How to Simplify, Add, and Subtract Rational Expressions?

High school students meet rational expressions in Algebra 2. Then, the concept appears on standardized tests, and in advanced math courses. It also helps model real-world situations, such as work rates, speed, and electrical resistance.

At first, you may find the rational expressions confusing because they combine fractions, variables, exponents, and polynomials in one problem. But they follow many of the same rules as the fractions we already know.

Today, we'll find out how to simplify, add, and subtract rational expressions step by step.

A Quick Review of Rational Expressions

A rational expression is a fraction where the numerator, the denominator, or both are polynomials instead of numbers. 

A polynomial is an expression made by adding or subtracting terms with numbers, variables, and whole-number exponents, such as:

• x + 3, 

• 2x² − 5x + 1,

• 4x − 7. 

Think of it this way, the same way \(\Large\frac{3}{4}\) is a ratio of two numbers, \(\Large\frac{x + 3}{x - 2}\) is a ratio of two expressions. Since rational expressions are still fractions, we can use the same basic rules we already know for simplifying, adding, and subtracting. 

The difference is that we are working with algebraic expressions instead of numbers alone.

What Is a Restricted Value in Rational Expressions?

A restricted value is the value that would make the denominator of a rational expression equal zero. That means if we substitute the value for x and it would make us divide by zero, we can’t use that value.

Let's identify the restricted value for our example, \(\Large\frac{x + 3}{x - 2}\). To do that, we need to find the x value that would make our denominator x – 2 equal 0:

x – 2 = 0 

We add 2 to both sides of the equation to isolate the variable that gives us x = 2. For this rational expression \(\Large\frac{x + 3}{x - 2}\), the restricted value is 2 because when we substitute 2 for x, our denominator equals 0 and division by zero is undefined.

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How to Simplify a Rational Expression

To simplify a rational expression, we factor the numerator and denominator, then cancel any factors they have in common.

We can think about this the same way we think about reducing a regular fraction. Take \(\Large\frac{8}{12}\):

1. We can rewrite it as: \(\Large\frac{4 × 2}{4 × 3}\)

2. Both the numerator and denominator have a factor of 4, so we can divide out that shared factor, which gives us: \(\Large\frac{8}{12}\) = \(\Large\frac{2}{3}\)

Let's try simplifying a rational: \(\Large\frac{3x + 6}{3x}\) step by step:

Step 1: Factor the Numerator and Denominator (If Needed)

We look for a number that divides evenly into both terms. Both 3x and 6 can be divided by 3, so 3 is our common factor. We divide each term by 3:

• 3x ÷ 3 = x

• 6 ÷ 3 = 2

We factor 3 out and get: 3x + 6 = 3(x + 2).

Now our expression reads \(\Large\frac{3(x + 2)}{3x}\). The denominator 3x is already a single factor (3 times x), so there’s nothing more to factor there. 

Step 2: Find the Restricted Values of the Rational Expression (If Any) 

We should identify the restricted values before cancelling factors so as not to lose the "hidden" restricted values that can disappear during simplification. 

To find the value of x that makes our denominator equal 0, we solve the equation: 3x = 0. x = 0, which means the restricted value for our expression is 0.

Step 3: Cancel the Common Factors 

Since the number 3 appears in both the numerator and denominator, we can cancel it, which leaves us: \(\Large\frac{x + 2}{x}\). 

Notice that we only canceled a factor, a number being multiplied, not a term being added or subtracted. The answer is x + 2x with a restriction x ≠ 0.

Simplifying a Rational Expression: A Solved Example

Now, we’ll put these four steps to work with this expression: \(\Large\frac{(4x^2 + 8x)}{(2x^2 + 6x)}\).

Step 1: Factor the Numerator and Denominator (If Needed)

For the numerator, 4x² + 8x, both terms can be divided by 4x, it is our common factor. So We factor it out by dividing each term by 4x:

• 4x² ÷ 4x = x 

• 8x ÷ 4x = 2 

So 4x² + 8x = 4x(x + 2).

For the denominator, 2x² + 6x, both terms can be divided by 2x. We factor it out by dividing each term by 2x: 

• 2x² ÷ 2x = x 

• 6x ÷ 2x = 3 

So 2x² + 6x = 2x(x + 3).

Our expression now reads: \(\Large\frac{4x(x + 2)}{2x(x + 3)}\).

Step 2: Find the Restricted Values (If Any)

To find the restricted values, we set the denominator equal to 0 and solve: 2x(x+3) = 0.

This means one of two things is true

• 2x = 0, which gives us x = 0

• x+3 = 0, which gives us x = −3

So the restricted values for our expression are 0 and −3.

Step 3: Cancel the Common Factors

Looking at \(\Large\frac{4x(x + 2)}{2x(x + 3)}\), we can see that both the numerator and denominator share a factor of 2x. We cancel that shared factor, which leaves us with \(\Large\frac{2(x + 2)}{(x + 3)}\) with restrictions x ≠ 0, x ≠ –3.

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How to Add and Subtract Rational Expressions

To add or subtract rational expressions, we find the least common denominator, or LCD, of the expressions, rewrite each with that denominator, then add or subtract the numerators, while keeping the denominator the same.

That means we follow similar steps as we would with numeric fractions such as \(\Large\frac{1}{4} + \Large\frac{1}{6}\).

1. First, we find the least common denominator by identifying the smallest number both 4 and 6 divide into evenly, which is 12. 

2. Next, we rewrite each fraction so it has 12 as its denominator. As 4 × 3 = 12, we multiply the top and bottom of \(\Large\frac{1}{4}\) by 3 and get \(\Large\frac{3}{12}\); as 6 × 2 = 12, we multiply the top and bottom of \(\Large\frac{1}{6}\) by 2 and get \(\Large\frac{2}{12}\) . 

3. Finally, we add the fractions. The denominators are the same, so we add the numerators and keep the denominator : \(\Large\frac{3}{12} + \Large\frac{2}{12} = \Large\frac{5}{12}\).

Rational expressions work in a similar way, with a few remarks:

• The LCD of rational expressions is built from polynomials instead of numbers 

• We still need to track the restricted values from each original denominator as we go. 

Let's try \(\Large\frac{1}{x} + \Large\frac{1}{x + 1}\):

Step 1: Identify the Restricted Values (If Any) 

We look at x and (x+1) separately. 

• For the first denominator, x, that value is 0. 

• For the second denominator, (x+1), we solve x+1 = 0 to get x = −1. 

So the restricted values for this expression are 0 and −1.

Step 2: Find the LCD of the Rational Expressions 

As x and (x+1) share no common factors, our LCD is their product: x(x+1).

Step 3: Rewrite Each Rational Expression With the LCD

We rewrite each fraction so it has x(x+1) as its denominator.

The first fraction, \(\Large\frac{1}{x}\), is missing a factor of (x+1), so we multiply its numerator and denominator by (x+1):

• 1 × (x + 1) = (x + 1), the numerator

• x × (x + 1) = x(x +1), the denominator

\(\Large\frac{1}{x}\) becomes \(\Large\frac{(x + 1)}{x(x + 1)}\). 

The second fraction, \(\Large\frac{1}{(x + 1)}\), is missing a factor of x, so we multiply its numerator and denominator by x:

• x × 1 = x, the numerator

• x × (x +1) = x(x + 1), the denominator

We get \(\Large\frac{x}{x(x + 1)}\). 

Step 4: Add or Subtract the Rational Expressions

When the denominators match, we can add the numerators:

 \(\Large\frac{(x + 1)}{x(x + 1)} + \Large\frac{x}{x(x + 1)} = \Large\frac{(x + 1) + x}{x(x + 1)} = \Large\frac{2x + 1}{x(x + 1)}\)

Our answer is \(\Large\frac{2x + 1}{x(x + 1)}\) with restrictions x ≠ 0, x ≠ –1.

Combining Rational Expressions: A Solved Subtraction Example

Now that we've seen the steps in action with addition, let's work through the same process with subtraction. We'll solve \(\Large\frac{x}{(5x + 15)} – \Large\frac{2}{(x + 3)}\).

Step 1: Find the Restricted Values of the Rational Expressions (If Any) 

To identify the restricted values, we look at each original denominator and ask what value of x would make it equal 0. 

For the first denominator, 5x + 15, we solve 5x + 15 = 0, which gives us x = −3. For the second denominator, x + 3, we solve x + 3 = 0 and get x = −3. So the restricted value for this expression is −3.

 Step 2: Find the LCD of the Rational Expressions

We start by factoring 5x + 15. Looking at both terms, 5 divides evenly into both 5x, giving us x, and 15, giving us 3, so 5x+15 breaks down into 5(x+3). 

As (x+3) shows in this factored form and is a denominator of the second expression, our LCD is 5(x+3).

Step 3: Rewrite Each Rational Expression With the LCD 

For the first term, \(\Large\frac{x}{(5x + 15)}\), we just need to write its denominator in the factored form x5(x + 3). 

The second term, \(\Large\frac{2}{(x + 3)}\), is missing a factor of 5, so we multiply its numerator and denominator by 5: \(\Large\frac{2}{(x + 3)}\) becomes \(\Large\frac{10}{5(x + 3)}\).

Step 4: Add or Subtract the Expressions

With both denominators matching, we can subtract the numerators: \(\Large\frac{x - 10}{5(x + 3)}\). Our answer is \(\Large\frac{x - 10}{5(x + 3)}\)) with restrictions x ≠ –3.

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Your Turn: Try Working With Rational Expressions Yourself

Now, try solving these problems on rational expressions on your own. You can check your work at the bottom of the page.

Problem 1: Simplify a Rational Expression

Simplify: \(\Large\frac{(10^2 + 15x)}{(2x^2 + 6x)}\).

Problem 2: Add Two Rational Expressions

Add: \(\Large\frac{3}{(x + 4)} + \Large\frac{2}{(x - 2)}\).

Problem 3: Subtract Two Rational Expressions

Subtract: \(\Large\frac{x}{(4x + 8)} − \Large\frac{1}{(x + 2)}\).

At Mathnasium, tutors use verbal, visual, mental, tactile, and written techniques to help students fully understand how to simplify, add, and subtract rational expressions.

How Mathnasium Helps Students With Rational Expressions (and Any Other Math Topic)

Mathnasium is a math-only learning center dedicated to helping K–12 students of all skill levels excel in math. We work with students to make math topics like rational expressions clear, approachable, and connected to the skills students need next.

Our specially trained tutors use a proprietary teaching approach, the Mathnasium Method™, to meet students where they are and guide them toward math mastery, step by step.

Each student's Mathnasium journey starts with a diagnostic assessment, which lets us identify their current skills and knowledge gaps, including fractions, factoring, or algebra, behind topics like rational expressions.

 From there, we create a personalized learning plan that builds the missing pieces step by step, using the same clear, example-led approach we used today.

Our approach also includes game-based activities and plenty of rewards to keep students motivated and engaged. Students work in a fun and caring group environment where they feel comfortable asking questions, making mistakes, and trying again.

Our tutors give students room to think through a problem before stepping in. They know when to guide, when to ask a better question, and when to let the student work through the problem. That balance helps build critical thinking and problem-solving skills, along with lasting independence they need for algebra and higher-level math.

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For students in and near Highland Park, TX, Mathnasium of Highland Park offers hands-on support with rational expressions, algebra, and any other math topic standing in the way of math confidence.

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Pssst! Check Your Answers Here

Ready to see how you did? Here are the worked solutions to the practice problems above.

Problem 1: Simplify a Rational Expression

Simplify: \(\Large\frac{(10x^2 + 15x)}{(2x^2 + 6x)}\).

1. After factoring \(\Large\frac{(10x^2 + 15x)}{(2x^2 + 6x)}\) becomes \(\Large\frac{5x(2x + 3)}{2x(x + 3)}\). 

2. Before canceling, we find the restricted values. The original denominator, 2x(x+3), equals zero when x = 0 or x = −3, so those are our restricted values.

3. Canceling the shared factor of x leaves us with \(\Large\frac{5(2x + 3)}{2(x + 3)}\). The answer is \(\Large\frac{5(2x + 3)}{2(x + 3)}\) with restrictions x ≠ 0, x ≠ –3.

Problem 2: Add Two Rational Expressions

Add: \(\Large\frac{3}{(x + 4)} + \Large\frac{2}{(x - 2)}\).

1. We solve x + 4 = 0 and x − 2 = 0, which gives us x = −4 and x = 2. Those are our restricted values. 

2. Since (x+4) and (x−2) share no common factors, our LCD is their product: (x+4)(x−2).

3. \(\Large\frac{3}{(x + 4)}\) is missing a factor of (x−2), so it becomes \(\Large\frac{3(x-2)}{(x + 4)(x - 2)} = \Large\frac{3x - 6}{(x + 4)(x - 2)}\). \(\Large\frac{2}{(x - 2)}\) is missing a factor of (x + 4), so it becomes \(\Large\frac{2(x + 4)}{(x+4)(x - 2)} = \Large\frac{2x + 8}{(x + 4)(x - 2)}\). 

4. Now, we can add the expressions. \(\Large\frac{3x - 6}{(x + 4)(x - 2)} + \Large\frac{2x + 8}{(x + 4)(x - 2)} = \Large\frac{3x - 6 + 2x + 8}{(x + 4)(x - 2)} = \Large\frac{(5x + 2)}{(x + 4)(x - 2)}\)The answer is \(\Large\frac{(5x + 2)}{(x + 4)(x - 2)}\) with restrictions x ≠ –4, x ≠ 2. 

Problem 3: Subtract Two Rational Expressions

Subtract: \(\Large\frac{x}{(4x + 8)} − \Large\frac{1}{(x + 2)}\).

1. We solve 4x + 8 = 0 and x + 2 = 0, which both give us x = −2. That's our restricted value.

2. Factoring 4x + 8 gives us 4(x + 2), and since (x + 2) already appears in that form, our LCD is 4(x + 2).

3. \(\Large\frac{x}{(4x + 8)}\) already has a denominator of 4(x+2), so we leave it as is. \(\Large\frac{1}{(x + 2)}\) is missing a factor of 4, so it becomes \(\Large\frac{4}{4(x + 2)} = \Large\frac{4}{4x + 8}\). 

4. We subtract the numerators that give us \(\Large\frac{(x - 4)}{(4x + 8)}\) with a restricted value x ≠ –2.