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Which of the following ordered pairs are solutions to the equation $y=4x-3$ ?
ⓐ $\left(0,3\right)$ ⓑ $\left(1,1\right)$ ⓒ $\left(\mathrm{-1},\mathrm{-1}\right)$
b
Which of the following ordered pairs are solutions to the equation $y=\mathrm{-2}x+6$ ?
ⓐ $\left(0,6\right)$ ⓑ $\left(1,4\right)$ ⓒ $\left(\mathrm{-2},\mathrm{-2}\right)$
a, b
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] .
$y=5x-1$ | ||
$x$ | $y$ | $\left(x,y\right)$ |
0 | $\mathrm{-1}$ | $\left(0,\mathrm{-1}\right)$ |
1 | 4 | $\left(1,4\right)$ |
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] .
$y=5x-1$ | ||
$x$ | $y$ | $\left(x,y\right)$ |
0 | $\mathrm{-1}$ | $\left(0,\mathrm{-1}\right)$ |
1 | 4 | $\left(1,4\right)$ |
2 | 9 | $\left(2,9\right)$ |
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.
Complete [link] to find three solutions to the equation $y=4x-2$ .
$y=4x-2$ | ||
$x$ | $y$ | $\left(x,y\right)$ |
0 | ||
$\mathrm{-1}$ | ||
2 |
Substitute
$x=0$ ,
$x=\mathrm{-1}$ , and
$x=2$ into
$y=4x-2$ .
The results are summarized in [link] .
$y=4x-2$ | ||
$x$ | $y$ | $\left(x,y\right)$ |
0 | $\mathrm{-2}$ | $\left(0,\mathrm{-2}\right)$ |
$\mathrm{-1}$ | $\mathrm{-6}$ | $(\mathrm{-1},\mathrm{-6})$ |
2 | 6 | $\left(2,6\right)$ |
Complete the table to find three solutions to this equation: $y=3x-1$ .
$y=3x-1$ | ||
$x$ | $y$ | $\left(x,y\right)$ |
0 | ||
$\mathrm{-1}$ | ||
2 |
$y=3x-1$ | ||
$x$ | $y$ | $\left(x,y\right)$ |
0 | $\mathrm{-1}$ | $\left(0,\mathrm{-1}\right)$ |
$\mathrm{-1}$ | $\mathrm{-4}$ | $\left(\mathrm{-1},\mathrm{-4}\right)$ |
2 | 5 | $\left(2,5\right)$ |
Complete the table to find three solutions to this equation: $y=6x+1$ .
$y=6x+1$ | ||
$x$ | $y$ | $\left(x,y\right)$ |
0 | ||
1 | ||
$\mathrm{-2}$ |
$y=6x+1$ | ||
$x$ | $y$ | $\left(x,y\right)$ |
0 | 1 | $\left(0,1\right)$ |
1 | 7 | $\left(1,7\right)$ |
$\mathrm{-2}$ | $\mathrm{-11}$ | $\left(\mathrm{-2},\mathrm{-11}\right)$ |
Complete [link] to find three solutions to the equation $5x-4y=20$ .
$5x-4y=20$ | ||
$x$ | $y$ | $\left(x,y\right)$ |
0 | ||
0 | ||
5 |
Substitute the given value into the equation
$5x-4y=20$ and solve for the other variable. Then, fill in the values in the table.
The results are summarized in [link] .
$5x-4y=20$ | ||
$x$ | $y$ | $\left(x,y\right)$ |
0 | $\mathrm{-5}$ | $\left(0,\mathrm{-5}\right)$ |
4 | 0 | $\left(4,0\right)$ |
8 | 5 | $\left(8,5\right)$ |
Complete the table to find three solutions to this equation: $2x-5y=20$ .
$2x-5y=20$ | ||
$x$ | $y$ | $\left(x,y\right)$ |
0 | ||
0 | ||
$\mathrm{-5}$ |
$2x-5y=20$ | ||
$x$ | $y$ | $\left(x,y\right)$ |
0 | $\mathrm{-4}$ | $\left(0,\mathrm{-4}\right)$ |
10 | 0 | $\left(10,0\right)$ |
$\mathrm{-5}$ | $\mathrm{-6}$ | $\left(\mathrm{-5},\mathrm{-6}\right)$ |
Complete the table to find three solutions to this equation: $3x-4y=12$ .
$3x-4y=12$ | ||
$x$ | $y$ | $\left(x,y\right)$ |
0 | ||
0 | ||
$\mathrm{-4}$ |
$3x-4y=12$ | ||
$x$ | $y$ | $\left(x,y\right)$ |
0 | $\mathrm{-3}$ | $\left(0,\mathrm{-3}\right)$ |
4 | 0 | $\left(4,0\right)$ |
$\mathrm{-4}$ | $\mathrm{-6}$ | $\left(\mathrm{-4},\mathrm{-6}\right)$ |
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 $y=\mathrm{-3}x+2$ .
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. | |||||
$\phantom{\rule{0.03em}{0ex}}y=\mathrm{-3}x+2$ | $\phantom{\rule{0.7em}{0ex}}y=\mathrm{-3}x+2$ | $\phantom{\rule{0.04em}{0ex}}y=\mathrm{-3}x+2$ | |||
$2\stackrel{?}{=}\mathrm{-3}\cdot 0+2$ | $\mathrm{-1}\stackrel{?}{=}\mathrm{-3}\cdot 1+2$ | $5\stackrel{?}{=}\mathrm{-3}\left(\mathrm{-1}\right)+2$ | |||
$2\stackrel{?}{=}0+2$ | $\mathrm{-1}\stackrel{?}{=}\mathrm{-3}+2$ | $5\stackrel{?}{=}3+2$ | |||
$2=2\u2713$ | $\mathrm{-1}=\mathrm{-1}\u2713$ | $5=5\u2713$ |
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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