Chapter 5.4, Problem 47E

To determine

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

Answer to Problem 47E

The solutions are: $\text{(3,0)}$ and $\text{(4,1)}$ .

Explanation of Solution

Given information:

The following equations:

  $\begin{array}{l}{y}^2=x-3\\ y=x-3\end{array}$

Formula used:

The following expansion formula:

  ${(a-b)}^2=a^2-2ab+b^2$

Calculation:

Substituting the value $y=x-3$ in the equation $y^2=x-3$ ,

  $\begin{array}{l}{y}^2=x-3[\text{Given]}\\ {\left(x-3\right)}^2=x-3[\text{Substitute (}x-3\text{) for }y\text{]}\\ x^2+9-6x=x-3[\text{Expand]}\\ x^2-7x+12=0[\text{Subtract }\left(x-3\right)\text{from both sides]}\\ (x-3)(x-4)=0[\text{Factor]}\\ x-3=0x-4=0[\text{Set each factor to 0]}\\ x=3x=4[\text{Simplify]}\end{array}$

Substituting $x=3$ in the equation $y=x-3$ ,

  $\begin{array}{l}y=x-3[\text{Given]}\\ y=3-3\text{[Substitute 3 for }x]\\ y=0[\text{Simplify]}\end{array}$

Substituting $x=4$ in the equation $y=x-3$ ,

  $\begin{array}{l}y=x-3[\text{Given]}\\ y=4-3\text{[Substitute 4 for }x]\\ y=1[\text{Simplify]}\end{array}$

Therefore, the solutions obtained are, $\text{(3,0)}$ and $\text{(4,1)}$ .

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 $y^2=x-3$ and $y=x-3$ 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.