Question

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

$$\begin{gathered} -2x+3y=3 \\ x+3y=12 \end{gathered}$$

Short answer

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

Step-by-step solution

Step 1. Given information.

Two linear equations are,

$$\begin{gathered} -2x+3y=3 \\ x+3y=12 \end{gathered}$$

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.

$$\begin{gathered} -2x+3y=3 \\ 3y=2x+3 \\ y=\frac{2}{3}\times x+\frac{3}{3} \\ y=\frac{2}{3}\times x+1 \\ \mathrm{Here}\mathrm{the}\mathrm{slope}\mathrm{is}m=\frac{2}{3} \\ \mathrm{And}\mathrm{the}\mathrm{y}-\mathrm{intercept}\mathrm{is}b=1 \end{gathered}$$

$\begin{array}{rcl}-2x+3y& =& 3\\ 3y& =& 3+2x\\ y& =& \frac{3}{3}+\frac{2}{3}x\\ y& =& \frac{2}{3}\times x+1\\ \mathrm{Here}\mathrm{the}\mathrm{slope}\mathrm{is}m& =& \frac{2}{3}\\ \mathrm{And}\mathrm{the}\mathrm{y}-\mathrm{intercept}\mathrm{is}b& =& 1\end{array}$

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

$$\begin{gathered} x+3y=12 \\ 3y=-x+12 \\ y=-\frac{1}{3}\times x+\frac{12}{3} \\ y=-\frac{1}{3}\times x+4 \\ \mathrm{Here}\mathrm{the}\mathrm{slope}\mathrm{is}m=-\frac{1}{3} \\ \mathrm{And}\mathrm{the}\mathrm{y}-\mathrm{intercept}\mathrm{is}b=4 \end{gathered}$$

Step 3. Graph obtained.

Step 4. Checking the equations

$$\begin{gathered} \mathrm{First}\mathrm{substitute}x=3, \\ y=3\mathrm{into}\mathrm{the}\mathrm{equation} \\ -2x+3y=3 \\ -2x+3y=3 \\ -2\times 3+3\left(3\right)=3 \\ -6+9=3 \\ 3=3 \\ \mathrm{This}\mathrm{is}\mathrm{true}. \end{gathered}$$

$$\begin{gathered} \mathrm{First}\mathrm{substitute}x=3, \\ y=3\mathrm{into}\mathrm{the}\mathrm{equation} \\ x+3y=12 \\ x+3y=12 \\ 3+3\left(3\right)=12 \\ 3+9=12 \\ 12=12 \\ \mathrm{This}\mathrm{is}\mathrm{true}. \end{gathered}$$