Chapter 7.1, Problem 18E

Solution

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

  $\begin{array}{c}{x}^2+y^2=16\\ 4x+7y=13\end{array}$

The solution of system of equation is $\left(-2.39,3.22\right)$ and $\left(3.99,-0.42\right)$ .

Given Information:

The given system of equation is,

  $\begin{array}{c}{x}^2+y^2=16\\ 4x+7y=13\end{array}$

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,

  $\begin{array}{c}{x}^2+y^2=16\\ 4x+7y=13\end{array}$

Solve the second equation for $x$ .

  $\begin{array}{c}4x=13-7y\\ x=\frac{1}{4}\left(13-7y\right)\end{array}$

Substitute second equation value $x=\frac{1}{4}\left(13-7y\right)$ in the first equation.

  $\begin{array}{c}{\left(\frac{13-7y}{4}\right)}^2+y^2=16\\ \frac{169}{16}-\frac{182}{16}y+\frac{49}{16}{y}^2+y^2=0\\ \frac{65}{16}{y}^2-\frac{182}{16}y-\frac{87}{16}=0\\ 65y^2-182y-87=0\end{array}$

Use the quadratic formula to solve the obtained quadratic equation.

  $\begin{array}{c}y=\frac{182\pm \sqrt{{\left(182\right)}^2-4\left(65\right)\left(-87\right)}}{2\left(65\right)}\\ =\frac{182\pm \sqrt{55744}}{130}\\ =\left(\frac{182+\sqrt{55744}}{130},\frac{182-\sqrt{55744}}{130}\right)\\ \approx \left(3.22,-0.42\right)\end{array}$

Now, find the value of $x$ by substituting $3.22$ for $y$ in the equation $x=\frac{1}{4}\left(13-7y\right)$ .

  $\begin{array}{c}x=\frac{1}{4}\left(13-7\left(3.22\right)\right)\\ \approx -2.39\end{array}$

Find the value of $x$ by substituting $-0.42$ for $y$ in the equation $x=\frac{1}{4}\left(13-7y\right)$ .

  $\begin{array}{c}x=\frac{1}{4}\left(13-7\left(-0.42\right)\right)\\ \approx 3.99\end{array}$

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

  

Hence, the required solution is $\left(-2.39,3.22\right)$ and $\left(3.99,-0.42\right)$ .