Simplifying Rational Expressions

Factor and cancel common factors.

$\frac{x^2 - 4}{x^2 + 4x + 4} = \frac{(x-2)(x+2)}{(x+2)^2} = \frac{x-2}{x+2}$

Critical: You can only cancel FACTORS (things being multiplied), never terms (things being added).

WRONG: $\frac{x + 3}{x + 5} \neq \frac{3}{5}$ ← Cannot cancel the $x$'s!

Operations with Rational Expressions

Multiplication

Factor, cancel, then multiply: $\frac{x^2-1}{x+3} \cdot \frac{x+3}{x-1} = \frac{(x+1)(x-1)}{x+3} \cdot \frac{x+3}{x-1} = x+1$

Division

Flip the second fraction and multiply: $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}$

Addition/Subtraction

Find a common denominator first: $\frac{1}{x} + \frac{2}{x+1} = \frac{x+1}{x(x+1)} + \frac{2x}{x(x+1)} = \frac{3x+1}{x(x+1)}$

Domain Restrictions

A rational expression is undefined when the denominator equals zero.

$\frac{x+3}{x^2-9} = \frac{x+3}{(x+3)(x-3)}$

Domain restrictions: $x \neq 3$ and $x \neq -3$

Even after simplifying to $\frac{1}{x-3}$, the restriction $x \neq -3$ still applies!

Solving Rational Equations

Strategy: Multiply both sides by the LCD to clear fractions.

$\frac{3}{x} + \frac{2}{x+1} = 1$

LCD = $x(x+1)$: $3(x+1) + 2x = x(x+1)$ $3x + 3 + 2x = x^2 + x$ $5x + 3 = x^2 + x$ $0 = x^2 - 4x - 3$

Always check for extraneous solutions! Plug your answers back in to make sure the denominators aren't zero.

SAT Question Types

Type 1: Simplify a Rational Expression

Factor and cancel.

Type 2: Find the Domain

Identify values that make the denominator zero.

Type 3: Add/Subtract Rational Expressions

Find common denominators and combine.

Type 4: Solve a Rational Equation

Clear fractions, solve, and check for extraneous solutions.

Common SAT Mistakes

1. Canceling terms instead of factors: $\frac{x+3}{x+5}$ cannot be simplified!
2. Forgetting domain restrictions after simplifying
3. Not checking for extraneous solutions — solutions that make a denominator zero must be rejected
4. Incorrect LCD — make sure to include all unique factors
5. Sign errors when distributing negatives in subtraction of rational expressions

Answer: $x - 3$, provided $x \neq -3$

Step 3: Add the numerators: $\frac{2(x+2) + 3(x-1)}{(x-1)(x+2)} = \frac{2x+4+3x-3}{(x-1)(x+2)} = \frac{5x+1}{(x-1)(x+2)}$

Answer: $\frac{5x+1}{(x-1)(x+2)}$

Step 3: Add the numerators: $\frac{2(x+2) + 3(x-1)}{(x-1)(x+2)} = \frac{2x+4+3x-3}{(x-1)(x+2)} = \frac{5x+1}{(x-1)(x+2)}$

Answer: $\frac{5x+1}{(x-1)(x+2)}$

Step 3: Solve: $x = 3 \quad \text{or} \quad x = -2$

Answer: The expression is undefined at $x = 3$ and $x = -2$.

Note: Even though the full expression simplifies (the numerator factors to $(x+5)(x-3)$, and $(x-3)$ cancels), $x = 3$ is still a restriction because it was in the original denominator.

Step 3: Solve: $x = 3 \quad \text{or} \quad x = -2$

Answer: The expression is undefined at $x = 3$ and $x = -2$.

Note: Even though the full expression simplifies (the numerator factors to $(x+5)(x-3)$, and $(x-3)$ cancels), $x = 3$ is still a restriction because it was in the original denominator.

Step 3: Check: $x = \frac{8}{3} \neq 2$, so it's valid. ✓

Verify: $\frac{4}{\frac{8}{3}-2} = \frac{4}{\frac{2}{3}} = 6$ and $\frac{\frac{8}{3}}{\frac{2}{3}} + 2 = 4 + 2 = 6$ ✓

Answer: $x = \frac{8}{3}$

Step 3: Check: $x = \frac{8}{3} \neq 2$, so it's valid. ✓

Verify: $\frac{4}{\frac{8}{3}-2} = \frac{4}{\frac{2}{3}} = 6$ and $\frac{\frac{8}{3}}{\frac{2}{3}} + 2 = 4 + 2 = 6$ ✓

Answer: $x = \frac{8}{3}$

Domain restrictions: $x \neq 1$ and $x \neq -1$

Step 2: Multiply every term by $(x+1)(x-1)$: $2(x-1) + 1(x+1) = 4$

Step 3: Distribute and solve: $2x - 2 + x + 1 = 4$ $3x - 1 = 4$ $3x = 5$ $x = \frac{5}{3}$

Step 4: Check: $x = \frac{5}{3} \neq \pm 1$, so it's valid. ✓

Answer: $x = \frac{5}{3}$

SAT Tip: Always factor the denominators first to find the LCD and identify domain restrictions.

Domain restrictions: $x \neq 1$ and $x \neq -1$

Step 2: Multiply every term by $(x+1)(x-1)$: $2(x-1) + 1(x+1) = 4$

Step 3: Distribute and solve: $2x - 2 + x + 1 = 4$ $3x - 1 = 4$ $3x = 5$ $x = \frac{5}{3}$

Step 4: Check: $x = \frac{5}{3} \neq \pm 1$, so it's valid. ✓

Answer: $x = \frac{5}{3}$

SAT Tip: Always factor the denominators first to find the LCD and identify domain restrictions.