7.3 Graphing linear equations in two variables  (Page 2/3)

$y-2x=-3$

To find the $y\text{-intercept}$ , let $x=0$ and $y=b$ .

$\begin{array}{rrr}\hfill b-2\left(0\right)& \hfill =& \hfill -3\\ \hfill b-0& \hfill =& \hfill -3\\ \hfill b& \hfill =& \hfill -3\end{array}$

Thus, we have the point $\left(0,-3\right)$ . So, if $x=0$ , $y=b=-3$ .

To find the $x\text{-intercept}$ , let $y=0$ and $x=a$ .

$$\begin{array}{rrrr}\hfill 0-2a& \hfill =& \hfill -3& \hfill \\ \hfill -2a& \hfill =& \hfill -3& \hfill \text{Divide by -2}\text{.}\\ \hfill a& \hfill =& \hfill \frac{-3}{-2}& \hfill \\ \hfill a& \hfill =& \hfill \frac{3}{2}& \hfill \end{array}$$

Thus, we have the point $\left(\frac{3}{2},0\right)$ . So, if $x=a=\frac{3}{2}$ , $y=0$ .

Construct a coordinate system, plot these two points, and draw a line through them. Keep in mind that every point on this line is a solution to the equation $y-2x=-3$ .

$-2x+3y=3$

To find the $y\text{-intercept}$ , let $x=0$ and $y=\mathrm{b}$ .

$\begin{array}{rrr}\hfill -2\left(0\right)+3b& \hfill =& \hfill 3\\ \hfill 0+3b& \hfill =& \hfill 3\\ \hfill 3b& \hfill =& \hfill 3\\ \hfill b& \hfill =& \hfill 1\end{array}$

Thus, we have the point $\left(0,1\right)$ . So, if $x=0$ , $y=b=1$ .

To find the $x\text{-intercept}$ , let $y=0$ and $x=a$ .

$\begin{array}{rrr}\hfill -2a+3\left(0\right)& \hfill =& \hfill 3\\ \hfill -2a+0& \hfill =& \hfill 3\\ \hfill -2a& \hfill =& \hfill 3\\ \hfill a& \hfill =& \hfill \frac{3}{-2}\\ \hfill a& \hfill =& \hfill -\frac{3}{2}\end{array}$

Thus, we have the point $\left(-\frac{3}{2},0\right)$ . So, if $x=a=-\frac{3}{2}$ , $y=0$ .

Construct a coordinate system, plot these two points, and draw a line through them. Keep in mind that all the solutions to the equation $-2x+3y=3$ are precisely on this line.

$4x+y=5$

To find the $y\text{-intercept}$ , let $x=0$ and $y=b$ .

$\begin{array}{rrr}\hfill 4\left(0\right)+b& \hfill =& \hfill 5\\ \hfill 0+b& \hfill =& \hfill 5\\ \hfill b& \hfill =& \hfill 5\end{array}$

Thus, we have the point $\left(0,5\right)$ . So, if $x=0$ , $y=b=5$ .

To find the $x\text{-intercept}$ , let $y=0$ and $x=a$ .

$\begin{array}{rrr}\hfill 4a+0& \hfill =& \hfill 5\\ \hfill 4a& \hfill =& \hfill 5\\ \hfill a& \hfill =& \hfill \frac{5}{4}\end{array}$

Thus, we have the point $\left(\frac{5}{4},0\right)$ . So, if $x=a=\frac{5}{4}$ , $y=0$ .

Construct a coordinate system, plot these two points, and draw a line through them.