## What are systems of linear equations?

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

and$x+y=5$ are both$2x-y=1$ **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

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

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:

In a system of equations, both equations are *simultaneously true*. In other words, since the first equation tells us that *also* equal to *substitute* for

From here, we can solve the equation

First, let's solve for

Now, we can substitute

The solution

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!

What is the solution

Since

For

Since

Since

What is the solution

Neither equation contains an isolated variable. However, we can isolate

Now, we can plug the expression

For

Since

Since

### Try it!

TRY: Follow the steps for substitution

In the system of equations above, the first equation tells us that

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

Next, we can solve the equation

.

Finally, since

.

## 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:

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

Notice that the *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.

- If two terms have the
*opposite*coefficients like in the system above ( and$-y$ ), we can$y$ *add*the two equations to cancel the terms. - If two terms have the
*same*coefficients, we can*subtract*the two equations to cancel the terms.

From here, we can solve the equation

First, let's solve for

Now, we can substitute

The solution

Sometimes, the system of equations does not have coefficients that readily cancel out. Consider this example:

In this case, we need to find ways to match a pair of coefficients by rewriting one of the equations. There are two ways to do this.

**Option 1:** We can set the system up for eliminating the *through addition* by multiplying both sides of the first equation by

From here, we can add the second equation to eliminate the

First, let's solve for

Now, we can plug

The solution

**Option 2:** We can also set the system up for eliminating the *through subtraction* by multiplying both sides of the second equation by

From here, we can subtract the equation from the first equation to eliminate the

First, let's solve for

Now, we can substitute

The solution

To solve a system of equations using elimination:

- Identify a pair of terms in the system that have both the same variable and coefficients with the same magnitude (ex:
and$2x$ , or$2x$ and$3y$ ). If necessary, rewrite one or both equations so that a pair of terms have both the same variable and coefficients with the same magnitude.$-3y$ - Add or subtract the two equations in the system to eliminate the terms identified in Step 1. This should result in a linear equation with only one variable.
- Solve the linear equation to obtain a value for the variable.
- Now that you have figured out the value of one variable, plug that value into either equation to find the value of the other variable.

### Let's look at some more examples!

What is the solution

The first equation contains the term

Adding each side of the two equations gives us:

Now, we can solve for

Since

Since

What is the solution

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

The system of equation is now:

From here, we can cancel the

Now, we can solve for

Since

Since

### Try it!

TRY: spot the mistake for elimination

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

## 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:
${x}=4y+1$ - You can isolate a variable in a single step:
$-3x{+y}=7$

**Elimination** is sometimes easier when:

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

### Try it!

TRY: Identify multiple ways to solve a system of equations

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

TRY: Identify multiple ways to solve a system of equations

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

### 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
instead of$5y$ ?$y$ - 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 **solution** to a system of linear equations is the point at which the lines representing the linear equations **intersect**.

Two lines in the **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, then they are parallel lines, and they will -intercepts$y$ *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*, then they will -intercept$y$ *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

, 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,

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

To determine the number of solutions a system of linear equations has using slope-intercept form,

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

#### Let's look at an example!

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

Since the first equation is equivalent to the second equation, the two equations describe the same line in the

**The system of equations above has infinitely many solutions.**

### Try it!

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

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

If

Practice: Solve a system of linear equations using elimination

The system of equations above has solution

Practice: Solve a system of linear equations

Which ordered pair

Practice: Determine the condition for no solution

The system of equations above has no solutions. If

## Things to remember

To determine the **number of solutions** a system of linear equations has using slope-intercept form,

- Rewrite both equations in slope-intercept form.
Compare the

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