Q 10 - Intermediate Algebra

Question

In the following exercises, solve the following systems of equations by graphing.  

$$\begin{gathered} y=\frac{2}{3} x-2 \\ y=-\frac{1}{3} x-5 \end{gathered}$$

Step-by-Step Solution

Answer

The solution of the system of linear equations is (-3,-4).

Step 1. Given information

Two linear equations are 

$$\begin{gathered} y=\frac{2}{3} x-2 \\ y=-\frac{1}{3} x-5 \end{gathered}$$

Step 2. Solving equations for the intersection points

First, solve both of these equations for y such that their slopes and y intercepts may be easily graphed.

And find the slope and y-intercept by solving the first equation for y.

$$\begin{gathered} y = \frac{2}{3}\times x - 2 \\ \mathrm{Here} \mathrm{the} \mathrm{slope} \mathrm{is} m =\frac{2}{3} \\ \mathrm{And} \mathrm{the} \mathrm{y}-\mathrm{intercept} \mathrm{is} b = - 2 \end{gathered}$$

Again find the slope and y-intercept by solving the Second equation for y. 

$$\begin{gathered} y = -\frac{1}{3}\times x - 5 \\ \mathrm{Here} \mathrm{the} \mathrm{slope} \mathrm{is} m = - 1/3 \\ \mathrm{And} \mathrm{the} \mathrm{y}-\mathrm{intercept} \mathrm{is} b = - 5 \end{gathered}$$

Step 3. Graph obtained

Step 4. Checking the equations

$$\begin{gathered} \mathbf{Check}\mathbf{:} \\ \mathrm{First} \mathrm{substitute} x=-3, \\ y=-4 \mathrm{into} \mathrm{the} \mathrm{equation} \\ y=\frac{2}{3}x-2 \\ -4=\frac{2}{3}(-3)-2 \\ -4=-2-2 \\ -4=4 \\ \mathrm{This} \mathrm{is} \mathrm{true}. \end{gathered}$$

$$\begin{gathered} \mathrm{Also} \mathrm{substitute} x = - 3, \\ y = - 4 \mathrm{into} \mathrm{the} \mathrm{equation} \\ y = - \frac{1}{3} \times x - 5 \\ y = - \frac{1}{3}\times x - 5 \\ - 4 = -\frac{1}{3} \times (- 3) - 5 \\ -4=1-5 \\ -4=4 \\ \mathrm{This} \mathrm{is} \mathrm{true}. \end{gathered}$$