Q.8 - Intermediate Algebra

Question

In the following exercises, for each ordered pair, decide

a) is the ordered pair a solution to the equation? b) is the point on the line?

$y=\frac{1}{3}x+2$

$A: \left(0,2\right); B: \left(3,3\right); C: \left(-3,2\right); D: \left(-6,0\right)$

Step-by-Step Solution

Answer

a) The ordered pairs $\left(0,2\right),\left(3,3\right),\left(-6,0\right)$ are solution to the equation and the ordered pair $\left(-3,2\right)$ is not a solution to the equation.

b) The points $\left(0,2\right),\left(3,3\right),\left(-6,0\right)$ are on the line and the points $\left(-3,2\right)$ is not on the line.

Step 1. Given Information

The given equation of the line is $y=\frac{1}{3}x+2$

Given ordered pairs are $\left(0,2\right),\left(3,3\right),\left(-3,2\right),\left(-6,0\right)$

a) It is to be checked if the ordered pairs are solution to the equation.

b) It is to be checked if the points are on the line.

Part (a) Step 1. Calculation

An ordered pair is a solution to an equation if it satisfies the equation when the values are plugged in.

When $\left(x,y\right)=\left(0,2\right)$:

$$\begin{gathered} y=\frac{1}{3}x+2 \\ 2\stackrel{?}{=}\frac{1}{3}\left(0\right)+2 \\ 2\stackrel{?}{=}0+2 \\ 2=2 \end{gathered}$$

When $\left(x,y\right)=\left(3,3\right)$:

$$\begin{gathered} y=\frac{1}{3}x+2 \\ 3\stackrel{?}{=}\frac{1}{3}\left(3\right)+2 \\ 3\stackrel{?}{=}1+2 \\ 3=3 \end{gathered}$$

When $\left(x,y\right)=\left(-3,2\right)$:

$$\begin{gathered} y=\frac{1}{3}x+2 \\ 2\stackrel{?}{=}\frac{1}{3}\left(-3\right)+2 \\ 2\stackrel{?}{=}-1+2 \\ 2\ne 1 \end{gathered}$$

When $\left(x,y\right)=\left(-6,0\right)$:

$$\begin{gathered} y=\frac{1}{3}x+2 \\ 0\stackrel{?}{=}\frac{1}{3}\left(-6\right)+2 \\ 0\stackrel{?}{=}-2+2 \\ 0=0 \end{gathered}$$

Part (a) Step 2. Conclusion

The ordered pairs $\left(0,2\right),\left(3,3\right),\left(-6,0\right)$ are solution to the equation and the ordered pair $\left(-3,2\right)$ is not a solution to the equation. 

Part (b) Step 1. Plotting the points

Plotting the points on a graph:

Part (b) Step 2. Conclusion

The points $\left(0,2\right),\left(3,3\right),\left(-6,0\right)$ are on the line and the points $\left(-3,2\right)$ is not on the line.