Q 12 - Intermediate Algebra

Question

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

$$\begin{gathered} -x+3y=3 \\ x+3y=3 \end{gathered}$$

Step-by-Step Solution

Answer

The solution of the system of linear equations is (0,1). 

Step 1. Given information

Two linear equations are 

$$\begin{gathered} -x+3y=3 \\ x+3y=3 \end{gathered}$$

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.

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

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

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

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

Step 3. Graph obtained

$$\begin{gathered} \mathrm{Check}: \\ \mathrm{First} \mathrm{substitute} x = 0, \\ y=1 \mathrm{into} \mathrm{the} \mathrm{equation} \\ - x+3y=3 \\ 0+3=3 \\ 3=3 \\ \mathrm{This} \mathrm{is} \mathrm{true}. \end{gathered}$$

$$\begin{gathered} \mathrm{Also} \mathrm{substitute} x = 0 , \\ y = 1 \mathrm{into} \mathrm{the} \mathrm{equation} \\ x + 3y = 3 \\ x + 3y = 3 \\ 0+3\left(1\right)=3 \\ 0+3=3 \\ 3=3 \\ \mathrm{This} \mathrm{is} \mathrm{true}. \end{gathered}$$

The solution of the system of linear equations is (0,1) 

$$\begin{gathered} \mathrm{Check}: \\ \mathrm{First} \mathrm{substitute} x = 0, \\ y=1 \mathrm{into} \mathrm{the} \mathrm{equation} \\ x+y=-4 \\ -2-2=-4 \\ -4=4 \\ \mathrm{This} \mathrm{is} \mathrm{true}. \end{gathered}$$