Chapter 5.4, Problem 49E

To determine

To solve: The given non-linear system. Also, justify the answer with a graph.

Answer to Problem 49E

The solutions are: $(0,-2)$ and $\text{(2,0)}$ .

Explanation of Solution

Given information:

The following equations:

  $\begin{array}{l}{x}^2+y^2=4\\ y=x-2\end{array}$

Formula used:

The following expansion formula:

  ${(a-b)}^2=a^2-2ab+b^2$

Calculation:

Substituting the value $y=x-2$ in the equation $x^2+y^2=4$ ,

  $\begin{array}{l}{x}^2+y^2=4[\text{Given]}\\ $ x^2+{(x-2)}^2=4[\text{Substitute (}x-2\text{) for }y\text{]}$x^2+x^2-4x+4=4[\text{Expand]}\\ \text{ 2}{x}^2-4x+4=4[\text{Add like terms]}\\ \text{ 2}{x}^2-4x=0[\text{Subtract 4 from both sides]}\\ 2x(x-2)=0[\text{Factor]}\\ \text{2}x=0x-2=0[\text{Set each factor to 0]}\\ x=0x=2[\text{Simplify]}\end{array}$

Substituting $x=0$ in the equation $y=x-2$ ,

  $\begin{array}{l}y=x-2[\text{Given]}\\ y=0-2\text{[Substitute 0 for }x]\\ y=-2[\text{Simplify]}\end{array}$

Substituting $x=2$ in the equation $y=x-2$ ,

  $\begin{array}{l}y=x-2[\text{Given]}\\ y=2-2\text{[Substitute 2 for }x]\\ y=0[\text{Simplify]}\end{array}$

Therefore, the solutions obtained are, $(0,-2)$ and $\text{(2,0)}$ .

It is verified on the graph as follows:

Graph:

Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Enter the equations $x^2+y^2=4$ and $y=x-2$ which is required to graph.

Step 3: Press GRAPH button to graph the function.

The graph is obtained as:

  

Using the $\boxed{\text{INTERSECT}}$ feature on the graphing calculator obtain the points of intersection.

Interpretation :

The points of intersection obtained are the solutions of the equations. Since the solutions obtained by graph and by calculation are the same, the solutions are verified.