Chapter 9.5, Problem 29IP

To determine

To graph the equation using slope and y-intercept.

Explanation of Solution

Given information:

The given equation is:

  $\text{3x}+\text{4y}=12$

Graph:

The equation is written as:

  $\begin{array}{l}\text{3x}+\text{4y}=12\\ \text{4y}=-3\text{x}+12\left[\text{Subtracting both sides by 3x}\right]\\ \text{y}=\frac{-\text{3x}}{4}-3\left[\text{Divide both sides by 4}\right]\\ \text{y}=-\frac{3}{4}\text{x}+\left(-3\right)\left[\text{Write the equation in the form y}=\text{mx}+\text{c}\right]\end{array}$

The slope of the equation is $-\frac{3}{4}$ and the y-intercept is $-3$ .

Graph the y-intercept at $\left(0,-3\right)$ .

Using the slope $-\frac{3}{4}$ locate another point on the line.

Move $\text{y}$ down by $\text{3 units}$ and move $\text{x}$ right by $4\text{ units}$ .

The other point is $\left(4,-6\right)$ .

Draw a line through the two points.

The graph is: