4.1 Use the rectangular coordinate system  (Page 4/12)

Complete a table of solutions to a linear equation in two variables

In the examples above, we substituted the x - and y -values of a given ordered pair to determine whether or not it was a solution to a linear equation. But how do you find the ordered pairs if they are not given? It’s easier than you might think—you can just pick a value for $x$ and then solve the equation for $y$ . Or, pick a value for $y$ and then solve for $x$ .

We’ll start by looking at the solutions to the equation $y=5x-1$ that we found in [link] . We can summarize this information in a table of solutions, as shown in [link] .

To find a third solution, we’ll let $x=2$ and solve for $y$ .

The ordered pair $\left(2,9\right)$ is a solution to $y=5x-1$ . We will add it to [link] .

We can find more solutions to the equation by substituting in any value of $x$ or any value of $y$ and solving the resulting equation to get another ordered pair that is a solution. There are infinitely many solutions of this equation.

Find solutions to a linear equation

To find a solution to a linear equation, you really can pick any number you want to substitute into the equation for $x$ or $y.$ But since you’ll need to use that number to solve for the other variable it’s a good idea to choose a number that’s easy to work with.

When the equation is in y -form, with the y by itself on one side of the equation, it is usually easier to choose values of $x$ and then solve for $y$ .

Find three solutions to the equation $y=\mathrm{-3}x+2$ .

Solution

We can substitute any value we want for $x$ or any value for $y$ . Since the equation is in y -form, it will be easier to substitute in values of $x$ . Let’s pick $x=0$ , $x=1$ , and $x=\mathrm{-1}$ .

Substitute the value into the equation.
Simplify.
Simplify.
Write the ordered pair.	(0, 2)	(1, −1)	(−1, 5)
Check.
$y=\mathrm{-3}x+2$	$y=\mathrm{-3}x+2$	$y=\mathrm{-3}x+2$
$2\stackrel{?}{=}\mathrm{-3}\cdot 0+2$	$\mathrm{-1}\stackrel{?}{=}\mathrm{-3}\cdot 1+2$	$5\stackrel{?}{=}\mathrm{-3}\left(\mathrm{-1}\right)+2$
$2\stackrel{?}{=}0+2$	$\mathrm{-1}\stackrel{?}{=}\mathrm{-3}+2$	$5\stackrel{?}{=}3+2$
$2=2✓$	$\mathrm{-1}=\mathrm{-1}✓$	$5=5✓$

So, $\left(0,2\right)$ , $\left(1,\mathrm{-1}\right)$ and $\left(\mathrm{-1},5\right)$ are all solutions to $y=\mathrm{-3}x+2$ . We show them in [link] .

$y=\mathrm{-3}x+2$
$x$	$y$	$(x,y)$
0	2	$\left(0,2\right)$
1	$\mathrm{-1}$	$\left(1,\mathrm{-1}\right)$
$\mathrm{-1}$	5	$\left(\mathrm{-1},5\right)$

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