Chapter 10.8, Problem 5CFU

To determine

To find: the solutions algebraically for the system of equations and then verify the answer by graphing each system. Round to the nearest tenth.

Answer to Problem 5CFU

  $\left(3,3\right)\text{ and }\left(-1,-1\right)$

Explanation of Solution

Given information:

  $\frac{{\left(x-1\right)}^2}{20}+\frac{{\left(y-1\right)}^2}{5}=1$..........(1)

  $x-y=0$..........(2)

Calculation:

Solve equation (2) as follows:

  $\begin{array}{c}x-y=0\\ x=y\left[\text{Add }y\text{ on each side}\right]\mathrm{...}\left(3\right)\end{array}$

Substitute the value of $x$ in equation (1) and solve $y$ as follows:

  $\begin{array}{c}\frac{{\left(y-1\right)}^2}{20}+\frac{{\left(y-1\right)}^2}{5}=1\\ {\left(y-1\right)}^2\left[\frac{1}{20}+\frac{1}{5}\right]=1\left[\text{Taking common terms}\right]\\ {\left(y-1\right)}^2=4\left[\text{Multiplying each side by 4}\right]\\ y-1=\pm 2\left[\text{Taking square root on each side}\right]\\ y=\pm 2+1\left[\text{Add 1 on each side}\right]\\ y=3,-1\end{array}$

Substitute each value of $y$ in equation (3) and solve $x$ as shown below.

  $\begin{array}{c}x=3\left[\text{For }y=3\right]\\ x=-1\left[\text{For }y=-1\right]\end{array}$

Thus, the solutions $\left(x,y\right)$ are $\left(3,3\right)\text{ and }\left(-1,-1\right)$ .

Now use graphing calculator to plot equation (1) and (2) on same screen as shown below.

  

The solution $\left(x,y\right)$ obtained from the graph i.e. $\left(3,3\right)\text{ and }\left(-1,-1\right)$ which are same as the calculated solutions. Thus, the solutions are verified.