Chapter 10.8, Problem 35E

To determine

To find: The solutions to the system of equations

Answer to Problem 35E

The solutions are $\text{(-4}\text{.505,2}\text{.168) and (4}\text{.905,-0}\text{.968)}$

Explanation of Solution

Given:

  $x^2+y^2=25$

  $x+3y=2$

Calculation:

Consider the system of equations

  $x^2+y^2=\mathrm{25.....}(1)$

  $x+3y=\mathrm{2.........}(2)$

Need to solve the system of equations by graphing

Clearly the graph of the first equation represents a circle, and the graph of the second equation represents a line

Before graph the equations, we must solve both the equations for $y$

Now solve the first equation for $y$

  $x^2+y^2=25$

  $y^2=25-x^2$ Isolate $y^2$ on LHS

  $y=\pm \sqrt{25-x^2}$ Take square roots

To graph the circle, draw the graphs for both functions

  $y=\sqrt{25-x^2}$ and $y=-\sqrt{25-x^2}$

Solve the second equation for $y$

  $x+3y=2$

  $3y=2-x$ Subtract $x$ from both sides

  $y=\frac{2}{3}-\frac{x}{3}$ Divide by $3$

Now using $TI-84$ calculator

Step 1: Press $\text{Y=}$ key and then enter the functions as $\text{Y1=}\sqrt{\left(25-X^2\right)}$

  $\text{Y2= -}\left(\sqrt{\left(25-X^2\right)}\right);\text{Y3= }\left(2÷3\right)-(x÷3)$

The display is as shown below

  

Step $2$ : set the window settings as follows

  

Step $3$ :Press GRAPHkey to view the graphs of several partial sums.

The display is as shown below

  

Step $4$ : To find the intersections, press the ${\text{2}}^{\text{nd}}\text{+TRACE}$ option and click on $5$ .

Use the up, down, left and right arrows to find the intersections as shown below.

  

From the graphs,

Observe that the point of intersections $\text{(-4}\text{.505,2}\text{.168) and (4}\text{.905,-0}\text{.968)}$

Thus the system has two solutions, rounded to two decimal places are $\text{(-4}\text{.505,2}\text{.168) and (4}\text{.905,-0}\text{.968)}$

Conclusion:

The solutions are $\text{(-4}\text{.505,2}\text{.168) and (4}\text{.905,-0}\text{.968)}$