Finding Solutions to Equations in Two Variables

Learning Outcomes

Determine whether an ordered pair is a solution of an equation
Complete a table of solutions for a linear equation

All the equations we solved so far have been equations with one variable. In almost every case, when we solved the equation we got exactly one solution. The process of solving an equation ended with a statement such as $x=4$ . Then we checked the solution by substituting back into the equation.

Here’s an example of a linear equation in one variable, and its one solution.

$\begin{array}{c}3x+5=17\hfill \\ \\ 3x=12\hfill \\ x=4\hfill \end{array}$

But equations can have more than one variable. Equations with two variables can be written in the general form $Ax+By=C$ . An equation of this form is called a linear equation in two variables.

Linear Equation

An equation of the form $Ax+By=C$ , where $A\text{ and }B$ are not both zero, is called a linear equation in two variables.

Notice that the word “line” is in linear.

Here is an example of a linear equation in two variables, $x$ and $y\text{:}$

$\color{red}{A}x+\color{blue}{B}y=\color{green}{C}$

$x+\color{blue}{4}y=\color{green}{8}$

$\color{red}{A=1},\color{blue}{B=4},\color{green}{C=8}$

Is $y=-5x+1$ a linear equation? It does not appear to be in the form $Ax+By=C$ . But we could rewrite it in this form.

$y=-5x+1$
Add $5x$ to both sides.	$y+5x=-5x+1+5x$
Simplify.	$y+5x=1$
Use the Commutative Property to put it in $Ax+By=C$ .	$\color{red}{A}x+\color{blue}{B}y=C$ $5x+y=1$

By rewriting $y=-5x+1$ as $5x+y=1$ , we can see that it is a linear equation in two variables because it can be written in the form $Ax+By=C$ .

Linear equations in two variables have infinitely many solutions. For every number that is substituted for $x$ , there is a corresponding $y$ value. This pair of values is a solution to the linear equation and is represented by the ordered pair $\left(x,y\right)$ . When we substitute these values of $x$ and $y$ into the equation, the result is a true statement because the value on the left side is equal to the value on the right side.

Solution to a Linear Equation in Two Variables

An ordered pair $\left(x,y\right)$ is a solution to the linear equation $Ax+By=C$ , if the equation is a true statement when the $x\text{-}$ and $y\text{-values}$ of the ordered pair are substituted into the equation.

Example

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

Show Solution

For this problem, you will use the substitution method. Substitute

x = - 2

$x=−2$ and

y = 4

$y=4$ 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 to the equation $4y+5x=3$ .

Answer

$(−2,4)$ is not a solution to 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

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

$x+4y=8$

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

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

$8=8\checkmark$

x = 2, y = - 4

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

$x+4y=8$

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

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

$-14\not=8$

x = - 4, y = 3

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

$x+4y=8$

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

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

$8=8\checkmark$

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

Show Solution

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

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

$y=5x-1$

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

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

$-1=-1\checkmark$

x = 1, y = 4

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

$y=5x-1$

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

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

$4=4\checkmark$

x = - 2, y = - 7

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

$y=5x-1$

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

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

$-7\not=-11$

(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

In the next video you will see more examples of how to determine whether an ordered pair is a solution to a linear equation.

Complete a Table of Solutions to a Linear Equation

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 to a 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 $x$ .

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

$y=5x - 1$
$x$	$y$	$\left(x,y\right)$
$0$	$-1$	$\left(0,-1\right)$
$1$	$4$	$\left(1,4\right)$

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

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

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

$y=5x - 1$
$x$	$y$	$\left(x,y\right)$
$0$	$-1$	$\left(0,-1\right)$
$1$	$4$	$\left(1,4\right)$
$2$	$9$	$\left(2,9\right)$

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 to 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 $5x - 4y=20\text{:}$

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

Show Solution

Solution

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

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

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

Show Solution

Solution

Step 1: Choose any value for one of the variables in the equation.

We can substitute any value we want for

x

$x$ or any value for

y

$y$ .

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

$0$ for

x

$x$ .

Simplify.

Divide both sides by $2$ .

3 x + 2 y = 6

$3x+2y=6$

$3\cdot\color{blue}{0}+2y=6$

$0+2y=6$

$2y=6$

$y=3$

Step 3: Write the solution as an ordered pair.

So, when

x = 0, y = 3

$x=0,y=3$ .

This solution is represented by the ordered pair

(0, 3)

$\left(0,3\right)$ .

Step 4: Check.

Substitute

x = 0, y = 3

$x=\color{blue}{0}, y=\color{red}{3}$ into the equation

3 x + 2 y = 6

$3x+2y=6$

Is the result a true equation?

Yes!

3 x + 2 y = 6

$3x+2y=6$

$3\cdot\color{blue}{0}+2\cdot\color{red}{3}\stackrel{?}{=}6$

$0+6\stackrel{?}{=}6$

$6=6\checkmark$

try it

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

example

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

Show Solution

Solution
To find solutions to $3x+2y=6$ , 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.	$y=\color{red}{0}$ $3x+2y=6$ $3x+2(\color{red}{0})=6$	$y=\color{blue}{1}$ $3x+2y=6$ $3(\color{blue}{1})+2y=6$	$y=\color{red}{-3}$ $3x+2y=6$ $3x+2(\color{red}{-3})=6$
Simplify. Solve.	$3x+0=6$ $3x=6$	$3+2y=6$ $2y=3$	$3x-6=6$ $3x=12$
	$x=2$	$y=\Large\frac{3}{2}$	$x=4$
Write the ordered pair.	$\left(2,0\right)$	$\left(1,\Large\frac{3}{2}\normalsize\right)$	$\left(4,-3\right)$

Check your answers.

(2, 0)

$\left(2,0\right)$

(1, \frac{3}{2})

$\left(1,\Large\frac{3}{2}\normalsize\right)$

(4, - 3)

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

3 x + 2 y = 6

$3x+2y=6$

$3\cdot\color{blue}{2}+2\cdot\color{red}{0}\stackrel{?}{=}6$

$6+0\stackrel{?}{=}6$

$6+6\checkmark$

3 x + 2 y = 6

$3x+2y=6$

$3\cdot\color{blue}{1}+2\cdot\color{red}{\Large\frac{3}{2}}\normalsize\stackrel{?}{=}6$

$3+3\stackrel{?}{=}6$

$6+6\checkmark$

3 x + 2 y = 6

$3x+2y=6$

$3\cdot\color{blue}{4}+2\cdot\color{red}{-3}\stackrel{?}{=}6$

$12+(-6)\stackrel{?}{=}6$

$6+6\checkmark$

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

$3x+2y=6$
$x$	$y$	$\left(x,y\right)$
$0$	$3$	$\left(0,3\right)$
$2$	$0$	$\left(2,0\right)$
$1$	$\Large\frac{3}{2}$	$\left(1,\Large\frac{3}{2}\normalsize\right)$
$4$	$-3$	$\left(4,-3\right)$

try it

Let’s find some solutions to another equation now.

example

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

Show Solution

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 point you find is a solution if it makes the equation true.

Module 15: Graphs