Solve Systems of Linear Equations with Two Variables

In Solving Linear Equations, we learned how to solve linear equations with one variable. Now we will work with two or more linear equations grouped together, which is known as a system of linear equations.

In this section, we will focus our work on systems of two linear equations in two unknowns. We will solve larger systems of equations later in this chapter.

An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations.

has an infinite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line.

To solve a system of two linear equations, we want to find the values of the variables that are solutions to both equations. In other words, we are looking for the ordered pairs

(x, y)

that make both equations true. These are called the solutions of a system of equations.

To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system.

Solve a System of Linear Equations by Graphing

In this section, we will use three methods to solve a system of linear equations. The first method we’ll use is graphing.

The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines. Then we can see all the points that are solutions to each equation. And, by finding what the lines have in common, we’ll find the solution to the system.

Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions.

Similarly, when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown.

Each time we demonstrate a new method, we will use it on the same system of linear equations. At the end of the section you’ll decide which method was the most convenient way to solve this system.

The steps to use to solve a system of linear equations by graphing are shown here.

In the next example, we’ll first re-write the equations into slope–intercept form as this will make it easy for us to quickly graph the lines.

In all the systems of linear equations so far, the lines intersected and the solution was one point. In the next two examples, we’ll look at a system of equations that has no solution and at a system of equations that has an infinite number of solutions.

Sometimes the equations in a system represent the same line. Since every point on the line makes both equations true, there are infinitely many ordered pairs that make both equations true. There are infinitely many solutions to the system.

When we graphed the second line in the last example, we drew it right over the first line. We say the two lines are coincident. Coincident lines have the same slope and same y-intercept.

The systems of equations in [link] and [link] each had two intersecting lines. Each system had one solution.

In [link], the equations gave coincident lines, and so the system had infinitely many solutions.

The systems in those three examples had at least one solution. A system of equations that has at least one solution is called a consistent system.

A system with parallel lines, like [link], has no solution. We call a system of equations like this inconsistent. It has no solution.

We also categorize the equations in a system of equations by calling the equations independent or dependent. If two equations are independent, they each have their own set of solutions. Intersecting lines and parallel lines are independent.

If two equations are dependent, all the solutions of one equation are also solutions of the other equation. When we graph two dependent equations, we get coincident lines.

Let’s sum this up by looking at the graphs of the three types of systems. See below and [link].

Lines	Intersecting	Parallel	Coincident
Number of solutions	1 point	No solution	Infinitely many
Consistent/inconsistent	Consistent	Inconsistent	Consistent
Dependent/ independent	Independent	Independent	Dependent

Without graphing, determine the number of solutions and then classify the system of equations.

ⓐ ${\begin{matrix} y = 3 x - 1 \\ 6 x - 2 y = 12 \end{matrix}$

ⓑ ${\begin{matrix} 2 x + y = - 3 \\ x - 5 y = 5 \end{matrix}$

ⓐ We will compare the slopes and intercepts of the two lines.

\begin{matrix} \begin{matrix} The first equation is already in slope-intercept form. \end{matrix} & \begin{matrix} {\begin{matrix} y & = & 3 x - 1 \\ 6 x - 2 y & = & 12 \end{matrix} \\ y = 3 x - 1 \end{matrix} \end{matrix}

\begin{matrix} \begin{matrix} Write the second equation in slope-intercept form. \\ Find the slope and intercept of each line. \end{matrix} & \begin{matrix} 6 x - 2 y & = & 12 \\ - 2 y & = & −6 x + 12 \\ \frac{−2 y}{−2} & = & \frac{−6 x + 12}{−2} \\ y & = & 3 x - 6 \\ y & = & 3 x - 1 & y & = & 3 x - 6 \\ m & = & 3 & m & = & 3 \\ b & = & −1 & b & = & −6 \end{matrix} \\ \begin{matrix} Since the slopes are the same and y -intercepts are \\ different, the lines are parallel. \end{matrix} \end{matrix}

A system of equations whose graphs are parallel lines has no solution and is inconsistent and independent.

ⓑ We will compare the slope and intercepts of the two lines.

\begin{matrix} {\begin{matrix} 2 x + y & = & −3 \\ x - 5 y & = & 5 \end{matrix} \\ \begin{matrix} Write both equations in slope–intercept form. \end{matrix} & \begin{matrix} 2 x + y & = & - 3 & x - 5 y & = & 5 \\ y & = & −2 x - 3 & −5 y & = & − x + 5 \\ \frac{- 5 y}{- 5} & = & \frac{- x + 5}{- 5} \\ y & = & \frac{1}{5} x - 1 \end{matrix} \\ \begin{matrix} Find the slope and intercept of each line. \end{matrix} & \begin{matrix} y & = & - 2 x - 3 & y & = & \frac{1}{5} x - 1 \\ m & = & - 2 & m & = & \frac{1}{5} \\ b & = & −3 & b & = & −1 \end{matrix} \\ Since the slopes are different, the lines intersect. \end{matrix}

A system of equations whose graphs are intersect has 1 solution and is consistent and independent.

Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between

−10

and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.

Solve a System of Equations by Substitution

We will first solve one of the equations for either x or y. We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.

Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!

After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.

Solve a System of Equations by Elimination

We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression.

The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there.

The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

This time we don’t see a variable that can be immediately eliminated if we add the equations.

we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by

−2 .

Minus 2 x minus 8y is minus 4 and 2 x plus 5y is minus 2.

Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.

Minus 2 x minus 8y is minus 4 and 2 x plus 5y is minus 2. Adding these, we get minus 3y equals minus 6.

Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.

Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution.

Now we’ll do an example where we need to multiply both equations by constants in order to make the coefficients of one variable opposites.

When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by the LCD of all the fractions in the equation.

When we solved the system by graphing, we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.

The same is true using substitution or elimination. If the equation at the end of substitution or elimination is a true statement, we have a consistent but dependent system and the system of equations has infinitely many solutions. If the equation at the end of substitution or elimination is a false statement, we have an inconsistent system and the system of equations has no solution.

Choose the Most Convenient Method to Solve a System of Linear Equations

When you solve a system of linear equations in in an application, you will not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.

Key Concepts

Practice Makes Perfect

Determine Whether an Ordered Pair is a Solution of a System of Equations

In the following exercises, determine if the following points are solutions to the given system of equations.

{\begin{matrix} 2 x - 6 y = 0 \\ 3 x - 4 y = 5 \end{matrix}

ⓐ $(3, 1)$

ⓑ $(−3, 4)$

ⓐ yes ⓑ no

{\begin{matrix} - 3 x + y = 8 \\ - x + 2 y = −9 \end{matrix}

ⓐ $(−5, −7)$

ⓑ $(−5, 7)$

{\begin{matrix} x + y = 2 \\ y = \frac{3}{4} x \end{matrix}

ⓐ $(\frac{8}{7}, \frac{6}{7})$

ⓑ $(1, \frac{3}{4})$

ⓐ yes ⓑ no

{\begin{matrix} 2 x + 3 y = 6 \\ y = \frac{2}{3} x + 2 \end{matrix}

ⓐ $(−6, 2)$

ⓑ $(−3, 4)$

Solve a System of Linear Equations by Graphing

In the following exercises, solve the following systems of equations by graphing.

{\begin{matrix} 3 x + y = −3 \\ 2 x + 3 y = 5 \end{matrix}

(−3, 2)

{\begin{matrix} - x + y = 2 \\ 2 x + y = −4 \end{matrix}

{\begin{matrix} y = x + 2 \\ y = −2 x + 2 \end{matrix}

(0, 2)

{\begin{matrix} y = x - 2 \\ y = −3 x + 2 \end{matrix}

{\begin{matrix} y = \frac{3}{2} x + 1 \\ y = - \frac{1}{2} x + 5 \end{matrix}

(2, 4)

{\begin{matrix} y = \frac{2}{3} x - 2 \\ y = - \frac{1}{3} x - 5 \end{matrix}

{\begin{matrix} x + y = −4 \\ - x + 2 y = −2 \end{matrix}

(−2, 2)

{\begin{matrix} - x + 3 y = 3 \\ x + 3 y = 3 \end{matrix}

{\begin{matrix} - 2 x + 3 y = 3 \\ x + 3 y = 12 \end{matrix}

(3, 3)

{\begin{matrix} 2 x - y = 4 \\ 2 x + 3 y = 12 \end{matrix}

{\begin{matrix} x + 3 y = −6 \\ y = - \frac{4}{3} x + 4 \end{matrix}

(6, −4)

{\begin{matrix} - x + 2 y = −6 \\ y = - \frac{1}{2} x - 1 \end{matrix}

{\begin{matrix} - 2 x + 4 y = 4 \\ y = \frac{1}{2} x \end{matrix}

no solution

{\begin{matrix} 3 x + 5 y = 10 \\ y = - \frac{3}{5} x + 1 \end{matrix}

{\begin{matrix} 4 x - 3 y = 8 \\ 8 x - 6 y = 14 \end{matrix}

no solution

{\begin{matrix} x + 3 y = 4 \\ - 2 x - 6 y = 3 \end{matrix}

{\begin{matrix} x = −3 y + 4 \\ 2 x + 6 y = 8 \end{matrix}

infinite solutions

{\begin{matrix} 4 x = 3 y + 7 \\ 8 x - 6 y = 14 \end{matrix}

{\begin{matrix} 2 x + y = 6 \\ - 8 x - 4 y = −24 \end{matrix}

infinite solutions

{\begin{matrix} 5 x + 2 y = 7 \\ - 10 x - 4 y = −14 \end{matrix}

Without graphing, determine the number of solutions and then classify the system of equations.

{\begin{matrix} y = \frac{2}{3} x + 1 \\ - 2 x + 3 y = 5 \end{matrix}

1 point, consistent and independent

{\begin{matrix} y = \frac{3}{2} x + 1 \\ 2 x - 3 y = 7 \end{matrix}

{\begin{matrix} 5 x + 3 y = 4 \\ 2 x - 3 y = 5 \end{matrix}

1 point, consistent and independent

{\begin{matrix} y = - \frac{1}{2} x + 5 \\ x + 2 y = 10 \end{matrix}

{\begin{matrix} 5 x - 2 y = 10 \\ y = \frac{5}{2} x - 5 \end{matrix}

infinite solutions, consistent, dependent

Solve a System of Equations by Substitution

In the following exercises, solve the systems of equations by substitution.

{\begin{matrix} 2 x + y = −4 \\ 3 x - 2 y = −6 \end{matrix}

{\begin{matrix} 2 x + y = −2 \\ 3 x - y = 7 \end{matrix}

(1, −4)

{\begin{matrix} x - 2 y = −5 \\ 2 x - 3 y = −4 \end{matrix}

{\begin{matrix} x - 3 y = −9 \\ 2 x + 5 y = 4 \end{matrix}

(−3, 2)

{\begin{matrix} 5 x - 2 y = −6 \\ y = 3 x + 3 \end{matrix}

{\begin{matrix} - 2 x + 2 y = 6 \\ y = −3 x + 1 \end{matrix}

(−1 / 2, 5 / 2)

{\begin{matrix} 2 x + 5 y = 1 \\ y = \frac{1}{3} x - 2 \end{matrix}

{\begin{matrix} 3 x + 4 y = 1 \\ y = - \frac{2}{5} x + 2 \end{matrix}

(−5, 4)

{\begin{matrix} 2 x + y = 5 \\ x - 2 y = −15 \end{matrix}

{\begin{matrix} 4 x + y = 10 \\ x - 2 y = −20 \end{matrix}

(0, 10)

{\begin{matrix} y = −2 x - 1 \\ y = - \frac{1}{3} x + 4 \end{matrix}

{\begin{matrix} y = x - 6 \\ y = - \frac{3}{2} x + 4 \end{matrix}

(4, −2)

{\begin{matrix} x = 2 y \\ 4 x - 8 y = 0 \end{matrix}

{\begin{matrix} 2 x - 16 y = 8 \\ - x - 8 y = −4 \end{matrix}

(4, 0)

{\begin{matrix} y = \frac{7}{8} x + 4 \\ - 7 x + 8 y = 6 \end{matrix}

{\begin{matrix} y = - \frac{2}{3} x + 5 \\ 2 x + 3 y = 11 \end{matrix}

none

Solve a System of Equations by Elimination

In the following exercises, solve the systems of equations by elimination.

{\begin{matrix} 5 x + 2 y = 2 \\ - 3 x - y = 0 \end{matrix}

{\begin{matrix} 6 x - 5 y = −1 \\ 2 x + y = 13 \end{matrix}

(6, 1)

{\begin{matrix} 2 x - 5 y = 7 \\ 3 x - y = 17 \end{matrix}

{\begin{matrix} 5 x - 3 y = −1 \\ 2 x - y = 2 \end{matrix}

(2, 3)

{\begin{matrix} 3 x - 5 y = −9 \\ 5 x + 2 y = 16 \end{matrix}

{\begin{matrix} 4 x - 3 y = 3 \\ 2 x + 5 y = −31 \end{matrix}

(−9, 3)

{\begin{matrix} 3 x + 8 y = −3 \\ 2 x + 5 y = −3 \end{matrix}

{\begin{matrix} 11 x + 9 y = −5 \\ 7 x + 5 y = −1 \end{matrix}

(9, 5)

{\begin{matrix} 3 x + 8 y = 67 \\ 5 x + 3 y = 60 \end{matrix}

{\begin{matrix} 2 x + 9 y = −4 \\ 3 x + 13 y = −7 \end{matrix}

(−3, 2)

{\begin{matrix} \frac{1}{3} x - y = −3 \\ x + \frac{5}{2} y = 2 \end{matrix}

{\begin{matrix} x + \frac{1}{2} y = \frac{3}{2} \\ \frac{1}{5} x - \frac{1}{5} y = 3 \end{matrix}

(−15 / 7, 24 / 7)

{\begin{matrix} x + \frac{1}{3} y = −1 \\ \frac{1}{3} x + \frac{1}{2} y = 1 \end{matrix}

{\begin{matrix} \frac{1}{3} x - y = −3 \\ \frac{2}{3} x + \frac{5}{2} y = 3 \end{matrix}

infinitely many

{\begin{matrix} 2 x + y = 3 \\ 6 x + 3 y = 9 \end{matrix}

{\begin{matrix} x - 4 y = −1 \\ - 3 x + 12 y = 3 \end{matrix}

infinitely many

{\begin{matrix} - 3 x - y = 8 \\ 6 x + 2 y = −16 \end{matrix}

{\begin{matrix} 4 x + 3 y = 2 \\ 20 x + 15 y = 10 \end{matrix}

infinitely many

Choose the Most Convenient Method to Solve a System of Linear Equations

In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination.

ⓐ ${\begin{matrix} 8 x - 15 y = −32 \\ 6 x + 3 y = −5 \end{matrix}$

ⓑ ${\begin{matrix} x = 4 y - 3 \\ 4 x - 2 y = −6 \end{matrix}$

ⓐ ${\begin{matrix} y = 7 x - 5 \\ 3 x - 2 y = 16 \end{matrix}$

ⓑ ${\begin{matrix} 12 x - 5 y = −42 \\ 3 x + 7 y = −15 \end{matrix}$

ⓐ substitution ⓑ elimination

ⓐ ${\begin{matrix} y = 4 x + 9 \\ 5 x - 2 y = −21 \end{matrix}$

ⓑ ${\begin{matrix} 9 x - 4 y = 24 \\ 3 x + 5 y = −14 \end{matrix}$

ⓐ ${\begin{matrix} 14 x - 15 y = −30 \\ 7 x + 2 y = 10 \end{matrix}$

ⓑ ${\begin{matrix} x = 9 y - 11 \\ 2 x - 7 y = −27 \end{matrix}$

ⓐ elimination ⓑ substituion

Writing Exercises

In a system of linear equations, the two equations have the same intercepts. Describe the possible solutions to the system.

Solve the system of equations by substitution and explain all your steps in words: ${\begin{matrix} 3 x + y = 12 \\ x = y - 8 \end{matrix} .$

Answers will vary.

Solve the system of equations by elimination and explain all your steps in words: ${\begin{matrix} 5 x + 4 y = 10 \\ 2 x = 3 y + 27 \end{matrix} .$

Solve the system of equations ${\begin{matrix} x + y = 10 \\ x - y = 6 \end{matrix}$

ⓐ by graphing ⓑ by substitution* * *

Answers will vary.

Self Check

After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Who can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.


Solve the first equation for y.
Find the slope and y-intercept.
Solve the second equation for y.
Find the slope and y-intercept.
Graph the lines.
Determine the point of intersection.	The lines intersect at $(−1, 2) .$
Check the solution in both equations.
	The solution is $(−1, 2) .$


To graph the first equation, we will use its slope and y-intercept.
To graph the second equation, we will use the intercepts.

Graph the lines.
Determine the points of intersection.	The lines are parallel. Since no point is on both lines, there is no ordered pair that makes both equations true. There is no solution to this system.


Find the slope and y-intercept of the first equation.
Find the intercepts of the second equation.

Graph the lines.
	The lines are the same! Since every point on the line makes both equations true, there are infinitely many ordered pairs that make both equations true. There are infinitely many solutions to this system.


Solve the first equation for y. Substitute $- 2 x + 2$ for y in the second equation.
Replace the y with $- 2 x + 2 .$
Solve the equation for x.
Substitute $x = \frac{5}{4}$ into $4 x + 2 y = 4$ to find y.
	The ordered pair is $(\frac{5}{4}, - \frac{1}{2}) .$
Check the ordered pair in both equations.
	The solution is $(\frac{5}{4}, - \frac{1}{2}) .$


Both equations are in standard form. To get opposite coefficients of y, we will multiply the first equation by 2 and the second equation by 3.
Simplify.
Add the two equations to eliminate y.
Solve for x.
Substitute $x = 0$ into one of the original equations.
Solve for y.
Write the solution as an ordered pair.	The ordered pair is $(0, −3) .$
Check that the ordered pair is a solution to both original equations.
	The solution is $(0, −3) .$


To clear the fractions, multiply each equation by its LCD.
Simplify.
Now we are ready to eliminate one of the variables. Notice that both equations are in standard form.
We can eliminate $y$ by multiplying the top equation by $- 4 .$
Simplify and add. Substitute $x = 3$ into one of the original equations.
Solve for $y$ .
Write the solution as an ordered pair.	The ordered pair is $(3, 6) .$
Check that the ordered pair is a solution to both original equations.
	The solution is $(3, 6) .$

Solve Systems of Linear Equations with Two Variables

Determine Whether an Ordered Pair is a Solution of a System of Equations

Solve a System of Linear Equations by Graphing

Solve a System of Equations by Substitution

Solve a System of Equations by Elimination

Choose the Most Convenient Method to Solve a System of Linear Equations

Key Concepts

Practice Makes Perfect

Writing Exercises

Self Check

Glossary