SAT Math – Advanced Math
Operations with Rational Expressions
Simplifying, adding, subtracting, multiplying, and dividing algebraic fractions
Rational expressions are fractions containing polynomials in the numerator and denominator. On the SAT, you'll simplify rational expressions, perform arithmetic operations, identify restrictions on variables, and solve complex fraction problems—skills essential for advanced algebra, calculus, and applications in physics and engineering.
Success requires understanding factoring to simplify, finding common denominators, recognizing restrictions (values that make denominators zero), and applying fraction arithmetic rules systematically. These operations aren't just algebraic manipulations—they're fundamental to analyzing rates (distance/time), concentrations (amount/volume), unit conversions, and any relationship expressed as a ratio of polynomial quantities.
Understanding Rational Expressions
What is a Rational Expression?
A rational expression is a fraction where numerator and denominator are polynomials.
Numerator: \(x^2 + 3x - 4\)
Denominator: \(2x - 6\)
Restriction: Denominator cannot equal zero (\(x \neq 3\))
Simplifying Rational Expressions
Factor numerator and denominator, then cancel common factors.
Cancel: Divide out common factors
Cannot cancel: Terms that are added/subtracted
Result: Simplified form with restrictions noted
Multiplying and Dividing
Follow same rules as numeric fractions.
Divide: Multiply by reciprocal (flip second fraction)
Simplify: Factor and cancel before multiplying
Efficiency: Cancel first to avoid large expressions
Adding and Subtracting
Require common denominator, just like numeric fractions.
Rewrite: Express each fraction with LCD
Combine: Add/subtract numerators only
Simplify: Factor and reduce if possible
Essential Rules and Operations
Simplifying Rational Expressions
\(\frac{P(x)}{Q(x)} = \frac{\text{factored } P(x)}{\text{factored } Q(x)}\) then cancel common factors
1. Factor numerator completely
2. Factor denominator completely
3. Cancel factors that appear in both (not terms!)
Multiplication
\(\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}\)
Factor first, cancel common factors, then multiply
Division
\(\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}\)
Multiply by reciprocal of second fraction (flip numerator and denominator)
Addition and Subtraction
\(\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}\) (same denominator)
\(\frac{a}{b} \pm \frac{c}{d} = \frac{ad \pm bc}{bd}\) (different denominators)
Find LCD for efficiency; combine numerators only
Common Pitfalls & Expert Tips
❌ Canceling terms instead of factors
Cannot cancel: \(\frac{x + 3}{x + 5}\). Terms connected by + or − don't cancel! Only factors (multiplication) can be canceled.
❌ Forgetting to flip when dividing
Division means multiply by reciprocal. \(\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}\), not \(\frac{ac}{bd}\)!
❌ Not distributing negative in subtraction
\(\frac{x - (2x - 5)}{3} = \frac{x - 2x + 5}{3}\), not \(\frac{x - 2x - 5}{3}\). Distribute the negative to ALL terms!
❌ Ignoring domain restrictions
Even after simplifying, original restrictions remain. If \(x = 2\) made original denominator zero, it's still restricted!
✓ Expert Tip: Factor everything first
Whether simplifying, multiplying, or dividing, factor completely first. This reveals cancellation opportunities before calculations get messy.
✓ Expert Tip: Use parentheses when subtracting
When subtracting fractions, put entire numerator in parentheses to ensure proper distribution of the negative sign.
✓ Expert Tip: Check by substitution
Verify simplification by substituting a value (not a restriction). Original and simplified should give same result.
Fully Worked SAT-Style Examples
Simplify: \(\frac{x^2 - 9}{x^2 + 5x + 6}\)
Solution:
Step 1: Factor numerator
\(x^2 - 9\) is difference of squares
\(x^2 - 9 = (x + 3)(x - 3)\)
Step 2: Factor denominator
Find factors of 6 that add to 5: 2 and 3
\(x^2 + 5x + 6 = (x + 2)(x + 3)\)
Step 3: Rewrite and cancel common factors
\(\frac{(x + 3)(x - 3)}{(x + 2)(x + 3)} = \frac{x - 3}{x + 2}\)
Cancel \((x + 3)\) from numerator and denominator
Restrictions:
\(x \neq -2\) and \(x \neq -3\)
(Values that make original denominator zero)
Answer: \(\frac{x - 3}{x + 2}\), \(x \neq -2, -3\)
Multiply: \(\frac{x^2 - 4}{x + 3} \cdot \frac{x + 3}{x + 2}\)
Solution:
Step 1: Factor where possible
\(x^2 - 4 = (x + 2)(x - 2)\)
Rewrite:
\(\frac{(x + 2)(x - 2)}{x + 3} \cdot \frac{x + 3}{x + 2}\)
Step 2: Cancel common factors before multiplying
\((x + 3)\) appears in numerator and denominator
\((x + 2)\) appears in numerator and denominator
Cancel both!
Step 3: Simplify
\(\frac{x - 2}{1} = x - 2\)
Answer: \(x - 2\)
Divide: \(\frac{x^2 - 1}{x} \div \frac{x + 1}{x^2}\)
Solution:
Step 1: Rewrite as multiplication by reciprocal
\(\frac{x^2 - 1}{x} \cdot \frac{x^2}{x + 1}\)
Step 2: Factor
\(x^2 - 1 = (x + 1)(x - 1)\)
\(\frac{(x + 1)(x - 1)}{x} \cdot \frac{x^2}{x + 1}\)
Step 3: Cancel common factors
Cancel \((x + 1)\) and one \(x\)
\(\frac{(x - 1) \cdot x}{1} = x(x - 1) = x^2 - x\)
Answer: \(x^2 - x\) or \(x(x - 1)\)
Add: \(\frac{3x}{x + 2} + \frac{6}{x + 2}\)
Solution:
Step 1: Same denominator—add numerators
\(\frac{3x + 6}{x + 2}\)
Step 2: Factor numerator
\(3x + 6 = 3(x + 2)\)
\(\frac{3(x + 2)}{x + 2}\)
Step 3: Cancel common factor
\(\frac{3(x + 2)}{x + 2} = 3\)
Answer: 3
Subtract: \(\frac{5}{x} - \frac{3}{x - 2}\)
Solution:
Step 1: Find LCD
LCD = \(x(x - 2)\)
Step 2: Rewrite with LCD
\(\frac{5(x - 2)}{x(x - 2)} - \frac{3x}{x(x - 2)}\)
Step 3: Subtract numerators (use parentheses!)
\(\frac{5(x - 2) - 3x}{x(x - 2)}\)
\(= \frac{5x - 10 - 3x}{x(x - 2)}\)
\(= \frac{2x - 10}{x(x - 2)}\)
Step 4: Factor and simplify if possible
\(\frac{2(x - 5)}{x(x - 2)}\)
Cannot simplify further
Answer: \(\frac{2(x - 5)}{x(x - 2)}\) or \(\frac{2x - 10}{x(x - 2)}\)
Simplify: \(\frac{\frac{1}{x} + \frac{1}{2}}{\frac{3}{x}}\)
Solution:
Step 1: Simplify numerator
LCD of numerator = 2x
\(\frac{1}{x} + \frac{1}{2} = \frac{2}{2x} + \frac{x}{2x} = \frac{2 + x}{2x}\)
Step 2: Rewrite main fraction
\(\frac{\frac{2 + x}{2x}}{\frac{3}{x}}\)
Step 3: Divide (multiply by reciprocal)
\(\frac{2 + x}{2x} \cdot \frac{x}{3} = \frac{(2 + x) \cdot x}{2x \cdot 3}\)
Step 4: Simplify
\(\frac{x(2 + x)}{6x} = \frac{2 + x}{6} = \frac{x + 2}{6}\)
Answer: \(\frac{x + 2}{6}\)
Add: \(\frac{2}{x - 3} + \frac{x}{x^2 - 9}\)
Solution:
Step 1: Factor denominators
\(x^2 - 9 = (x + 3)(x - 3)\)
\(\frac{2}{x - 3} + \frac{x}{(x + 3)(x - 3)}\)
Step 2: LCD = \((x + 3)(x - 3)\)
First fraction needs \((x + 3)\) in denominator
\(\frac{2(x + 3)}{(x + 3)(x - 3)} + \frac{x}{(x + 3)(x - 3)}\)
Step 3: Add numerators
\(\frac{2(x + 3) + x}{(x + 3)(x - 3)} = \frac{2x + 6 + x}{(x + 3)(x - 3)}\)
\(= \frac{3x + 6}{(x + 3)(x - 3)}\)
Step 4: Factor and simplify
\(\frac{3(x + 2)}{(x + 3)(x - 3)}\)
Cannot simplify further
Answer: \(\frac{3(x + 2)}{(x + 3)(x - 3)}\)
Simplify: \(\frac{x}{x + 1} - \frac{1}{x - 1} \cdot \frac{x - 1}{x + 1}\)
Solution:
Step 1: Simplify multiplication first
\(\frac{1}{x - 1} \cdot \frac{x - 1}{x + 1} = \frac{1}{x + 1}\)
Step 2: Rewrite problem
\(\frac{x}{x + 1} - \frac{1}{x + 1}\)
Step 3: Same denominator—subtract numerators
\(\frac{x - 1}{x + 1}\)
Answer: \(\frac{x - 1}{x + 1}\)
Operations Quick Reference
Multiply
Factor, cancel, then multiply
Divide
Flip second, then multiply
Add/Subtract
Find LCD, combine numerators
Simplify
Factor, cancel common factors
Rational Expressions: Extending Fraction Operations to Algebra
Rational expressions represent one of algebra's most practical applications—whenever you divide one quantity by another, you're working with a ratio. The SAT tests these skills because they're essential for calculus (limits of rational functions, derivatives of quotients), physics (combined resistance, lens equations), chemistry (reaction rates, concentrations), and engineering (gear ratios, efficiency calculations). Master the systematic approach: factor everything first to reveal cancellation opportunities, maintain restrictions throughout simplification, convert division to multiplication by reciprocal, find LCD methodically for addition and subtraction, and use parentheses religiously when subtracting to ensure proper sign distribution. Understanding that \(\frac{x^2-9}{x+3}\) simplifies to \(x-3\) (with restriction \(x \neq -3\)) demonstrates algebraic maturity—recognizing that mathematical equivalence isn't absolute but conditional based on domain. These operations transform complex real-world relationships into manageable calculations: combined work rates (where individual rates are fractions), average speeds (total distance divided by total time), mixing problems (concentration as amount per volume), and optimization scenarios. The fluency you develop with rational expressions—factoring instantly, canceling confidently, finding common denominators efficiently—distinguishes computational competence from true algebraic understanding.
