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$ax+by=c$
is said to be in general form .
We must stipulate that $a$ and $b$ cannot both equal zero at the same time, for if they were we would have
$0x+0y=c$
$0=c$
This statement is true only if $c=0$ . If $c$ were to be any other number, we would get a false statement.
Now, we have the following:
The graphing of all ordered pairs that solve a linear equation in two variables produces a straight line.
This implies,
The graph of a linear equation in two variables is a straight line.
From these statements we can conclude,
If an ordered pair is a solution to a linear equations in two variables, then it lies on the graph of the equation.
Also,
Any point (ordered pairs) that lies on the graph of a linear equation in two variables is a solution to that equation.
When we want to graph a linear equation, it is certainly impractical to graph infinitely many points. Since a straight line is determined by only two points, we need only find two solutions to the equation (although a third point is helpful as a check).
Graph the following equations using the intercept method.
$y-2x=-3$
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 $\left(0,\text{\hspace{0.17em}}-3\right)$ . 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 $\left(0,\text{\hspace{0.17em}}1\right)$ . 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 $\left(-\frac{3}{2},\text{\hspace{0.17em}}0\right)$ . 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 $\left(0,\text{\hspace{0.17em}}5\right)$ . 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 $\left(\frac{5}{4},\text{\hspace{0.17em}}0\right)$ . So, if $x=a=\frac{5}{4}$ , $y=0$ .
Construct a coordinate system, plot these two points, and draw a line through them.
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