Rational Expressions on the SAT: Rules, Examples and Practice
Rational expressions are algebraic expressions that contain fractions involving variables.
On the Digital SAT, rational expressions are included within the Advanced Math content domain. College Board includes skills such as rewriting simple rational expressions and solving simple rational equations within this area.
The important skills are not about memorizing a large number of formulas.
You need to know how to:
• identify values that make an expression undefined
• factor numerators and denominators
• simplify rational expressions
• add and subtract rational expressions
• multiply and divide rational expressions
• solve rational equations
• check restrictions
• recognize when a canceled factor still creates a restriction
The examples below focus on the algebraic reasoning you need for SAT Math.
Main Ideas to Know
Before working with rational expressions, remember these rules:
A denominator cannot equal zero.
Factor before canceling whenever possible.
You can cancel factors, not individual terms.
Use a common denominator when adding or subtracting rational expressions.
When dividing by a rational expression, multiply by its reciprocal.
Restrictions from the original expression still apply after simplification.
When solving a rational equation, check the answer against the original restrictions.
These ideas cover most of the algebra involved in this topic.
What Is a Rational Expression?
A rational expression is an algebraic expression written as one polynomial divided by another polynomial.
Examples include:
(x + 3)⁄(x − 2)
(2x² − 5x + 1)⁄(x + 4)
6⁄(x² − 9)
The variable may appear in the numerator, the denominator, or both.
Rational expressions follow many of the same rules as ordinary fractions.
You can:
• simplify them
• add them
• subtract them
• multiply them
• divide them
The important difference is that some values of the variable can make the denominator equal to zero.
Those values are not allowed.
The Most Important Rule: A Denominator Cannot Be Zero
Division by zero is undefined.
Therefore, any value that makes a denominator equal to zero must be excluded.
Consider:
5⁄(x − 7)
Set the denominator equal to zero:
x − 7 = 0
Therefore:
x = 7
So the expression is undefined when x = 7.
The restriction is:
x ≠ 7
Another example
Consider:
(x + 1)⁄(x² − 4)
Factor the denominator:
x² − 4 = (x − 2)(x + 2)
The denominator is zero when:
x = 2
or
x = −2
Therefore:
x ≠ 2, −2
These excluded values are called domain restrictions.
A useful SAT habit is to check the denominators before doing complicated algebra.
Simplifying Rational Expressions
A reliable method is:
Factor the numerator.
Factor the denominator.
Find common factors.
Cancel common factors.
Keep the restrictions from the original expression.
Consider:
(x² − 9)⁄(x² + x − 12)
Factor the numerator:
x² − 9 = (x − 3)(x + 3)
Factor the denominator:
x² + x − 12 = (x + 4)(x − 3)
Therefore:
[(x − 3)(x + 3)]⁄[(x + 4)(x − 3)]
Cancel the common factor:
(x + 3)⁄(x + 4)
So the simplified expression is:
(x + 3)⁄(x + 4)
However, the original denominator was:
(x + 4)(x − 3)
Therefore:
x ≠ −4
and
x ≠ 3
The restriction x ≠ 3 still matters even though the factor (x − 3) was canceled.
This is one of the most important ideas in rational expressions.
Why You Cannot Cancel Terms
Consider:
(x + 5)⁄x
You cannot cancel the x.
Why?
Because the numerator is a sum:
x + 5
The x is not a factor of the entire numerator.
Compare that with:
x(x + 5)⁄x
Here, x is a factor of the entire numerator, so it can be canceled:
x + 5
provided that:
x ≠ 0
Remember:
Cancel factors, not terms.
Adding Rational Expressions
When rational expressions have different denominators, you need a common denominator.
Consider:
3⁄x + 2⁄(x + 1)
A common denominator is:
x(x + 1)
Rewrite the first fraction:
3⁄x = 3(x + 1)⁄[x(x + 1)]
Rewrite the second fraction:
2⁄(x + 1) = 2x⁄[x(x + 1)]
Now combine the numerators:
[3(x + 1) + 2x]⁄[x(x + 1)]
Expand:
(3x + 3 + 2x)⁄[x(x + 1)]
Therefore:
(5x + 3)⁄[x(x + 1)]
The original denominators give the restrictions:
x ≠ 0
and
x ≠ −1
Subtracting Rational Expressions
The same common-denominator method works for subtraction.
Consider:
5⁄(x + 2) − 1⁄(x − 3)
A common denominator is:
(x + 2)(x − 3)
Rewrite both fractions:
5(x − 3)⁄[(x + 2)(x − 3)]
and
(x + 2)⁄[(x + 2)(x − 3)]
Now subtract:
[5(x − 3) − (x + 2)]⁄[(x + 2)(x − 3)]
Expand the numerator:
(5x − 15 − x − 2)⁄[(x + 2)(x − 3)]
Therefore:
(4x − 17)⁄[(x + 2)(x − 3)]
The restrictions are:
x ≠ −2
and
x ≠ 3
Be especially careful with the negative sign before the second numerator.
Multiplying Rational Expressions
Multiplication is often easier because you can factor and cancel before multiplying.
Consider:
(x + 2)⁄(x − 1) × (x − 1)⁄(x + 5)
The factor (x − 1) appears in both the numerator and denominator.
Cancel it:
(x + 2)⁄(x + 5)
However, the original expression contained the denominators:
x − 1
and
x + 5
Therefore:
x ≠ 1
and
x ≠ −5
So the simplified result is:
(x + 2)⁄(x + 5)
with the original restrictions still applying.
Dividing Rational Expressions
When dividing rational expressions, multiply by the reciprocal of the second expression.
Consider:
(x² − 4)⁄(x + 3) ÷ (x − 2)⁄(x + 1)
Change division to multiplication:
(x² − 4)⁄(x + 3) × (x + 1)⁄(x − 2)
Factor:
x² − 4 = (x − 2)(x + 2)
Therefore:
[(x − 2)(x + 2)]⁄(x + 3) × (x + 1)⁄(x − 2)
Cancel (x − 2):
(x + 2)(x + 1)⁄(x + 3)
Now consider the restrictions.
From the original denominators:
x + 3 ≠ 0
so:
x ≠ −3
and:
x + 1 ≠ 0
so:
x ≠ −1
There is another restriction.
The expression being divided by is:
(x − 2)⁄(x + 1)
A divisor cannot equal zero.
Therefore:
x − 2 ≠ 0
so:
x ≠ 2
The complete set of restrictions is therefore:
x ≠ −3, −1, 2
This is an important difference between ordinary simplification and division of rational expressions.
Solving Rational Equations
A rational equation contains one or more rational expressions and asks you to find the value of the variable.
Consider:
2⁄(x − 3) = 4⁄(x + 1)
First identify the restrictions:
x ≠ 3
and
x ≠ −1
Now cross-multiply:
2(x + 1) = 4(x − 3)
Expand:
2x + 2 = 4x − 12
Rearrange:
14 = 2x
Therefore:
x = 7
Check the original denominators:
7 − 3 = 4
and:
7 + 1 = 8
Neither denominator is zero.
Therefore:
x = 7
is a valid solution.
When an Algebraic Answer Is Not Allowed
Consider:
x⁄(x − 5) = 5⁄(x − 5)
The denominator tells us:
x ≠ 5
If the numerators are compared, we get:
x = 5
But x = 5 is not allowed because it makes the original denominator zero.
Therefore, the equation has:
no solution
This is why restrictions should be identified before or during the solution process.
Clearing Denominators
For some rational equations, multiplying every term by a common denominator is easier than cross-multiplication.
Consider:
2⁄x + 1⁄(x + 2) = 3
The common denominator is:
x(x + 2)
The restrictions are:
x ≠ 0
and
x ≠ −2
Multiply every term by x(x + 2):
2(x + 2) + x = 3x(x + 2)
Expand:
2x + 4 + x = 3x² + 6x
Combine like terms:
3x + 4 = 3x² + 6x
Rearrange:
3x² + 3x − 4 = 0
The resulting quadratic can then be solved using an appropriate method.
Any value obtained must still satisfy the original restrictions.
Complex Fractions
A complex fraction contains a fraction within a fraction.
For example:
(1⁄x + 1)⁄(1⁄x − 1)
One efficient method is to multiply both the numerator and denominator by x:
[(1⁄x + 1)x]⁄[(1⁄x − 1)x]
Simplify:
(1 + x)⁄(1 − x)
Therefore:
(x + 1)⁄(1 − x)
However, restrictions must still be considered.
The original expression contains:
1⁄x
so:
x ≠ 0
The entire original denominator is:
1⁄x − 1
This cannot equal zero.
Set it equal to zero:
1⁄x − 1 = 0
Therefore:
1⁄x = 1
so:
x = 1
Thus the original expression requires:
x ≠ 0, 1
Equivalent Rational Expressions
Some SAT questions ask you to rewrite an expression into an equivalent form.
For example:
(x + 1)⁄(x + 1) − 4⁄(4x − 5)
A common denominator is:
(x + 1)(4x − 5)
Rewrite the first fraction:
(x + 1)(4x − 5)⁄[(x + 1)(4x − 5)]
Rewrite the second fraction:
4(x + 1)⁄[(x + 1)(4x − 5)]
Subtract:
[(x + 1)(4x − 5) − 4(x + 1)]⁄[(x + 1)(4x − 5)]
Factor (x + 1):
[(x + 1)(4x − 5 − 4)]⁄[(x + 1)(4x − 5)]
Simplify:
[(x + 1)(4x − 9)]⁄[(x + 1)(4x − 5)]
Cancel the common factor:
(4x − 9)⁄(4x − 5)
The original expression still has restrictions from its original denominators.
This kind of algebra is closely connected to the SAT skill of rewriting rational expressions into equivalent forms.
Common SAT Rational Expression Mistakes
1. Forgetting a denominator restriction
If you see:
3⁄(x − 4)
remember:
x ≠ 4
2. Canceling terms instead of factors
You cannot cancel the x in:
(x + 5)⁄x
because x is not a factor of the entire numerator.
3. Adding denominators
This is incorrect:
1⁄x + 1⁄(x + 2) = 2⁄(2x + 2)
The denominators must not be added.
Find a common denominator instead.
4. Losing a negative sign
Consider:
−(x + 3)⁄(x − 2)
The negative sign applies to the entire numerator.
It can also be written as:
(−x − 3)⁄(x − 2)
Be particularly careful when subtracting rational expressions.
5. Forgetting an original restriction
Suppose:
[(x − 4)(x + 2)]⁄[(x − 4)(x + 5)]
simplifies to:
(x + 2)⁄(x + 5)
The original expression still requires:
x ≠ 4
and:
x ≠ −5
6. Accepting a forbidden solution
A value that makes an original denominator zero cannot be accepted as a solution.
Always check the final answer against the original equation.
A Fast SAT Method
When you encounter a rational-expression question, use this sequence.
Step 1: Check the denominators
Ask:
Which values make a denominator equal to zero?
Write those restrictions down.
Step 2: Factor when useful
Factoring can reveal common factors and restrictions.
Step 3: Identify the operation
Determine whether the question asks you to:
• simplify
• add
• subtract
• multiply
• divide
• solve an equation
Step 4: Apply the appropriate rule
Do not use the same procedure for every rational-expression problem.
Step 5: Check your result
Make sure your answer does not violate an original restriction.
This process helps reduce careless algebra errors.
Rational Expressions and Advanced Math on the SAT
Rational expressions are included within the SAT's Advanced Math domain.
College Board describes Advanced Math as covering nonlinear expressions and equations involving topics such as quadratic, exponential, polynomial, rational, radical, and other nonlinear relationships.
The Advanced Math domain accounts for approximately 35% of SAT Math.
That percentage refers to the entire Advanced Math domain.
It does not mean that 35% of the Math section consists of rational-expression questions.
College Board does not specify a fixed number of rational-expression questions that will appear on every SAT.
Therefore, preparation should focus on mastering the underlying algebra rather than trying to predict an exact number of questions.
Rational Expressions: Quick Review
Denominator rule
Denominator ≠ 0
Multiplication
a⁄b × c⁄d = ac⁄bd
Division
a⁄b ÷ c⁄d = a⁄b × d⁄c
The divisor itself cannot equal zero.
Addition
Find a common denominator before combining the numerators.
Subtraction
Find a common denominator before subtracting the numerators.
Simplification
Factor first and cancel common factors.
Restrictions
Restrictions come from the original expression.
Rational equations
Check every proposed solution against the original restrictions.
Final SAT Strategy
Rational expressions become much easier when you recognize that most questions rely on a small group of recurring skills:
factor → find restrictions → choose the correct operation → simplify → check
The goal is not to memorize complicated shortcuts.
Instead, become comfortable with the basic algebra.
Before entering an answer, ask yourself:
Did I make a denominator equal to zero?
Did I cancel a term instead of a factor?
Did I lose a negative sign?
Did I use a common denominator correctly?
Did I remember the restrictions from the original expression?
Does my final answer actually work in the original equation?
These quick checks can prevent many avoidable mistakes.
Rational expressions also connect naturally with other SAT Math skills, especially factoring, nonlinear equations, equivalent expressions, and algebraic manipulation.
Once those skills become automatic, rational-expression questions become much more manageable.
