}{3}x+4\end{array}\right.$

Step-by-Step Solution

Answer

The solution of the system of the linear equation is (6,-4).

Step 1. Given information.

Given the system of linear equations,

$$\begin{gathered} x+3y=-6 \\ y=-\frac{4}{3}x+4 \end{gathered}$$

Step 2. Solve the first equation for y , and find the slope and y-intercept.

Write the equation in slope-intercept form.

$$\begin{gathered} x+3y=-6 \\ 3y=-x-6 \\ y=-\frac{1}{3}x-\frac{6}{3} \\ y=-\frac{1}{3}-2 \end{gathered}$$

Here the slope is $m=-\frac{1}{3}$and the y-intercept is $b= -2$.

Step 3. Solve the second equation for y and the slope and intercept of y.

The equation is given in the standard form. So,

$y=-\frac{4}{3}x+4$

The slope is $m=-\frac{4}{3}$and the y-intercept $b= 4.$

Step 4. Graphs for both the lines.

The obtained graph form the given equation is,

Step 5. Check the solution.

First substitute $x=6,y=-4$into the equation$x+3y=-6$.

$\begin{array}{rcl}x+3y& =& -6\\ 6+3(-4)& =& -6\\ 6-12& =& -6\\ -6& =& -6\end{array}$

This is true.

Also substitute x=6, y=-4 into the equation $y=-\frac{4}{3}x+4$.

$\begin{array}{rcl}-4& =& -\frac{4}{3}\left(6\right)+4\\ -4& =& -8+4\\ -4& =& -4\end{array}$

This is true.

Therefore the solution of the system of the linear equation is (6,-4).