Q 14 - Intermediate Algebra

Question

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

$$\begin{gathered} 2x-y=4 \\ 2x+3y=12 \end{gathered}$$

Step-by-Step Solution

Answer

The solution of the system of linear equations is (3,2).

Step 1. Given information

The linear equations are

$$\begin{gathered} 2x-y=4 \\ 2x+3y=12 \end{gathered}$$

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.

$$\begin{gathered} 2x - y = 4 \\ - y = - 2x + 4 \\ y = 2x - 4 \\ \mathrm{Here} \mathrm{the} \mathrm{slope} \mathrm{is} m = 2 \\ \mathrm{And} \mathrm{the} \mathrm{y}-\mathrm{intercept} \mathrm{is} b = - 4 \end{gathered}$$

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

$$\begin{gathered} 2x + 3y = 12 \\ 3y = - 2x + 12 \\ y = -\frac{2}{3} \times x +\frac{12}{3} \\ y = -\frac{2}{3} x + 4 \\ \mathrm{Here} \mathrm{the} \mathrm{slope} \mathrm{is} m = -\frac{2}{3} \\ \mathrm{And} \mathrm{the} \mathrm{y}-\mathrm{intercept} \mathrm{is} b = 4 \end{gathered}$$

Step 3. Graph obtained

The solution of the system of linear equations is (3,2).

Step 4. Checking the equations

$$\begin{gathered} \mathrm{First} \mathrm{substitute} x = 3, \\ y = 2 \mathrm{into} \mathrm{the} \mathrm{equation} \\ 2x-y=4 \\ 2\left(3\right)-2=4 \\ 6-2=4 \\ 4=4 \\ \mathrm{This} \mathrm{is} \mathrm{true}. \end{gathered}$$

$$\begin{gathered} \mathrm{Also} \mathrm{substitute} x = 3, \\ y = 2 \mathrm{into} \mathrm{the} \mathrm{equation} \\ 2x+3y=12 \\ 2x+3y=12 \\ 2\left(3\right)+3\left(2\right)=12 \\ 6+6=12 \\ 12=12 \\ \mathrm{This} \mathrm{is} \mathrm{true}. \end{gathered}$$