6.2: Solving Linear Equations in Two Variables

Learning Objectives

Determine whether a given ordered pair is a solution of a given linear equation.
Find solutions of a linear equation.
Complete a table of solutions.

Key words

Ordered pair solution: a solution written in the form $\left (x,y\right )$

Finding Solutions of Linear Equations in Two Variables

When an equation has two variables, any solution will be an ordered pair with a value for each variable.

Solution to a Linear Equation in Two Variables

An ordered pair $\left(x,y\right)$ is a solution of the linear equation $ax+by=c$ , if the equation is a true statement when the $x$ – and $y$ -values of the ordered pair are substituted into the equation.

Example

Determine whether $(−2,4)$ is a solution of the equation $4y+5x=3$ .

Solution

Substitute $x=−2$ and $y=4$ into the equation:

$\begin{array}{r}4y+5x=3\\4\left(4\right)+5\left(−2\right)=3\end{array}$

Evaluate.

$\begin{array}{r}16+\left(−10\right)=3\\6=3\end{array}$

The statement is not true, so $(−2,4)$ is not a solution.

Answer

$(−2,4)$ is not a solution of the equation $4y+5x=3$ .

example

Determine which ordered pairs are solutions of the equation $x+4y=8\text{:}$

1. $\left(0,2\right)$

2. $\left(2,-4\right)$

3. $\left(-4,3\right)$

Solution

Substitute the $x\text{- and }y\text{-values}$ from each ordered pair into the equation and determine if the result is a true statement.

(0, 2)

$\left(0,2\right)$

(2, - 4)

$\left(2,-4\right)$

(- 4, 3)

$\left(-4,3\right)$

x = 0, y = 2

$x=\color{blue}{0}, y=\color{red}{2}$

x + 4 y = 8

$x+4y=8$

$\color{blue}{0}+4\cdot\color{red}{2}\stackrel{?}{=}8$

$0+8\stackrel{?}{=}8$

$8=8\checkmark$

x = 2, y = - 4

$x=\color{blue}{2}, y=\color{red}{-4}$

x + 4 y = 8

$x+4y=8$

$\color{blue}{2}+4(\color{red}{-4})\stackrel{?}{=}8$

$2+(-16)\stackrel{?}{=}8$

$-14\not=8$

x = - 4, y = 3

$x=\color{blue}{-4}, y=\color{red}{3}$

x + 4 y = 8

$x+4y=8$

$\color{blue}{-4}+4\cdot\color{red}{3}\stackrel{?}{=}8$

$-4+12\stackrel{?}{=}8$

$8=8\checkmark$

(0, 2)

$\left(0,2\right)$ is a solution.

(2, - 4)

$\left(2,-4\right)$ is not a solution.

(- 4, 3)

$\left(-4,3\right)$ is a solution.

try it

example

Determine which ordered pairs are solutions of the equation. $y=5x - 1\text{:}$

1. $\left(0,-1\right)$

2. $\left(1,4\right)$

3. $\left(-2,-7\right)$

Solution

Substitute the $x\text{-}$ and $y\text{-values}$ from each ordered pair into the equation and determine if it results in a true statement.

(0, - 1)

$\left(0,-1\right)$

(1, 4)

$\left(1,4\right)$

(- 2, - 7)

$\left(-2,-7\right)$

x = 0, y = - 1

$x=\color{blue}{0}, y=\color{red}{-1}$

y = 5 x - 1

$y=5x-1$

$\color{red}{-1}\stackrel{?}{=}5(\color{blue}{0})-1$

$-1\stackrel{?}{=}0-1$

$-1=-1\checkmark$

x = 1, y = 4

$x=\color{blue}{1}, y=\color{red}{4}$

y = 5 x - 1

$y=5x-1$

$\color{red}{4}\stackrel{?}{=}5(\color{blue}{1})-1$

$4\stackrel{?}{=}5-1$

$4=4\checkmark$

x = - 2, y = - 7

$x=\color{blue}{-2}, y=\color{red}{-7}$

y = 5 x - 1

$y=5x-1$

$\color{red}{-7}\stackrel{?}{=}5(\color{blue}{-2})-1$

$-7\stackrel{?}{=}-10-1$

$-7\not=-11$

(0, - 1)

$\left(0,-1\right)$ is a solution.

(1, 4)

$\left(1,4\right)$ is a solution.

(- 2, - 7)

$\left(-2,-7\right)$ is not a solution.

try it

The video shows more examples of how to determine whether an ordered pair is a solution of a linear equation.

Complete a Table of Solutions

In the previous examples, we substituted the $x\text{- and }y\text{-values}$ of a given ordered pair to determine whether or not it was a solution of a given linear equation. But how do we find the ordered pairs if they are not given? 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 $x$ .

Let’s consider the equation $y=5x - 1$ . The easiest value to choose for $x$ or $y$ is zero:

$\begin{equation}\begin{aligned}y & =5x-1 \;\;\;\;\;\;\;\;\;\;\text{Substitute}\;x=0\\y & = 5(0)-1\\y & = -1\end{aligned}\end{equation}$ So, $x=0,\;y=-1$ is a solution, which as an ordered pair is $\left (0,\,-1\right )$ .

$\begin{equation}\begin{aligned}y & =5x-1 \;\;\;\;\;\;\;\;\;\text{Substitute}\;y=0\\0 & = 5x-1\;\;\;\;\;\;\;\;\;\text{Solve for}\;x\\1 & = 5x\\ \frac{1}{5} & =x\end{aligned}\end{equation}$ So, $x=\frac{1}{5},\;y=0$ is a solution, which as an ordered pair is $\left (\frac{1}{5},\,0\right )$ .

We can continue to find more solutions by choosing different values of $x$ and $y$ .

Suppose $x=2$ :

	$y=5x - 1$
Substitute $x=2$	$y=5(\color{blue}{2})-1$
Multiply.	$y=10 - 1$
Simplify.	$y=9$

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

We can write our solutions in a table:

$y=5x - 1$
$x$	$y$	$\left(x,y\right)$
$0$	$-1$	$\left(0,-1\right)$
$\frac{1}{5}$	$0$	$\left(\frac{1}{5},0\right)$
$2$	$9$	$\left(2,9\right)$

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

Complete the table to find three solutions of the equation $y=4x - 2\text{:}$

$y=4x - 2$
$x$	$y$	$\left(x,y\right)$
$0$
$-1$
$2$

Solution

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

$x=\color{blue}{0}$	$x=\color{blue}{-1}$	$x=\color{blue}{2}$
$y=4x - 2$	$y=4x - 2$	$y=4x - 2$
$y=4\cdot{\color{blue}{0}}-2$	$y=4(\color{blue}{-1})-2$	$y=4\cdot{\color{blue}{2}}-2$
$y=0 - 2$	$y=-4 - 2$	$y=8 - 2$
$y=-2$	$y=-6$	$y=6$
$\left(0,-2\right)$	$\left(-1,-6\right)$	$\left(2,6\right)$

The results are summarized in the table.

$y=4x - 2$
$x$	$y$	$\left(x,y\right)$
$0$	$-2$	$\left(0,-2\right)$
$-1$	$-6$	$\left(-1,-6\right)$
$2$	$6$	$\left(2,6\right)$

try it

example

Complete the table to find three solutions to the equation $5x - 4y=20\text{:}$

$5x - 4y=20$
$x$	$y$	$\left(x,y\right)$
$0$
	$0$
	$5$

Solution

The figure shows three algebraic substitutions into an equation. The first substitution is x = 0, with 0 shown in blue. The next line is 5 x- 4 y = 20. The next line is 5 times 0, shown in blue - 4 y = 20. The next line is 0 - 4 y = 20. The next line is - 4 y = 20. The next line is y = -5. The last line is

The results are summarized in the table.

$5x - 4y=20$
$x$	$y$	$\left(x,y\right)$
$0$	$-5$	$\left(0,-5\right)$
$4$	$0$	$\left(4,0\right)$
$8$	$5$	$\left(8,5\right)$

try it

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 $1,100,-1,000, -\frac{4}{5}, 2.6$ , 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

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

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 $x=0$ . What is the value of $y$ if $x=0$ ?
Step 2: Substitute that value into the equation.Solve for the other variable.	Substitute $0$ for $x$ .Simplify. Divide both sides by $2$ .	$3x+2y=6$ $3\cdot\color{blue}{0}+2y=6$ $0+2y=6$ $2y=6$ $y=3$
Step 3: Write the solution as an ordered pair.	So, when $x=0,y=3$ .	This solution is represented by the ordered pair $\left(0,3\right)$ .
Step 4: Check.	Substitute $x=\color{blue}{0}, y=\color{red}{3}$ into the equation $3x+2y=6$ Is the result a true equation? Yes!	$3x+2y=6$ $3\cdot\color{blue}{0}+2\cdot\color{red}{3}\stackrel{?}{=}6$ $0+6\stackrel{?}{=}6$ $6=6\checkmark$

try it

example

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

Solution

$x-4y=8$	$x-4y=8$	$x-4y=8$
Choose a value for $x$ or $y$ .	$x=\color{blue}{0}$	$y=\color{red}{0}$	$y=\color{red}{3}$
Substitute it into the equation.	$\color{blue}{0}-4y=8$	$x-4\cdot\color{red}{0}=8$	$x-4\cdot\color{red}{3}=8$
Solve.	$-4y=8$ $y=-2$	$x-0=8$ $x=8$	$x-12=8$ $x=20$
Write the ordered pair.	$\left(0,-2\right)$	$\left(8,0\right)$	$\left(20,3\right)$

So $\left(0,-2\right),\left(8,0\right)$ , and $\left(20,3\right)$ are three solutions to the equation $x - 4y=8$ .

$x - 4y=8$
$x$	$y$	$\left(x,y\right)$
$0$	$-2$	$\left(0,-2\right)$
$8$	$0$	$\left(8,0\right)$
$20$	$3$	$\left(20,3\right)$

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

Chapter 6: Linear Equations in Two Variables