Chapter 7.1, Problem 17E

Solution

To calculate: The solution of the system of equation by algebraically.

  $\begin{array}{c}{x}^2+y^2=9\\ x-3y=-1\end{array}$

The solution of system of equation is $\left(2.72,1.24\right)$, and $\left(-2.92,-0.64\right)$ .

Given Information:

The given system of equation is,

  $\begin{array}{c}{x}^2+y^2=9\\ x-3y=-1\end{array}$

Concept Used:

Substitute one equation value to in other equation and make the one variable equation and solve.

If the general equation is $a{x}^2+bx+c=0$ then root of the equation is defined as,

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Where, $a$, $b$, and $c$ are constant real numbers.

Calculation:

Consider the given system of equations,

  $\begin{array}{c}{x}^2+y^2=9\\ x-3y=-1\end{array}$

Substitute second equation value $x=3y-1$ in the first equation.

  $\begin{array}{c}{\left(3y-1\right)}^2+y^2=9\\ 9y^2-6y+1+y^2=9\\ 10y^2-6y-8=0\\ 5y^2-3y-4=0\end{array}$

Use the quadratic formula to solve the obtained quadratic equation.

  $\begin{array}{c}y=\frac{3\pm \sqrt{{\left(-3\right)}^2-4\left(5\right)\left(-4\right)}}{2\left(5\right)}\\ =\frac{3\pm \sqrt{89}}{10}\\ =\left(\frac{3+\sqrt{89}}{10},\frac{3-\sqrt{89}}{10}\right)\\ \approx \left(1.24,-0.64\right)\end{array}$

Now, find the value of $x$ by substituting $1.24$ for $y$ in the equation $x=3y-1$ .

  $\begin{array}{c}x=3\left(1.24\right)-1\\ \approx 2.72\end{array}$

Find the value of $x$ by substituting $-0.64$ for $y$ in the equation $x=3y-1$ .

  $\begin{array}{c}x=3\left(-0.64\right)-1\\ \approx -2.92\end{array}$

Now, to support the solution, use the graphing calculator to draw the graph of the system.

  

Hence, the required solution is $\left(2.72,1.24\right)$ and $\left(-2.92,-0.64\right)$ .