Chapter 13, Problem 21RP

(a)

To determine

To find: The intersection of the graph of the equation in each of the system.

Answer to Problem 21RP

The intersection point of the two equation is $\left(2,7\right)$ .

Explanation of Solution

Given:

The given equation is $\{\begin{array}{l}y=4x-1\\ y=2x+3\end{array}$ .

Calculation:

Consider the given equation is,

  $\{\begin{array}{l}y=4x-1\\ y=2x+3\end{array}$

Substitute 2^nd equation in the first one as,

  $\begin{array}{l}y=2x+3\\ 4x+1=2x+3\\ 2x=4\\ x=2\end{array}$

Then, the value of the $y$ is,

  $\begin{array}{c}y=2x+3\\ =2\left(2\right)+3\\ =7\end{array}$

The intersection point of the two equation is $\left(2,7\right)$ .

(b)

To determine

To find: The intersection of the graph of the equation in each of the system.

Answer to Problem 21RP

The intersection point of the two equation is $\left(7,-1\right)$ .

Explanation of Solution

Given:

The given equation is $\{\begin{array}{l}x-3y=10\\ 2x+y=13\end{array}$ .

Calculation:

Consider the given equation is,

  $\{\begin{array}{l}x-3y=10\\ 2x+y=13\end{array}$

Solve the first equation as,

  $\begin{array}{l}x-3y=10\\ x=10+3y\end{array}$

Substitute the above value in the second equation as,

  $\begin{array}{l}2x+y=13\\ 2\left(10+3y\right)+y=13\\ 20+6y+y=13\\ 20+7y=13\end{array}$

Solve further as,

  $\begin{array}{l}20+7y=13\\ 7y=-7\\ y=-1\end{array}$

Then, the value of the $x$ is,

  $\begin{array}{c}x=10+3y\\ =10+3\left(-1\right)\\ =7\end{array}$

The intersection point of the two equation is $\left(7,-1\right)$ .

(c)

To determine

To find: The intersection of the graph of the equation in each of the system.

Answer to Problem 21RP

The intersection point of the two equation is $\left(-3,4\right),\left(5,4\right)$ .

Explanation of Solution

Given:

The given equation is $\{\begin{array}{l}y=4\\ {\left(x-1\right)}^2+{\left(y-5\right)}^2=17\end{array}$ .

Calculation:

Consider the given equation is,

  $\{\begin{array}{l}y=4\\ {\left(x-1\right)}^2+{\left(y-5\right)}^2=17\end{array}$

Substitute the first equation in the second equation as,

  $\begin{array}{l}{\left(x-1\right)}^2+{\left(4-5\right)}^2=17\\ x^2-2x+1+1=17\\ x^2-2x=15\\ \left(x+3\right)\left(x-5\right)=0\end{array}$

Solve further as,

  $x=-3,5$

The intersection point of the two equation is $\left(-3,4\right),\left(5,4\right)$ .