Equivalent
Expressions
Master the art of recognizing and creating expressions that look different but always produce the same value: a core SAT Math skill.
SAT Math
What Are Equivalent Expressions?
Two algebraic expressions are equivalent if they produce the same output for every possible value of the variable. They may look completely different on paper, but they are mathematically identical.
Why This Matters on the SAT
The SAT frequently asks you to identify which expression among the answer choices is equivalent to a given expression. These questions test your ability to simplify, factor, and expand algebraic expressions fluently.
Simplifying
Combine like terms and reduce fractions to their simplest form
Factoring
Rewrite a sum or difference as a product of factors
Expanding
Distribute and multiply out products into a sum of terms
The Core Principle
Key Properties That Create Equivalence
1
Distributive Property: \( a(b + c) = ab + ac \)
2
Combining Like Terms: \( 3x + 5x = 8x \)
3
Commutative Property: \( a + b = b + a \) and \( ab = ba \)
4
Factoring: \( x^2 - 9 = (x+3)(x-3) \)
5
Exponent Rules: \( x^a \cdot x^b = x^{a+b} \)
Equivalent vs. Equal
| Equivalent Expressions | Equations (Equal) | |
|---|---|---|
| Definition | Same value for all inputs | True for specific values only |
| Example | \( 2(x+1) = 2x + 2 \) | \( 2x + 2 = 10 \) |
| Variables | Identity: holds for every \( x \) | Conditional: only \( x = 4 \) |
| SAT Question | "Which is equivalent to...?" | "What is the value of \( x \)?" |
How to Find Equivalent Expressions
1
Identify the operation: Is the expression asking you to expand, factor, or simplify?
2
Apply algebraic properties: Distribute, combine like terms, or factor as needed.
3
Simplify completely: Reduce all terms to their simplest form.
4
Match with choices: Compare your result to the answer choices.
5
Verify with substitution: Plug in a value (like \( x = 1 \)) to confirm both expressions give the same result.
Example 1: Expanding
1
Distribute the 3: \( 3 \cdot 2x - 3 \cdot 5 + 4x = \) \( 6x - 15 + 4x \)
2
Combine like terms: \( 6x + 4x - 15 = \) \( 10x - 15 \)
3
Verify with \( x = 1 \): Original: \( 3(2-5)+4 = 3(-3)+4 = -5 \). Result: \( 10(1)-15 = -5 \) ✓
Example 2: Factoring
1
Find two numbers that multiply to \( -14 \) and add to \( -5 \).
2
Test pairs: \( (-7) \times 2 = -14 \) and \( -7 + 2 = -5 \). ✓
3
Write the factors: \( (x - 7)(x + 2) \)
4
Verify with \( x = 0 \): Original: \( 0 - 0 - 14 = -14 \). Factored: \( (-7)(2) = -14 \) ✓
Example 3: Rational Expressions
1
Factor the numerator: \( x^2 - 4 \) is a difference of squares: \( (x+2)(x-2) \)
2
Rewrite: \( \dfrac{(x+2)(x-2)}{x+2} \)
3
Cancel common factor \( (x+2) \): Result is \( x - 2 \), where \( x \neq -2 \)
Example 4: Exponent Rules
1
Divide coefficients: \( \dfrac{4}{2} = \) \( 2 \)
2
Subtract exponents (same base): \( x^{5-2} = \) \( x^3 \) and \( y^{3-1} = \) \( y^2 \)
3
Combine: \( 2x^3 y^2 \)
The Substitution Strategy
When you are stuck or want to double-check your answer, plug in a simple value for the variable and see which answer choice gives the same result as the original expression.
1
Pick a simple value: Try \( x = 2 \) or \( x = -1 \). Avoid 0 and 1 (too many things equal 0 or 1).
2
Evaluate the original expression with your chosen value.
3
Evaluate each answer choice with the same value.
4
Eliminate any choice that gives a different result. If two choices match, try a second value.
Special Products to Memorize
Difference of Squares
\( a^2 - b^2 = (a+b)(a-b) \)
Perfect Square (Sum)
\( (a+b)^2 = a^2 + 2ab + b^2 \)
Perfect Square (Diff)
\( (a-b)^2 = a^2 - 2ab + b^2 \)
Common Mistakes
✕ Distributing Only Partially
Writing \( 3(x + 4) = 3x + 4 \) instead of \( 3x + 12 \). You must distribute to every term inside the parentheses.
✕ Sign Errors with Negatives
Writing \( -(x - 5) = -x - 5 \) instead of \( -x + 5 \). A negative sign distributes to both terms, flipping both signs.
✕ Squaring a Binomial Incorrectly
Writing \( (x+3)^2 = x^2 + 9 \) and forgetting the middle term \( 2 \cdot x \cdot 3 = 6x \). The correct expansion is \( x^2 + 6x + 9 \).
Technique Summary
| Technique | When to Use | Example |
|---|---|---|
| Distribute | Product with a sum/difference | \( a(b+c) \to ab + ac \) |
| Combine Like Terms | Same variable, same exponent | \( 3x + 7x \to 10x \) |
| Factor (GCF) | All terms share a common factor | \( 6x + 9 \to 3(2x+3) \) |
| Factor (Quadratic) | Trinomial \( ax^2 + bx + c \) | \( x^2+5x+6 \to (x+2)(x+3) \) |
| Cancel Common Factors | Rational expressions | \( \dfrac{x^2-4}{x+2} \to x-2 \) |
| Substitution Check | Verifying or stuck | Plug in \( x = 2 \) to test |
Question 1 of 5
Which expression is equivalent to \( 4(3x - 2) - 5x \)?
Distribute \( 4 \): \( 4 \cdot 3x - 4 \cdot 2 = 12x - 8 \). Then subtract \( 5x \): \( 12x - 8 - 5x = 7x - 8 \). Choice (A) forgot to distribute \( 4 \) to the \( -2 \); choice (D) forgot to subtract \( 5x \).
Question 2 of 5
Which expression is equivalent to \( x^2 + 10x + 25 \)?
This is a perfect square trinomial. Since \( 25 = 5^2 \) and \( 10x = 2 \cdot x \cdot 5 \), it factors as \( (x+5)^2 \). Choice (B) would give \( x^2 - 25 \), a difference of squares.
Question 3 of 5
Which expression is equivalent to \( \dfrac{9x^2 - 1}{3x + 1} \)?
Factor the numerator as a difference of squares: \( 9x^2 - 1 = (3x+1)(3x-1) \). Cancel the common factor \( (3x+1) \): \( \dfrac{(3x+1)(3x-1)}{3x+1} = 3x - 1 \) (where \( x \neq -\frac{1}{3} \)).
Question 4 of 5
If \( a = 3 \) and \( b = -2 \), which expression is NOT equivalent to \( 2a^2 - 3b \)?
Evaluate the original expression: \( 2(3)^2 - 3(-2) = 2(9) - (-6) = 18 + 6 = 24 \). Choices A, B, and C all correctly evaluate to 24. Choice (D) evaluates to \( 18 - 6 = 12 \). This is incorrect and likely results from substituting \( b=2 \) instead of \( b=-2 \).
Question 5 of 5
Which expression is equivalent to \( (2x + 3)(x - 4) \)?
Use FOIL: First: \( 2x \cdot x = 2x^2 \). Outer: \( 2x \cdot (-4) = -8x \). Inner: \( 3 \cdot x = 3x \). Last: \( 3 \cdot (-4) = -12 \). Combine: \( 2x^2 - 8x + 3x - 12 = 2x^2 - 5x - 12 \). Choice (C) has the wrong sign on \( 5x \); choice (A) omits the middle term.