Chapter 13.6, Problem 12PSB

(a)

To determine

To find: The solution set for each of the system.

Answer to Problem 12PSB

The required solution set is $\left\{\left(3,-4\right),\left(3,4\right)\right\}$ .

Explanation of Solution

Given:

The given system of equation is,

  $[\begin{array}{l}{x}^2+y^2=25\\ x=3\end{array}$

Calculation:

Consider the given system of equation is,

  $[\begin{array}{l}{x}^2+y^2=25\\ x=3\end{array}$

Substitute second equation in the second equation as,

  $\begin{array}{l}{\left(3\right)}^2+y^2=25\\ y^2=25-9\\ y=\pm 4\end{array}$

Thus, the required solution set is $\left\{\left(3,-4\right),\left(3,4\right)\right\}$ .

(b)

To determine

To find: The solution set for each of the system.

Answer to Problem 12PSB

The required solution set is $\left\{\left(-4,-3\right),\left(-4,3\right),\left(4,-3\right),\left(4,3\right)\right\}$ .

Explanation of Solution

Given:

The given system of equation is,

  $[\begin{array}{l}{x}^2+y^2=25\\ x^2-y^2=7\end{array}$

Calculation:

Consider the given system of equation is,

  $[\begin{array}{l}{x}^2+y^2=25\\ x^2-y^2=7\end{array}$

Add both the equations as,

  $\begin{array}{l}{x}^2-y^2+x^2+y^2=7+25\\ 2x^2=32\\ x^2=16\\ x=\pm 4\end{array}$

Substitute $x=\pm 4$ in the first equation as,

  $\begin{array}{l}{\left(4\right)}^2+y^2=25\\ y^2=9\\ y=\pm 3\end{array}$

Thus, the required solution set is $\left\{\left(-4,-3\right),\left(-4,3\right),\left(4,-3\right),\left(4,3\right)\right\}$ .

(c)

To determine

To find: The solution set for each of the system.

Answer to Problem 12PSB

The required solution set is $\left(5,3\right)\left(3,5\right)$ .

Explanation of Solution

Given:

The given system of equation is,

  $[\begin{array}{l}{x}^2+y^2=34\\ x+y=8\end{array}$

Calculation:

Consider the given system of equation is,

  $[\begin{array}{l}{x}^2+y^2=34\\ x^2+y^2=8\end{array}$

Consider the second equation as,

  $y=8-x$

Substitute the above value in equation one as,

  $\begin{array}{l}{\left(8-x\right)}^2+y^2=34\\ x^2+64-16x+x^2=34\\ x^2-8x+15=0\\ \left(x-5\right)\left(x-3\right)=0\end{array}$

Then,

  $x=5,3$

Then, the value of $y$ is,

  $\begin{array}{c}y=8-5\\ =3\end{array}$ or $\begin{array}{c}y=8-3\\ =5\end{array}$

Thus, the required solution set is $\left\{\left(5,3\right)\left(3,5\right)\right\}$ .

(d)

To determine

To find: The solution set for each of the system.

Answer to Problem 12PSB

The required solution set is $\left\{\left(8,6\right),\left(8,-6\right),\left(-8,6\right),\left(-8,-6\right)\right\}$ .

Explanation of Solution

Given:

The given system of equation is,

  $[\begin{array}{l}\left|y\right|=6\\ x^2-y^2=7\end{array}$

Calculation:

Consider the given system of equation is,

  $[\begin{array}{l}\left|y\right|=6\\ x^2-y^2=7\end{array}$

Consider the first equation as,

  $y=\pm 6$

Substitute the above value in equation one as,

  $\begin{array}{l}{x}^2-{\left(\pm 6\right)}^2=7\\ x=\pm 8\end{array}$

Thus, the required solution set is $\left\{\left(8,6\right),\left(8,-6\right),\left(-8,6\right),\left(-8,-6\right)\right\}$ .