Chapter 2.1, Problem 34E

To determine

To find: the system of equations.

Answer to Problem 34E

  $\begin{array}{l}y=4x-19\\ y=-\frac{1}{4}x-2\end{array}$

Explanation of Solution

Given information:

The solution of system of equations $\left(x,y\right)$ is $\left(4,-3\right)$ .

Slope of first equation $\left(m_1\right)$ is $4$ .

Slope of second equation $\left(m_2\right)$ is $-\frac{1}{4}$ .

Calculation:

Let’s write the equation in slope intercept form that is $y=mx+b$

Set the first equation in slope intercept form as follows:

  $y=4x+b_1$

Put the values of $\left(x,y\right)$ in first equation and solve it as shown below.

  $\begin{array}{c}-3=4\left(4\right)+b_1\\ -3=16+b_1\\ b_1=-3-16\\ =-19\end{array}$

So, the first equation can be rewritten as $y=4x-19$ .

Set the second equation in slope intercept form as follows:

  $y=-\frac{1}{4}x+b_2$

Put the values of $\left(x,y\right)$ in second equation and solve it as shown below.

  $\begin{array}{c}-3=-\frac{1}{4}\left(4\right)+b_2\\ -3=-1+b_2\\ b_2=-3+1\\ =-2\end{array}$

So, the second equation can be rewritten as, $y=-\frac{1}{4}x-2$ .

Therefore the system of equations for different slopes is as follows:

  $\begin{array}{c}y=4x-19\\ y=-\frac{1}{4}x-2\end{array}$