Rational expression operations

1. Factor the numerator: 4x + 12 = 4(x + 3).
2. Rewrite: \dfrac{4(x + 3)}{4}.
3. The 4 is now a factor of the whole numerator and of the denominator, so it cancels.
4. Result: x + 3.

Another way to see it: split into two fractions, \dfrac{4x}{4} + \dfrac{12}{4} = x + 3. The 4 divides both terms.

Answer: x + 3

Check: Let x = 2: \dfrac{8 + 12}{4} = \dfrac{20}{4} = 5, and 2 + 3 = 5 ✓.

Tip: You can only cancel a number that divides the entire numerator. If it only divides one term, it does not cancel.

1. The numerator is a difference of squares: x^2 - 25 = x^2 - 5^2 = (x - 5)(x + 5).
2. Rewrite: \dfrac{(x - 5)(x + 5)}{x - 5}.
3. Cancel the common factor (x - 5), which is allowed because x \ne 5.
4. Result: x + 5.

Answer: x + 5

Check: Let x = 7: \dfrac{49 - 25}{7 - 5} = \dfrac{24}{2} = 12, and 7 + 5 = 12 ✓.

Tip: Look for a factor of the numerator that matches the denominator exactly. Then the whole factor cancels — not just the numbers inside it.

1. Factor the numerator: x^2 + 6x = x(x + 6).
2. Write the fraction: \dfrac{x(x + 6)}{3x}.
3. The factor x is in both the numerator and the denominator, so cancel it: \dfrac{x + 6}{3}.
4. Nothing else cancels: 3 does not divide the whole numerator x + 6.

Answer: \dfrac{x + 6}{3}

Check: Let x = 3: original \dfrac{9 + 18}{9} = \dfrac{27}{9} = 3; answer \dfrac{3 + 6}{3} = 3 ✓.

Tip: Dividing a sum by a monomial means dividing every term: \dfrac{x^2}{3x} + \dfrac{6x}{3x} = \dfrac{x}{3} + 2, which is the same as \dfrac{x + 6}{3}.

1. A rational expression is undefined when the denominator is 0.
2. Set the denominator equal to 0: x^2 - 9 = 0.
3. Factor: (x - 3)(x + 3) = 0.
4. So x - 3 = 0 or x + 3 = 0, giving x = 3 or x = -3.
5. The numerator (x + 1) does not affect where the expression is undefined.

Answer: x = 3 and x = -3

Check: At x = 3: denominator 9 - 9 = 0 ✓. At x = -3: 9 - 9 = 0 ✓. At x = -1: denominator 1 - 9 = -8, so the expression is \dfrac{0}{-8} = 0, which is defined.

Tip: x^2 = 9 has two solutions, 3 and -3. Do not forget the negative root.

1. Multiply numerators and denominators: \dfrac{3x(x + 2)}{6(x + 2)}.
2. Cancel the common factor (x + 2): \dfrac{3x}{6}.
3. Simplify the numbers: \dfrac{3x}{6} = \dfrac{x}{2}.

Answer: \dfrac{x}{2}

Check: Let x = 4: \dfrac{12}{6} \cdot \dfrac{6}{6} = 2 \cdot 1 = 2, and \dfrac{4}{2} = 2 ✓.

Tip: In multiplication, you can cancel a factor in any numerator against the same factor in any denominator. You do not need a common denominator to multiply.

1. Keep, change, flip: \dfrac{x^2 - 4}{x + 3} \cdot \dfrac{x + 3}{x - 2}.
2. Factor: x^2 - 4 = (x - 2)(x + 2).
3. Write it out: \dfrac{(x - 2)(x + 2)}{x + 3} \cdot \dfrac{x + 3}{x - 2}.
4. Cancel (x + 3) and (x - 2).
5. Result: x + 2.

Answer: x + 2

Check: Let x = 4: \dfrac{12}{7} \div \dfrac{2}{7} = \dfrac{12}{7} \cdot \dfrac{7}{2} = 6, and 4 + 2 = 6 ✓.

Tip: In division, only the fraction after the \div sign is flipped. Then it is an ordinary multiplication problem.

1. LCD = (x + 3)(x - 1).
2. First fraction: multiply top and bottom by (x - 1): \dfrac{2(x - 1)}{(x + 3)(x - 1)}.
3. Second fraction: multiply top and bottom by (x + 3): \dfrac{5(x + 3)}{(x + 3)(x - 1)}.
4. Add numerators: 2x - 2 + 5x + 15 = 7x + 13.
5. Result: \dfrac{7x + 13}{(x + 3)(x - 1)}.

Answer: \dfrac{7x + 13}{(x + 3)(x - 1)}

Check: Let x = 2: \dfrac{2}{5} + \dfrac{5}{1} = 5.4; formula \dfrac{27}{5 \cdot 1} = 5.4 ✓.

Tip: Distribute carefully: 2(x - 1) is 2x - 2, not 2x - 1. Each numerator gets multiplied by the whole missing factor.

1. The denominators are the same, so subtract the numerators: \dfrac{(3x + 1) - (x - 5)}{x + 2}.
2. Distribute the minus sign: -(x - 5) = -x + 5.
3. Combine: 3x + 1 - x + 5 = 2x + 6.
4. Result: \dfrac{2x + 6}{x + 2}. (It can also be written \dfrac{2(x + 3)}{x + 2}; nothing cancels.)

Answer: \dfrac{2x + 6}{x + 2}

Check: Let x = 1: \dfrac{4}{3} - \dfrac{-4}{3} = \dfrac{8}{3}, and \dfrac{2 + 6}{3} = \dfrac{8}{3} ✓.

Tip: A minus sign in front of a fraction changes the sign of every term in that numerator. Writing the parentheses makes the sign change automatic.

1. The LCD of 2x and 3x is 6x.
2. Rewrite the first fraction: \dfrac{5}{2x} = \dfrac{5 \cdot 3}{2x \cdot 3} = \dfrac{15}{6x}.
3. Rewrite the second: \dfrac{1}{3x} = \dfrac{1 \cdot 2}{3x \cdot 2} = \dfrac{2}{6x}.
4. Subtract numerators: \dfrac{15 - 2}{6x} = \dfrac{13}{6x}.

Answer: \dfrac{13}{6x}

Check: Let x = 1: \dfrac{5}{2} - \dfrac{1}{3} = \dfrac{15}{6} - \dfrac{2}{6} = \dfrac{13}{6} ✓.

Tip: It works exactly like \dfrac{5}{2} - \dfrac{1}{3} with numbers. Never subtract the denominators.

1. Rewrite as multiplication by the reciprocal: \dfrac{x^2 + 3x}{x - 5} \cdot \dfrac{x^2 - 25}{x}.
2. Factor: x^2 + 3x = x(x + 3) and x^2 - 25 = (x - 5)(x + 5).
3. Product: \dfrac{x(x + 3)(x - 5)(x + 5)}{(x - 5) \cdot x}.
4. Cancel x and (x - 5): (x + 3)(x + 5).

Answer: (x + 3)(x + 5)

Check: Let x = 1: \dfrac{4}{-4} \div \dfrac{1}{-24} = -1 \cdot (-24) = 24, and (1 + 3)(1 + 5) = 24 ✓.

Tip: "Keep, change, flip": keep the first fraction, change \div to \times, flip the second. Factor before canceling.

1. Factor the numerator: 6 - 2x = -2(x - 3). (Check: -2 \cdot x = -2x and -2 \cdot (-3) = 6.)
2. The fraction is \dfrac{-2(x - 3)}{x - 3}.
3. Cancel (x - 3): the result is -2.

Answer: -2

Check: Let x = 5: \dfrac{6 - 10}{5 - 3} = \dfrac{-4}{2} = -2 ✓. Let x = 0: \dfrac{6}{-3} = -2 ✓.

Tip: (3 - x) and (x - 3) are opposites: \dfrac{3 - x}{x - 3} = -1. Whenever you see reversed subtraction, factor out -1.

1. Step 1: (x + 4)(x - 2) = x^2 + 2x - 8 ✓ and (x - 2)(x + 2) = x^2 - 4 ✓.
2. Step 2: canceling the common factor (x - 2) is correct ✓.
3. Step 3: Kim crossed out the x in x + 4 and in x + 2. But x is a term, not a factor of the whole numerator and denominator, so it cannot be canceled. ✗
4. The simplified form is \dfrac{x + 4}{x + 2}.

Answer: Step 3 is wrong; the answer is \dfrac{x + 4}{x + 2}.

Check: Let x = 1: original \dfrac{1 + 2 - 8}{1 - 4} = \dfrac{-5}{-3} = \dfrac{5}{3}; \dfrac{1 + 4}{1 + 2} = \dfrac{5}{3} ✓; but Kim's 2 is not \dfrac{5}{3}.

Tip: Cancel only things that are multiplied by everything else. Pieces joined by + or - stay put.

1. Factor: x^2 - 4 = (x - 2)(x + 2), so the LCD is (x - 2)(x + 2).
2. The first fraction is missing (x + 2): \dfrac{x(x + 2)}{(x - 2)(x + 2)} = \dfrac{x^2 + 2x}{(x - 2)(x + 2)}.
3. Subtract: \dfrac{x^2 + 2x - 8}{(x - 2)(x + 2)}.
4. Factor the numerator: two numbers that multiply to -8 and add to 2 are 4 and -2, so x^2 + 2x - 8 = (x + 4)(x - 2).
5. Cancel (x - 2): \dfrac{x + 4}{x + 2}.

Answer: \dfrac{x + 4}{x + 2}

Check: Let x = 4: \dfrac{4}{2} - \dfrac{8}{12} = 2 - \dfrac{2}{3} = \dfrac{4}{3}, and \dfrac{4 + 4}{4 + 2} = \dfrac{8}{6} = \dfrac{4}{3} ✓.

Tip: GED-hard rational questions often have one more step after combining: factor the new numerator and look for a factor to cancel.

1. LCD = (x - 2)(x + 4).
2. First fraction: \dfrac{3(x + 4)}{(x - 2)(x + 4)} = \dfrac{3x + 12}{(x - 2)(x + 4)}.
3. Second fraction: \dfrac{2(x - 2)}{(x - 2)(x + 4)} = \dfrac{2x - 4}{(x - 2)(x + 4)}.
4. Subtract: (3x + 12) - (2x - 4) = 3x + 12 - 2x + 4 = x + 16.
5. Result: \dfrac{x + 16}{(x - 2)(x + 4)}.

Answer: \dfrac{x + 16}{(x - 2)(x + 4)}

Check: Let x = 3: \dfrac{3}{1} - \dfrac{2}{7} = \dfrac{19}{7}, and \dfrac{3 + 16}{(1)(7)} = \dfrac{19}{7} ✓.

Tip: Subtracting (2x - 4) turns -4 into +4. This is the most common error on unlike-denominator subtraction.

1. LCD = x(x + 3).
2. \dfrac{1}{x} = \dfrac{x + 3}{x(x + 3)} and \dfrac{1}{x + 3} = \dfrac{x}{x(x + 3)}.
3. Add numerators: (x + 3) + x = 2x + 3.
4. Result: \dfrac{2x + 3}{x(x + 3)}.

Answer: \dfrac{2x + 3}{x(x + 3)}

Check: If x = 3: the pumps take 3 and 6 hours, so together they fill \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{1}{2} of the tank per hour. Formula: \dfrac{2(3) + 3}{3 \cdot 6} = \dfrac{9}{18} = \dfrac{1}{2} ✓.

Tip: Rates add; times do not. Working together, the pumps do \tfrac{1}{x} + \tfrac{1}{x + 3} of the job each hour.

1. Factor each part:
   • x^2 - 9 = (x - 3)(x + 3)
   • x^2 + 5x + 6 = (x + 2)(x + 3)
   • x^2 - 3x = x(x - 3)
2. Product: \dfrac{(x - 3)(x + 3)(x + 2)}{(x + 2)(x + 3) \cdot x(x - 3)}.
3. Cancel (x - 3), (x + 3) and (x + 2).
4. What remains: \dfrac{1}{x}.

Answer: \dfrac{1}{x}

Check: Let x = 1: \dfrac{-8}{12} \cdot \dfrac{3}{-2} = \dfrac{-24}{-24} = 1, and \dfrac{1}{1} = 1. Let x = 4: \dfrac{7}{42} \cdot \dfrac{6}{4} = \dfrac{42}{168} = \dfrac{1}{4} ✓.

Tip: When every binomial cancels, a 1 stays in the numerator, and any leftover factor (here x) stays in the denominator.

1. Same denominator: \dfrac{(x^2 + 2x) - (3x + 2)}{x^2 - 1}.
2. Distribute: x^2 + 2x - 3x - 2 = x^2 - x - 2.
3. Factor the numerator: numbers that multiply to -2 and add to -1 are -2 and 1, so x^2 - x - 2 = (x - 2)(x + 1).
4. Factor the denominator: x^2 - 1 = (x - 1)(x + 1).
5. Cancel (x + 1): \dfrac{x - 2}{x - 1}.

Answer: \dfrac{x - 2}{x - 1}

Check: Let x = 3: \dfrac{15}{8} - \dfrac{11}{8} = \dfrac{4}{8} = \dfrac{1}{2}, and \dfrac{3 - 2}{3 - 1} = \dfrac{1}{2} ✓.

Tip: After combining, always try to factor the new numerator. On the GED, the answer is usually the fully simplified form.

1. Step 1: the denominators match, so subtracting the whole numerators in parentheses is correct ✓.
2. Step 2: -(x - 5) should be -x + 5. Luis wrote -x - 5, so the minus sign was not distributed to the -5. ✗
3. Correct Step 2: \dfrac{x + 1 - x + 5}{x - 3}.
4. Correct Step 3: \dfrac{6}{x - 3}.
5. Luis's Step 3 follows correctly from his Step 2, but Step 2 is where the error started.

Answer: Step 2 is the first error; the answer should be \dfrac{6}{x - 3}.

Check: Let x = 4: \dfrac{5}{1} - \dfrac{-1}{1} = 6, and \dfrac{6}{4 - 3} = 6 ✓; Luis's answer gives -4.

Tip: Subtracting a binomial flips both of its signs: -(x - 5) = -x + 5.

1. Numerator: a \cdot c = 2 \cdot (-3) = -6; numbers that multiply to -6 and add to -5 are -6 and 1.
2. Split: 2x^2 - 6x + x - 3 = 2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3).
3. Denominator: 4x^2 - 1 = (2x - 1)(2x + 1).
4. Fraction: \dfrac{(2x + 1)(x - 3)}{(2x - 1)(2x + 1)}.
5. Cancel (2x + 1): \dfrac{x - 3}{2x - 1}.

Answer: \dfrac{x - 3}{2x - 1}

Check: Let x = 2: \dfrac{8 - 10 - 3}{16 - 1} = \dfrac{-5}{15} = -\dfrac{1}{3}, and \dfrac{2 - 3}{4 - 1} = -\dfrac{1}{3} ✓.

Tip: Never cancel x^2 terms or numbers that are only parts of a sum. Factor completely, then cancel identical factors.

1. Ana: she canceled the x terms. Test x = 1: \dfrac{4}{6} = \dfrac{2}{3}, but \dfrac{3}{5} \ne \dfrac{2}{3} ✗.
2. Ben: factor: \dfrac{x(x + 3)}{x(x + 5)}. The common factor x cancels, giving \dfrac{x + 3}{x + 5} ✓.
3. Cara: she added tops and bottoms. Test x = 1, y = 1: 1 + 1 = 2, but \dfrac{2}{2} = 1 ✗. The correct sum is \dfrac{y + x}{xy}.
4. Only Ben's claim is true.

Answer: Only Ben is correct.

Check: Ben at x = 1: \dfrac{4}{6} = \dfrac{2}{3} and \dfrac{4}{6} = \dfrac{2}{3} ✓.

Tip: Plugging in small numbers is a powerful way to test any algebra claim. One counterexample proves a claim false.

1. Rate = \dfrac{\text{area}}{\text{time}} = \dfrac{2x^2 + 6x}{x - 4} \div \dfrac{x + 3}{x - 4}.
2. Multiply by the reciprocal: \dfrac{2x^2 + 6x}{x - 4} \cdot \dfrac{x - 4}{x + 3}.
3. Factor: 2x^2 + 6x = 2x(x + 3).
4. Product: \dfrac{2x(x + 3)(x - 4)}{(x - 4)(x + 3)}.
5. Cancel (x + 3) and (x - 4): rate = 2x.

Answer: 2x square yards per hour

Check: Let x = 5: area = \dfrac{50 + 30}{1} = 80 yd², time = \dfrac{8}{1} = 8 h, rate = 10 yd²/h, and 2(5) = 10 ✓.

Tip: For any rate (miles per hour, dollars per item), divide the amount by the time or count. With rational expressions, that means: multiply by the reciprocal.