Chapter 3.1, Problem 19PPS

To determine

To write:The x- and y-intercepts of the graph of the linear equation.

  

Answer to Problem 19PPS

The required answer is: x-intercept $=3$ and y- intercept $=4$

Explanation of Solution

Given information:

The given equation: $4x+3y=12$

Calculation:

To find the x- intercept, let $y=0$

  $\begin{array}{l}4x+3y=12\\ 4x+3(0)=12\end{array}$

(Replace y with 0)

  $\begin{array}{l}4x=12\\ \frac{4x}{4}=\frac{12}{4}\end{array}$

(Divide by 4 on both sides)

  $x=3$

(Simplify)

The x- intercept is 3, this means the graph intersects the x-axis at (3, 0)

To find the y- intercept,

  $letx=0$

  $\begin{array}{l}4x+3y=12\\ 4(0)+3y=12\end{array}$

(Replace x with 0)

  $\begin{array}{l}3y=12\\ \frac{3y}{3}=\frac{12}{3}\end{array}$

(Divide by 3 on both sides)

  $y=4$

(Simplify)

The y- intercept is 4, this means the graph intersects the y-axis at (0, 4)

Therefore, the x- intercept is 3 and y- intercept is 4 of the graph of the linear equation $4x+3y=12$