Chapter 10.8, Problem 36E

To determine

To find: The solutions to the system of equations

Answer to Problem 36E

The solutions of the system of equations are $(2,3.61),(2,-3.61)$

Explanation of Solution

Given:

  $x^2+y^2=17$

  $x^2-2x+y^2=13$

Calculation:

Consider the system of equations

  $x^2+y^2=17$

  $x^2-2x+y^2=13$

Need to solve the system of equations by graphing method

Solve for $y$ in terms of $x$

  $x^2+y^2=17$

  $y^2=-x^2+17$ Subtract $x^2$ on both sides

  $y=\sqrt{-x^2+17}$

The second equation is

  $x^2-2x+y^2=13$

  $y^2=-(x^2-2x)+13$ Subtract $x^2-2x$ on both sides

  $y^2=-x^2+2x+13$

  $y=\sqrt{-x^2+2x+13}$

The system of equation becomes

  $y=\sqrt{-x^2+17}$

  $y=\sqrt{-x^2+2x+13}$

Need to sketch the graph of the equations and find the intersect points

Enter the equations in graphing calculator as follows.

  $\begin{array}{l}Y1=\sqrt{-x^2+17}\\ Y2=-\sqrt{-x^2+17}\\ Y3=\sqrt{-x^2+2x+13}\\ Y4=-\sqrt{-x^2+2x+13}\end{array}$

  

Set the windows as follows.

  

To get the graph of the equations, press GRAPH button

  

From the graph of the system of equations, observe that

The graphs of the equations intersect in quadrant I and quadrant IV

So, there are two intersect points

To round to the two decimal places, set the graphing calculator as follows.

Press MODE then set FLOAT up to $2$ then press ENTER

  

Need to find the intersections, press the ${\text{2}}^{\text{nd}}\text{+TRACE}$ option and click on $5$ .

Use the left and right buttons to confirm the first and second curves by pressing ENTER

  

From the display, the intersect point in the quadrant I is $(2,3.61)$

Need to find the intersect point in the quadrant IV as above

  

From the display, the point of intersections in quadrant IV is $(2,-3.61)$

Therefore, the intersect points are $(2,3.61),(2,-3.61)$

Conclusion:

The solutions of the system of equations are $(2,3.61),(2,-3.61)$