Chapter 13.3, Problem 2PSA

(a)

To determine

To find: The point of intersection of the graphs of each system.

Answer to Problem 2PSA

The point of intersection is $\left(4,\pm 3\right)$ .

Explanation of Solution

Given:

The given system of equation is:

  $\begin{array}{c}x=4\left(1\right)\\ x^2+y^2=25\left(2\right)\end{array}$

Calculation:

The point of intersection is calculated as:

  $\begin{array}{c}{4}^2+y^2=25\left[\text{substitute}x=4\text{in equation}\left(2\right)\right]\\ y^2=9\\ y=\pm 3\end{array}$

Therefore the point of intersection is $\left(4,\pm 3\right)$ .

(b)

To determine

To find: The point of intersection of the graphs of each system.

Answer to Problem 2PSA

The point of intersection is $\left(1,3\right)$ .

Explanation of Solution

Given:

The given system of equation is:

  $\begin{array}{c}y=3\left(1\right)\\ \left|y-2\right|=x\left(2\right)\end{array}$

Calculation:

The point of intersection is calculated as:

  $\begin{array}{c}\left|3-2\right|=x\left[\text{substitute}y=3\text{in equation}\left(2\right)\right]\\ 1=x\end{array}$

Therefore the point of intersection is $\left(1,3\right)$ .