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

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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.

## The intercept method of graphing

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).

## Intercepts

When a linear equation in two variables is given in general from, $ax+by=c$ , often the two most convenient points (solutions) to fine are called the Intercepts: these are the points at which the line intercepts the coordinate axes. Of course, a horizontal or vertical line intercepts only one axis, so this method does not apply. Horizontal and vertical lines are easily recognized as they contain only one variable. (See Sample Set $\text{C}$ .)

## $y\text{-Intercept}$

The point at which the line crosses the $y\text{-axis}$ is called the $y\text{-intercept}$ . The $x\text{-value}$ at this point is zero (since the point is neither to the left nor right of the origin).

## $x\text{-Intercept}$

The point at which the line crosses the $x\text{-axis}$ is called the $x\text{-intercept}$  and the $y\text{-value}$ at that point is zero. The $y\text{-intercept}$ can be found by substituting the value 0 for $x$ into the equation and solving for $y$ . The $x\text{-intercept}$ can be found by substituting the value 0 for $y$ into the equation and solving for $x$ .

## Intercept method

Since we are graphing an equation by finding the intercepts, we call this method the intercept method

## Sample set a

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 $y=b$ .

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

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.

#### Questions & Answers

can someone help me with some logarithmic and exponential equations.
sure. what is your question?
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
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