Solving systems of linear equations | Lesson (article)

What are systems of linear equations?

A system of linear equations is usually a set of two linear equations with two variables.

$x + y = 5$ ‍ and $2 x - y = 1$ ‍ are both linear equations with two variables.
When considered together, they form a system of linear equations.

A linear equation with two variables has an infinite number of solutions (for example, consider how $(0, 5)$ ‍, $(1, 4)$ ‍, $(2, 3)$ ‍, etc. are all solutions to the equation $x + y = 5$ ‍). However, systems of two linear equations with two variables can have a single solution that satisfies both solutions.

$(2, 3)$ ‍ is the only solution to both $x + y = 5$ ‍ and $2 x - y = 1$ ‍.

In this lesson, we'll:

Look at two ways to solve systems of linear equations algebraically: substitution and elimination.
Look at systems of linear equations graphically to help us understand when systems of linear equations have one solution, no solutions, or infinitely many solutions.
Explore algebraic methods of identifying the number of solutions that exist for systems with two linear equations.

You can learn anything. Let's do this!

How do I solve systems of linear equations by substitution?

Systems of equations with substitution

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Systems of equations with substitution: 2y=x+7 & x=y-4

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How does substitution work?

Our goal when solving a system of equations is to reduce two equations with two variables down to a single equation with one variable. Since each equation in the system has two variables, one way to reduce the number of variables in an equation is to substitute an expression for a variable.

Consider the following example:

$\begin{aligned} x & = 2 y \\ x + y & = 3 \end{aligned}$ ‍

In a system of equations, both equations are simultaneously true. In other words, since the first equation tells us that $x$ ‍ is equal to $2 y$ ‍, the $x$ ‍ in the second equation is also equal to $2 y$ ‍. Therefore, we can plug in $2 y$ ‍ as a substitute for $x$ ‍ in the second equation:

$\begin{aligned} x + y & = 3 \\ (2 y) + y & = 3 \\ 3 y & = 3 \end{aligned}$ ‍

From here, we can solve the equation $3 y = 3$ ‍, then use the value of $y$ ‍ to calculate $x$ ‍.

First, let's solve for $y$ ‍:

$\begin{aligned} 3 y & = 3 \\ \frac{3 y}{3} & = \frac{3}{3} \\ y & = 1 \end{aligned}$ ‍

Now, we can substitute $1$ ‍ for $y$ ‍ into either equation in the system. Using $x = 2 y$ ‍:

$\begin{aligned} x & = 2 y \\ x & = 2 (1) \\ x & = 2 \end{aligned}$ ‍

The solution $(x, y)$ ‍ to the system is $(2, 1)$ ‍.

To solve a system of equations using substitution:

Isolate one of the two variables in one of the equations.
Substitute the expression that is equal to the isolated variable from Step 1 into the other equation. This should result in a linear equation with only one variable.
Solve the linear equation for the remaining variable.
Use the solution of Step 3 to calculate the value of the other variable in the system by using one of the original equations.

Let's look at some more examples!

$\begin{aligned} y & = 3 x - 1 \\ 4 x + y & = - 8 \end{aligned}$ ‍

What is the solution $(x, y)$ ‍ to the system of equations above?

Since $y$ ‍ is already written in terms of $x$ ‍ in the first equation, let's start there! We can plug the expression $3 x - 1$ ‍ into the second equation as a substitute for $y$ ‍, and solve for $x$ ‍. Then, we can use $x$ ‍ to calculate $y$ ‍.

For $y = 3 x - 1$ ‍:

$\begin{aligned} 4 x + y & = - 8 \\ 4 x + (3 x - 1) & = - 8 \\ 4 x + 3 x - 1 & = - 8 \\ 7 x - 1 & = - 8 \\ 7 x - 1 + 1 & = - 8 + 1 \\ 7 x & = - 7 \\ \frac{7 x}{7} & = \frac{- 7}{7} \\ x & = - 1 \end{aligned}$ ‍

Since $x = - 1$ ‍, we can plug $- 1$ ‍ into the first equation for $x$ ‍ to calculate $y$ ‍:

$\begin{aligned} y & = 3 x - 1 \\ = 3 (- 1) - 1 \\ = - 3 - 1 \\ = - 4 \end{aligned}$ ‍

Since $x = - 1$ ‍ and $y = - 4$ ‍, $(x, y) = (- 1, - 4)$ ‍.

$(- 1, - 4)$ ‍ is the solution to the system of equations above.

$\begin{aligned} x - 2 y & = - 2 \\ 7 x - 3 y & = 19 \end{aligned}$ ‍

What is the solution $(x, y)$ ‍ to the system of equations above?

Neither equation contains an isolated variable. However, we can isolate $x$ ‍ in the first equation by adding $2 y$ ‍ to both sides of the equation:

$\begin{aligned} x - 2 y & = - 2 \\ x - 2 y + 2 y & = - 2 + 2 y \\ x & = 2 y - 2 \end{aligned}$ ‍

Now, we can plug the expression $2 y - 2$ ‍ into the second equation as a substitute for $x$ ‍, and solve for $y$ ‍. Then, we can use $y$ ‍ to calculate $x$ ‍.

For $x = 2 y - 2$ ‍:

$\begin{aligned} 7 x - 3 y & = 19 \\ 7 (2 y - 2) - 3 y & = 19 \\ 14 y - 14 - 3 y & = 19 \\ 11 y - 14 & = 19 \\ 11 y - 14 + 14 & = 19 + 14 \\ 11 y & = 33 \\ \frac{11 y}{11} & = \frac{33}{11} \\ y & = 3 \end{aligned}$ ‍

Since $y = 3$ ‍, we can plug $3$ ‍ into the equation $x = 2 y - 2$ ‍ to calculate $x$ ‍:

$\begin{aligned} x & = 2 y - 2 \\ = 2 (3) - 2 \\ = 6 - 2 \\ = 4 \end{aligned}$ ‍

Since $x = 4$ ‍ and $y = 3$ ‍, $(x, y) = (4, 3)$ ‍.

$(4, 3)$ ‍ is the solution to the system of equations above.

Try it!

TRY: Follow the steps for substitution

$\begin{aligned} y & = 2 - x \\ 4 x + y & = 2 \end{aligned}$ ‍

In the system of equations above, the first equation tells us that $y$ ‍ is equal to $2 - x$ ‍. This means we can replace the $y$ ‍ in the second equation with

to reduce the equation to a linear equation with a single variable.

Next, we can solve the equation $3 x + 2 = 2$ ‍ for $x$ ‍. The value of $x$ ‍ is

Finally, since $y = 2 - x$ ‍, we can substitute the value of $x$ ‍ into the equation to calculate $y$ ‍. The value of $y$ ‍ is

How do I solve systems of linear equations by elimination?

System of equations with elimination

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Systems of equations with elimination: x+2y=6 & 4x-2y=14

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How does elimination work?

Our goal when solving a system of equations is to reduce two equations with two variables down to a single equation with one variable. Since each equation in the system has two variables, one way to reduce the number of variables is to add or subtract the two equations in the system to cancel out, or eliminate, one of the variables.

Consider the following system of equations:

$\begin{aligned} 3 x - y & = 7 \\ 2 x + y & = 8 \end{aligned}$ ‍

Recall that when we're solving equations, we can perform the same operations to both sides of the equation and maintain the equality. Since the second equation tells us that $2 x + y$ ‍ is equal to $8$ ‍, we can add $2 x + y$ ‍ to the left side of the first equation, add $8$ ‍ to the right side of the first equation, and maintain the equality:

$\begin{aligned} 3 x - y & = 7 \\ + 2 x + y & = 8 \\ 5 x + 0 & = 15 \end{aligned}$ ‍

Notice that the $y$ ‍-terms cancel out and are eliminated as a result of adding the two equations. When solving systems of equations using elimination, we're always looking for opportunities to cancel out terms.

Let's look at some more examples!

$\begin{aligned} 3 x + 2 y & = 10 \\ 2 x - 2 y & = 5 \end{aligned}$ ‍

What is the solution $(x, y)$ ‍ to the system of equations above?

The first equation contains the term $2 y$ ‍, and the second equation contains the term $- 2 y$ ‍. We can cancel the $y$ ‍-terms by adding the two equations. Then, we can solve for $x$ ‍. Finally, we can use the value of $x$ ‍ to calculate $y$ ‍.

Adding each side of the two equations gives us:

$\begin{aligned} 3 x + 2 y & = 10 \\ + 2 x - 2 y & = 5 \\ 5 x + 0 & = 15 \end{aligned}$ ‍

Now, we can solve for $x$ ‍:

$\begin{aligned} 5 x & = 15 \\ \frac{5 x}{5} & = \frac{15}{5} \\ x & = 3 \end{aligned}$ ‍

Since $x = 3$ ‍, we can plug $3$ ‍ into either equation in the system to calculate $y$ ‍. Using $3 x + 2 y = 10$ ‍:

$\begin{aligned} 3 x + 2 y & = 10 \\ 3 (3) + 2 y & = 10 \\ 9 + 2 y & = 10 \\ 9 + 2 y - 9 & = 10 - 9 \\ 2 y & = 1 \\ \frac{2 y}{2} & = \frac{1}{2} \\ y & = \frac{1}{2} \end{aligned}$ ‍

Since $x = 3$ ‍ and $y = \frac{1}{2}$ ‍, $(x, y) = (3, \frac{1}{2})$ ‍.

$(3, \frac{1}{2})$ ‍ is the solution to the system of equations above.

$\begin{aligned} 3 x + 2 y & = 10 \\ 4 x - y & = 6 \end{aligned}$ ‍

What is the solution $(x, y)$ ‍ to the system of equations above?

At first glance, the two equations in the system do not have any terms that can be readily eliminated. However, notice that the first equation contains the term $+ 2 y$ ‍ and the second equation contains the term $- y$ ‍. If we multiply both sides of the second equation by $2$ ‍, we get:

$\begin{aligned} 4 x - y & = 6 \\ 2 \cdot (4 x - y) & = 2 \cdot 6 \\ 8 x - 2 y & = 12 \end{aligned}$ ‍

The system of equation is now:

$\begin{aligned} 3 x + 2 y & = 10 \\ 8 x - 2 y & = 12 \end{aligned}$ ‍

From here, we can cancel the $y$ ‍-terms by adding the two equations, solve for $x$ ‍, and use the value of $x$ ‍ to calculate $y$ ‍.

$\begin{aligned} 3 x + 2 y & = 10 \\ + 8 x - 2 y & = 12 \\ 11 x + 0 & = 22 \end{aligned}$ ‍

Now, we can solve for $x$ ‍:

$\begin{aligned} 11 x & = 22 \\ \frac{11 x}{11} & = \frac{22}{11} \\ x & = 2 \end{aligned}$ ‍

Since $x = 2$ ‍, we can plug $2$ ‍ into either equation in the system to calculate $y$ ‍. Using $3 x + 2 y = 10$ ‍:

$\begin{aligned} 3 x + 2 y & = 10 \\ 3 (2) + 2 y & = 10 \\ 6 + 2 y & = 10 \\ 6 + 2 y - 6 & = 10 - 6 \\ 2 y & = 4 \\ \frac{2 y}{2} & = \frac{4}{2} \\ y & = 2 \end{aligned}$ ‍

Since $x = 2$ ‍ and $y = 2$ ‍, $(x, y) = (2, 2)$ ‍.

$(2, 2)$ ‍ is the solution to the system of equations above.

Try it!

TRY: spot the mistake for elimination

$\begin{aligned} 5 x + 3 y & = 34 \\ 3 x + 3 y & = 30 \end{aligned}$ ‍

The solution steps to the system of equations above are shown in order below. Select the earliest mistake in the solution steps, if any.

Choose 1 answer:

When do I use substitution, and when do I use elimination?

It's up to you!

All systems of linear equations can be solved with either substitution or elimination. On test day, you should use whichever method you're more comfortable with.

Substitution is sometimes easier when:

A variable is already isolated: $x = 4 y + 1$ ‍
You can isolate a variable in a single step: $- 3 x + y = 7$ ‍

Elimination is sometimes easier when:

Both equations contain an identical term: $2 x + 3 y = 11$ ‍ and $2 x + 7 y = 23$ ‍
The equations contain opposite terms: $2 x + 2 y = 7$ ‍ and $5 x - 2 y = 14$ ‍
An equation contains a term that is an integer multiple of a term in the other equation: $3 x + 4 y = 26$ ‍ and $5 x + 2 y = 20$ ‍.

Try it!

TRY: Identify multiple ways to solve a system of equations

$\begin{aligned} 7 x + y & = 20 \\ 3 x & = y \end{aligned}$ ‍

Consider the system of equations above. Which of the following can be used to solve the system of equations, and how?

Choose 2 answers:

(Choice A)
Substitution: we can plug $3 x$ ‍ into the first equation as a substitute for $y$ ‍ and solve for $x$ ‍.
(Choice B)
Substitution: we can plug $y$ ‍ into the first equation as a substitute for $7 x$ ‍ and solve for $y$ ‍.
(Choice C)
Elimination: we can rewrite the second equation as $3 x + y = 0$ ‍ and subtract the two equations to eliminate the $y$ ‍-terms and solve for $x$ ‍.
(Choice D)
Elimination: we can rewrite the second equation as $3 x - y = 0$ ‍ and add the two equations to eliminate the $y$ ‍-terms and solve for $x$ ‍.

TRY: Identify multiple ways to solve a system of equations

$\begin{aligned} 3 x + 5 y & = 17 \\ 2 x - 10 y & = 38 \end{aligned}$ ‍

Consider the system of equations above. Which of the following can be used to solve the system of equations, and how?

Choose 2 answers:

(Choice A)
Substitution: we can isolate $3 x$ ‍ and plug in $17 - 5 y$ ‍ as a substitute for $x$ ‍ in the second equation to solve for $y$ ‍.
(Choice B)
Substitution: we can isolate $5 y$ ‍, and rewrite the second equation as $2 x - 2 (5 y) = 38$ ‍. Then, we can plug $17 - 3 x$ ‍ in as a substitute for $5 y$ ‍ in the second equation to solve for $x$ ‍.
(Choice C)
Elimination: we can multiply both sides of the first equation by $2$ ‍, then add the two equations to eliminate the $y$ ‍-terms and solve for $x$ ‍.
(Choice D)
Elimination: we can multiply both sides of the second equation by $3$ ‍, then subtract the two equations to eliminate the $x$ ‍-terms and solve for $y$ ‍.

Self-reflection

Even though we can solve the systems of equations above using either substitution or elimination, ask yourself these questions:

For each system, which solution method came to mind first?
How comfortable am I with isolating variables?
How comfortable am I with multiplying both sides of an equation by a constant?
How comfortable am I with adding and subtracting two linear equations?
How comfortable am I with making more complex substitutions, e.g., substituting for $5 y$ ‍ instead of $y$ ‍?
How comfortable am I with finding least common multiples and using them to set up eliminations?

Your answers to these questions should inform which method you use.

What is the relationship between lines and the number of solutions to systems of linear equations?

Linear systems by graphing

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Number of solutions to a system of equations

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Intersections and number of solutions

A linear equation can be represented by a line in the $x y$ ‍-plane. The solution to a system of linear equations is the point at which the lines representing the linear equations intersect.

Two lines in the $x y$ ‍-plane can intersect once, never intersect, or completely overlap. Each of these scenarios corresponds to a different number of solutions to the system of equations the two lines represent.

If the two lines have two different slopes, then they will intersect once. Therefore, the system of equations has exactly one solution.
If the two lines have the same slope but different $y$ ‍-intercepts, then they are parallel lines, and they will never intersect. Therefore, we can say that the system of equations has no solutions.
If the two lines have the same slope and the same $y$ ‍-intercept, then they will completely overlap—they are the same line!. When this is the case, we say that the system has infinitely many solutions.

Try it!

TRY: Determine the number of solutions from a graph

The two lines in the $x y$ ‍-plane above represent a system of linear equations. Because the two lines have

, they

, and the system they represent has

How do I determine the number of solutions for systems of linear equations?

How to determine the number of solutions to a system of equations algebraically

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Number of solutions to a system of equations algebraically

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How do I identify the number of solutions?

In the previous section, we covered the graphical method of determining the number of solutions to a system of linear equations. However, when we don't have the aid of a graph, we can determine the number of solutions algebraically.

One way to do it is to rewrite both equations in slope-intercept form, $y = m x + b$ ‍. This allows us to compare the slopes of the lines, $m$ ‍, and their $y$ ‍-intercepts, $b$ ‍, to determine the number of solutions.

If the two equations have different $m$ ‍-values, then the system has one solution.
If the two equations have the same $m$ ‍-value but different $b$ ‍-values, then the system has no solution.
If the two equations have both the same $m$ ‍-value and the same $b$ ‍-value, then the system has infinitely many solutions.

To determine the number of solutions a system of linear equations has using slope-intercept form, $y = m x + b$ ‍:

Rewrite both equations in slope-intercept form.
Compare the $m$ ‍- and $b$ ‍-values of the equations to determine the number of solutions.

Let's look at an example!

$\begin{aligned} \frac{1}{3} y & = x + 2 \\ y & = 3 x + 6 \end{aligned}$ ‍

How many solutions does the system of equations above have?

We can rewrite the first equation in slope-intercept form by multiplying both sides of the equation by $3$ ‍:

$\begin{aligned} 3 \cdot \frac{1}{3} y & = 3 \cdot (x + 2) \\ y & = 3 x + 6 \end{aligned}$ ‍

Since the first equation is equivalent to the second equation, the two equations describe the same line in the $x y$ ‍-plane! The two lines completely overlap and share an infinite number of points together.

The system of equations above has infinitely many solutions.

Try it!

TRY: Identify the number of solutions in slope-intercept form

$\begin{aligned} y & = 3 x - 1 \\ y & = - 3 x - 1 \end{aligned}$ ‍

In the system of equations above, the two equations have

and

. Therefore, the system has

Your turn!

Practice: Solve a system of linear equations using substitution

$\begin{aligned} y & = x - 2 \\ 3 x + 5 y & = 30 \end{aligned}$ ‍

If $(x, y)$ ‍ is a solution to the system of equations above, what is the value of $x$ ‍ ?

Choose 1 answer:

(Choice A)
$3$ ‍
(Choice B)
$5$ ‍
(Choice C)
$8$ ‍
(Choice D)
$15$ ‍

Practice: Solve a system of linear equations using elimination

$\begin{aligned} 4 x + 2 y & = 2 \\ 2 x - y & = - 9 \end{aligned}$ ‍

The system of equations above has solution $(x, y)$ ‍. What is the value of $y$ ‍ ?

Practice: Solve a system of linear equations

$\begin{aligned} 2 x - y & = 12 \\ 3 x - 7 y & = 7 \end{aligned}$ ‍

Which ordered pair $(x, y)$ ‍ satisfies the system of equations shown above?

Choose 1 answer:

(Choice A)
$(0, - 1)$ ‍
(Choice B)
$(4, - 4)$ ‍
(Choice C)
$(7, 2)$ ‍
(Choice D)
$(10, 8)$ ‍

Practice: Determine the condition for no solution

$\begin{aligned} a x - 3 y & = 21 \\ b x - 3 y & = 11 \end{aligned}$ ‍

The system of equations above has no solutions. If $a$ ‍ and $b$ ‍ are positive constants, what is the value of $\frac{a}{b}$ ‍ ?

Things to remember

To determine the number of solutions a system of linear equations has using slope-intercept form, $y = m x + b$ ‍:

Rewrite both equations in slope-intercept form.
Compare the $m$ ‍- and $b$ ‍-values of the equations to determine the number of solutions.
If the two equations have different $m$ ‍-values, then the system has one solution.
If the two equations have the same $m$ ‍-value but different $b$ ‍-values, then the system has no solution.
If the two equations have both the same $m$ ‍-value and the same $b$ ‍-value, then the system has infinitely many solutions.

Solving systems of linear equations | Lesson (article) | Khan Academy (2024)

What are systems of linear equations?

How do I solve systems of linear equations by substitution?

Systems of equations with substitution

How does substitution work?

Let's look at some more examples!

Try it!

How do I solve systems of linear equations by elimination?

System of equations with elimination

How does elimination work?

Let's look at some more examples!

Try it!

When do I use substitution, and when do I use elimination?

It's up to you!

Try it!

Self-reflection

What is the relationship between lines and the number of solutions to systems of linear equations?

Linear systems by graphing

Intersections and number of solutions

Try it!

How do I determine the number of solutions for systems of linear equations?

How to determine the number of solutions to a system of equations algebraically

How do I identify the number of solutions?

Let's look at an example!

Try it!

Your turn!

Things to remember

References