Chapter 7.1, Problem 65E

Solution

a.

To calculate: The solution of equation $\frac{x^2}{4}+\frac{y^2}{9}=1$ in the terms for y in terms of x to obtained two function.

The obtained matrix is $y\approx -3.62x+1645.64,$ and $y\approx 22.41x+794.77$ .

Given Information:

The equation is $\frac{x^2}{4}+\frac{y^2}{9}=1$ .

Calculation:

Consider the given equation,

  $\frac{x^2}{4}+\frac{y^2}{9}=1$

Now, solve the equation.

  $\begin{array}{c}\frac{x^2}{4}+\frac{y^2}{9}=1\\ \frac{y^2}{9}=1-\frac{x^2}{4}\\ y^2=\frac{1}{9}\left(1-\frac{x^2}{4}\right)\\ y=\pm \frac{1}{3}\sqrt{\left(1-\frac{x^2}{4}\right)}\end{array}$

So, the functions are defined as,

  $y=\frac{1}{3}\sqrt{\left(1-\frac{x^2}{4}\right)},y=-\frac{1}{3}\sqrt{\left(1-\frac{x^2}{4}\right)}$

Therefore, the obtained functions is $y=\frac{1}{3}\sqrt{\left(1-\frac{x^2}{4}\right)},y=-\frac{1}{3}\sqrt{\left(1-\frac{x^2}{4}\right)}$ .

b.

To calculate: The solution of system of the equation by using graph.

The obtained result is $\left(0.687,0.3131\right),\left(1.259,-0.259\right)$ .

Given Information:

The obtained system of equation is $y=\frac{1}{3}\sqrt{\left(1-\frac{x^2}{4}\right)},y=-\frac{1}{3}\sqrt{\left(1-\frac{x^2}{4}\right)}$ and $x+y=1$ .

Calculation:

Consider the given information,

  $y=\frac{1}{3}\sqrt{\left(1-\frac{x^2}{4}\right)},y=-\frac{1}{3}\sqrt{\left(1-\frac{x^2}{4}\right)}$ and $x+y=1$ .

Use the graphing calculator to graph system and find the solution.

  

The obtained solution is $\left(0.687,0.3131\right),\left(1.259,-0.259\right)$ .

Hence, the obtained result is $\left(0.687,0.3131\right),\left(1.259,-0.259\right)$ .

c.

To calculate: The solution of system of the equation by substitution method.

The obtained result is $\left(0.687,0.3131\right),\left(1.259,-0.259\right)$ .

Given Information:

The obtained system of equation is $y=\frac{1}{3}\sqrt{\left(1-\frac{x^2}{4}\right)},y=-\frac{1}{3}\sqrt{\left(1-\frac{x^2}{4}\right)}$ and $x+y=1$ .

Calculation:

Consider the given information,

  $y=\frac{1}{3}\sqrt{\left(1-\frac{x^2}{4}\right)},y=-\frac{1}{3}\sqrt{\left(1-\frac{x^2}{4}\right)}$ and $x+y=1$ .

Substitute $y=1-x$ in the equation $y=\frac{1}{3}\sqrt{\left(1-\frac{x^2}{4}\right)}$ .

  $\begin{array}{c}1-x=\frac{1}{3}\sqrt{\left(1-\frac{x^2}{4}\right)}\\ {\left(1-x\right)}^2={\left(\frac{1}{3}\sqrt{\left(1-\frac{x^2}{4}\right)}\right)}^2\\ 1+x^2-2x=\frac{1}{9}\left(1-\frac{x^2}{4}\right)\\ 37x^2-72x+32=0\end{array}$

Use the quadratic formula to find the solution.

  $\begin{array}{c}37x^2-72x+32=0\\ x=\frac{72\pm \sqrt{{\left(72\right)}^2-4\left(37\right)\left(32\right)}}{2\left(37\right)}\\ \frac{72\pm \text{21}\text{.16601}}{74}\\ =\left(1.259,0.68694\right)\end{array}$

From the obtained solution find the value of $y$ .

  $\begin{array}{c}y=1-1.259\\ =0.259\left(x=1.259\right)\end{array}$

And,

  $\begin{array}{c}y=1-0.6869\\ =0.3131\left(x=0.6869\right)\end{array}$

The obtained solution is $\left(0.687,0.3131\right),\left(1.259,-0.259\right)$ .

Hence, the obtained result is $\left(0.687,0.3131\right),\left(1.259,-0.259\right)$ .