6.2 Finding y for a Given x

Finding y for a Given x

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Complete a Table of Solutions to a Linear Equation

How do we find an ordered pair that is a solution to a linear equation? One way is to choose a value for $x$ and then solve the equation for $y .$ Or, choose a value for $y$ and then solve for $x .$

We’ll start by looking at some solutions to the equation $y = 5 x - 1$ . We can summarize this information in a table of solutions.

$y = 5 x - 1$
$x$	$y$	$(x, y)$
$0$	$−1$	$open parenthesis 0, −1 close parenthesis$
$1$	$4$	$open parenthesis 1, 4 close parenthesis$

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

	$y = 5 x - 1$

Multiply.	$y = 10 - 1$
Simplify.	$y equals 9$

The ordered pair is a solution to $y = 5 x - 1$ . We will add it to the table.

$y = 5 x - 1$
$x$	$y$	$(x, y)$
$0$	$−1$	$open parenthesis 0, −1 close parenthesis$
$1$	$4$	$open parenthesis 1, 4 close parenthesis$
$2$	$9$	$open parenthesis 2, 9 close parenthesis$

We can find more solutions to the equation by substituting 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 an infinite number of solutions for this equation.

Example 1

Complete the table to find three solutions to the equation $y = 4 x - 2 :$

$y = 4 x - 2$
$x$	$y$	$(x, y)$
$0$
$−1$
$2$

Show solution

Substitute $x = 0, x = −1,$ and $x equals 2$ into $y = 4 x - 2 .$


$y = 4 x - 2$	$y = 4 x - 2$	$y = 4 x - 2$

$y = 0 - 2$	$y = −4 - 2$	$y = 8 - 2$
$y equals −2$	$y equals −6$	$y equals 6$
$open parenthesis 0, −2 close parenthesis$	$open parenthesis −1, −6 close parenthesis$	$open parenthesis 2, 6 close parenthesis$

The results are summarized in the table.

$y = 4 x - 2$
$x$	$y$	$(x, y)$
$0$	$−2$	$open parenthesis 0, −2 close parenthesis$
$−1$	$−6$	$open parenthesis −1, −6 close parenthesis$
$2$	$6$	$open parenthesis 2, 6 close parenthesis$

Find Solutions to Linear Equations in Two Variables

To find a solution to a linear equation, we can choose any number we want to substitute into the equation for either $x$ or $y .$ We could choose $1, 100, 1,000,$ or any other value we want. But it’s a good idea to choose a number that’s easy to work with. We’ll usually choose $0$ as one of our values.

Example 3

Find a solution to the equation $3 x + 2 y = 6 .$

Show solution

Step 1: Choose any value for one of the variables in the equation.		We can substitute any value we want for $x$ or any value for $y .$ Let's pick $x equals 0.$ What is the value of $y$ if $x equals 0$ ?
Step 2: Substitute that value into the equation. Solve for the other variable.	Substitute $0$ for $x .$ Simplify. Divide both sides by 2.
Step 3: Write the solution as an ordered pair.	So, when $x = 0, y = 3 .$	This solution is represented by the ordered pair $open parenthesis 0, 3 close parenthesis.$
Step 4: Check.	Is the result a true equation? Yes!

A linear equation in two variables has infinitely many solutions, and we’ve just found one of them. Let’s find some other solutions to the equation $3 x + 2 y = 6 .$

Example 4

Find three more solutions to the equation $3 x + 2 y = 6 .$

Show solution

To find solutions to $3 x + 2 y = 6,$ choose a value for $x$ or $y .$ Remember, we can choose any value we want for $x$ or $y .$ Here we chose $1$ for $x,$ and $0$ and $−3$ for $y .$

Substitute it into the equation.
Simplify. Solve.
		$y = \frac{3}{2}$
Write the ordered pair.	$open parenthesis 2, 0 close parenthesis$	$(1, \frac{3}{2})$	$open parenthesis 4, −3 close parenthesis$

Check your answers.

$open parenthesis 2, 0 close parenthesis$	$(1, \frac{3}{2})$	$open parenthesis 4, −3 close parenthesis$

So $(2, 0), (1, \frac{3}{2})$ and $open parenthesis 4, −3 close parenthesis$ are all solutions to the equation $3 x + 2 y = 6 .$ In the previous example, we found that $open parenthesis 0, 3 close parenthesis$ is a solution, too. We can list these solutions in a table.

$3 x + 2 y = 6$
$x$	$y$	$(x, y)$
$0$	$3$	$open parenthesis 0, 3 close parenthesis$
$2$	$0$	$open parenthesis 2, 0 close parenthesis$
$1$	$\frac{3}{2}$	$(1, \frac{3}{2})$
$4$	$−3$	$open parenthesis 4, −3 close parenthesis$

Let’s find some solutions to another equation now.

Example 5

Find three solutions to the equation $x - 4 y = 8 .$

Show solution


Choose a value for $x$ or $y .$
Substitute it into the equation.
Solve.
Write the ordered pair.	$open parenthesis 0, −2 close parenthesis$	$open parenthesis 8, 0 close parenthesis$	$open parenthesis 20, 3 close parenthesis$

So $open parenthesis 0, −2 close parenthesis, open parenthesis 8, 0 close parenthesis,$ and $open parenthesis 20, 3 close parenthesis$ are three solutions to the equation $x - 4 y = 8 .$

$x - 4 y = 8$
$x$	$y$	$(x, y)$
$0$	$−2$	$open parenthesis 0, −2 close parenthesis$
$8$	$0$	$open parenthesis 8, 0 close parenthesis$
$20$	$3$	$open parenthesis 20, 3 close parenthesis$

Remember, there are an infinite number of solutions to each linear equation. Any point you find is a solution if it makes the equation true.

Practice Makes Perfect

Find Solutions to Linear Equations in Two Variables

In the following exercises, complete the table to find solutions to each linear equation.

$y = 2 x - 4$

$x$	$y$	$(x, y)$
$−1$
$0$
$2$

Show solution

$x$	$y$	$(x, y)$
$−1$	$−6$	$open parenthesis −1, −6 close parenthesis$
$0$	$−4$	$open parenthesis 0, −4 close parenthesis$
$2$	$0$	$open parenthesis 2, 0 close parenthesis$

$y = 3 x - 1$

$x$	$y$	$(x, y)$
$−1$
$0$
$2$

$y = - x + 5$

$x$	$y$	$(x, y)$
$−2$
$0$
$3$

Show solution

$x$	$y$	$(x, y)$
$−2$	$7$	$open parenthesis −2, 7 close parenthesis$
$0$	$5$	$open parenthesis 0, 5 close parenthesis$
$3$	$2$	$open parenthesis 3, 2 close parenthesis$

$y = \frac{1}{3} x + 1$

$x$	$y$	$(x, y)$
$0$
$3$
$6$

$y = - \frac{3}{2} x - 2$

$x$	$y$	$(x, y)$
$−2$
$0$
$2$

Show solution

$x$	$y$	$(x, y)$
$−2$	$1$	$open parenthesis −2, 1 close parenthesis$
$0$	$−2$	$open parenthesis 0, −2 close parenthesis$
$2$	$−5$	$open parenthesis 2, −5 close parenthesis$

$x + 2 y = 8$

$x$	$y$	$(x, y)$
$0$
$4$
	$0$

Corequisite Skills for Introductory Statistics is a custom course edition compiled for Introductory Statistics corequisite courses from OpenStax Contemporary Mathematics, Elementary Algebra 2e, Intermediate Algebra 2e, Prealgebra 2e. Content by OpenStax is reproduced as published, apart from any corrections recorded for this edition. Sections are renumbered to follow the course outline. The course outline is adapted from Larry Green's Support Course for Elementary Statistics (LibreTexts, https://stats.libretexts.org/Bookshelves/Introductory_Statistics/Support_Course_for_Elementary_Statistics), used under CC BY 4.0 (https://creativecommons.org/licenses/by/4.0/). Lessons 2.4 and 6.7 are written for this edition by XYZ Homework. Every OpenStax lesson is abridged to the subsections and practice exercises this course uses. Lessons are headed with their course titles, not the source section titles. Access the complete originals free at openstax.org. License: CC-BY-NC-SA-4.0.

$y = 6 x + 1$
$x$	$y$	$(x, y)$
$0$	$1$	$open parenthesis 0, 1 close parenthesis$
$1$	$7$	$open parenthesis 1, 7 close parenthesis$
$−2$	$−11$	$open parenthesis −2, −11 close parenthesis$

$5 x - 4 y = 20$
$x$	$y$	$(x, y)$
$0$	$−5$	$open parenthesis 0, −5 close parenthesis$
$4$	$0$	$open parenthesis 4, 0 close parenthesis$
$8$	$5$	$open parenthesis 8, 5 close parenthesis$

$3 x - 4 y = 12$
$x$	$y$	$(x, y)$
$0$	$−3$	$open parenthesis 0, −3 close parenthesis$
$4$	$0$	$open parenthesis 4, 0 close parenthesis$
$−4$	$−6$	$open parenthesis −4, −6 close parenthesis$