Algebra Quiz: Operating With Rational Expressions
20 questions · exam conditions
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Operating With Rational ExpressionsQuestion 1 of 20

Multiply and simplify the rational expressions (the result will also be a rational expression): x2−9x2−4⋅x+2x−3\frac{x^2-9}{x^2-4}\cdot\frac{x+2}{x-3}

x+3x−2\frac{x+3}{x-2}
x2−9(x−2)(x−3)\frac{x^2-9}{(x-2)(x-3)}
x+3x+2\frac{x+3}{x+2}
(x+3)(x+2)(x−2)(x−3)\frac{(x+3)(x+2)}{(x-2)(x-3)}
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Algebra Quiz: Operating With Rational Expressions

Practice Operating With Rational Expressions in Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Operating With Rational Expressions, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Multiply and simplify the rational expressions (the result will also be a rational expression): x2−9x2−4⋅x+2x−3\frac{x^2-9}{x^2-4}\cdot\frac{x+2}{x-3}

  1. x+3x−2\frac{x+3}{x-2} (correct answer)
  2. x2−9(x−2)(x−3)\frac{x^2-9}{(x-2)(x-3)}
  3. x+3x+2\frac{x+3}{x+2}
  4. (x+3)(x+2)(x−2)(x−3)\frac{(x+3)(x+2)}{(x-2)(x-3)}
Explanation: This question tests your understanding of how to multiply rational expressions—algebraic fractions that work just like regular fractions but with variables. Multiplying rational expressions works just like multiplying numeric fractions: multiply the numerators together and multiply the denominators together, giving (a/b)·(c/d) = (ac)/(bd). But here's the smart way: factor first, cancel common factors, THEN multiply—it keeps the numbers smaller and the result already simplified! Let's factor everything first: (x²-9)/(x²-4) · (x+2)/(x-3) = [(x+3)(x-3)]/[(x+2)(x-2)] · (x+2)/(x-3). Now we can cancel: the (x-3) cancels from numerator and denominator, and the (x+2) cancels from numerator and denominator, leaving us with (x+3)/(x-2). Choice A correctly performs the multiplication and simplifies to (x+3)/(x-2) by factoring and canceling common factors. Great work! Choice B shows the unsimplified form without canceling, while choices C and D have incorrect denominators from factoring errors. The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. This prevents working with huge expressions and catches opportunities to simplify. Example: [(x²-4)/x]·[x/(x+2)] looks messy, but factor x²-4 to get [(x+2)(x-2)/x]·[x/(x+2)], cancel the (x+2) and x, leaving just (x-2). So much cleaner!

Question 2

Subtract the rational expressions and simplify (the difference is a rational expression, showing closure under subtraction): xx−2−3x−2\frac{x}{x-2}-\frac{3}{x-2}

  1. −3x−2\frac{-3}{x-2}
  2. x−3x−2\frac{x-3}{x-2} (correct answer)
  3. x+3x−2\frac{x+3}{x-2}
  4. x−3x\frac{x-3}{x}
Explanation: This question tests your understanding of how to subtract rational expressions—algebraic fractions that work just like regular fractions but with variables. Adding or subtracting rational expressions requires a common denominator, just like with 1/2 - 1/3: we find the LCD (least common denominator), rewrite each fraction with that denominator, then add or subtract the numerators while keeping the denominator the same. For x/(x-2) - 3/(x-2), we're lucky—the denominators are already the same! When denominators match, we simply subtract the numerators: (x - 3)/(x-2). Choice A correctly performs the subtraction to get (x-3)/(x-2) by subtracting numerators while keeping the common denominator. Great work! Choice C shows just -3 in the numerator—remember we're subtracting 3 from x, not replacing x with the subtraction! The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. For adding/subtracting with the same denominator, it's even simpler: just combine the numerators! It's exactly like 5/7 - 2/7 = 3/7, just with variables in place of numbers.

Question 3

Multiply and simplify by factoring first (the result is a rational expression): x2+3xx2−9⋅x−3x\frac{x^2+3x}{x^2-9}\cdot\frac{x-3}{x}

  1. x+3x2−9\frac{x+3}{x^2-9}
  2. x+3x\frac{x+3}{x} (correct answer)
  3. x+3x−3\frac{x+3}{x-3}
  4. x+3x+3\frac{x+3}{x+3}
Explanation: This question tests your understanding of how to multiply rational expressions—algebraic fractions that work just like regular fractions but with variables. Multiplying rational expressions works just like multiplying numeric fractions: multiply the numerators together and multiply the denominators together, giving (a/b)·(c/d) = (ac)/(bd). But here's the smart way: factor first, cancel common factors, THEN multiply—it keeps the numbers smaller and the result already simplified! Let's factor everything first: (x²+3x)/(x²-9) · (x-3)/x = [x(x+3)]/[(x+3)(x-3)] · (x-3)/x. Now we can cancel: the (x+3) cancels from numerator and denominator in the first fraction, the (x-3) cancels between fractions, and x cancels between fractions, leaving us with (x+3)/x. Choice B correctly performs the multiplication and simplifies to (x+3)/x by factoring and canceling common factors. Great work! Choice A has the wrong denominator x²-9, choice C incorrectly gives (x+3)/(x-3), and choice D would mean everything canceled to 1. The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. This prevents working with huge expressions and catches opportunities to simplify. Example: [(x²-4)/x]·[x/(x+2)] looks messy, but factor x²-4 to get [(x+2)(x-2)/x]·[x/(x+2)], cancel the (x+2) and x, leaving just (x-2). So much cleaner!

Question 4

Perform the indicated operation and simplify. (Rational expressions are closed under multiplication, so the result is a rational expression.) 3x⋅x2x+2\frac{3}{x} \cdot \frac{x^2}{x+2}

  1. 3x+2\frac{3}{x+2}
  2. 3x2x+2\frac{3x^2}{x+2}
  3. 3xx(x+2)\frac{3x}{x(x+2)}
  4. 3xx+2\frac{3x}{x+2} (correct answer)
Explanation: This question tests your understanding of how to multiply rational expressions—algebraic fractions that work just like regular fractions but with variables. Multiplying rational expressions works just like multiplying numeric fractions: multiply the numerators together and multiply the denominators together, giving ab⋅cd=acbd\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}. But here's the smart way: factor first, cancel common factors, THEN multiply—it keeps the numbers smaller and the result already simplified! To multiply, combine: 3⋅x2x⋅(x+2)=3x2x2+2x\frac{3 \cdot x^2}{x \cdot (x+2)} = \frac{3x^2}{x^2 + 2x}, then cancel a common x from numerator and denominator (factoring x from denominator): 3xx+2\frac{3x}{x + 2}. Choice A correctly performs the multiplication and simplifies to 3xx+2\frac{3x}{x+2} by canceling the common x factor. Great work! For example, choice B might be the product before simplifying, but don't forget to cancel common factors afterward—it's like 23⋅34=612=12\frac{2}{3} \cdot \frac{3}{4} = \frac{6}{12} = \frac{1}{2}, not stopping at 612\frac{6}{12}. The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. This prevents working with huge expressions and catches opportunities to simplify. Example: x2−4x⋅xx+2\frac{x^2 - 4}{x} \cdot \frac{x}{x+2} looks messy, but factor x2−4x^2 - 4 to get (x+2)(x−2)x⋅xx+2\frac{(x+2)(x-2)}{x} \cdot \frac{x}{x+2}, cancel the (x+2)(x+2) and xx, leaving just x−2x-2. So much cleaner!

Question 5

Add and simplify (the sum of two rational expressions is a rational expression): 3x+2x+1\frac{3}{x}+\frac{2}{x+1}

  1. 5x(x+1)\frac{5}{x(x+1)}
  2. 5x+3x(x+1)\frac{5x+3}{x(x+1)} (correct answer)
  3. 5x+2x(x+1)\frac{5x+2}{x(x+1)}
  4. 3x+1+2x\frac{3}{x+1}+\frac{2}{x}
Explanation: This question tests your understanding of how to add rational expressions—algebraic fractions that work just like regular fractions but with variables. Adding or subtracting rational expressions requires a common denominator, just like with 1/2 + 1/3: we find the LCD (least common denominator), rewrite each fraction with that denominator, then add or subtract the numerators while keeping the denominator the same. The tricky part is finding the LCD when denominators have variables! For 3/x + 2/(x+1), the LCD is x(x+1) since these factors share no common factors. We rewrite: 3/x = 3(x+1)/[x(x+1)] = (3x+3)/[x(x+1)] and 2/(x+1) = 2x/[x(x+1)]. Now we can add: (3x+3)/[x(x+1)] + 2x/[x(x+1)] = (3x+3+2x)/[x(x+1)] = (5x+3)/[x(x+1)]. Choice B correctly adds the fractions and simplifies to (5x+3)/[x(x+1)] by finding the LCD and combining numerators. Great work! Choice A incorrectly adds to get just 5 in the numerator, C has an arithmetic error getting 5x+2, and D shows the original unsimplified expression. For adding/subtracting: (1) Factor all denominators to see what you're working with, (2) Find LCD by taking each factor to its highest power, (3) Multiply numerator and denominator of each fraction by what's needed to get LCD, (4) Add/subtract numerators, (5) Simplify if possible. It's exactly like 1/6 + 1/4: LCD = 12, rewrite as 2/12 + 3/12 = 5/12, just with variables!

Question 6

3x−1−1x+1\frac{3}{x-1} - \frac{1}{x+1}

  1. 2x+4x2−1\frac{2x+4}{x^2-1} (correct answer)
  2. 4x2−1\frac{4}{x^2-1}
  3. 2x2−1\frac{2}{x^2-1}
  4. 2x+2x2−1\frac{2x+2}{x^2-1}
Explanation: This question tests your understanding of how to subtract rational expressions—algebraic fractions that work just like regular fractions but with variables. Adding or subtracting rational expressions requires a common denominator, just like with 12+13\frac{1}{2} + \frac{1}{3}: we find the LCD (least common denominator), rewrite each fraction with that denominator, then add or subtract the numerators while keeping the denominator the same. The tricky part is finding the LCD when denominators have variables! The denominators are (x−1)(x-1) and (x+1)(x+1), so LCD is (x−1)(x+1)=x2−1(x-1)(x+1) = x^2 - 1; rewrite as [3(x+1)−1(x−1)]/(x2−1)=(3x+3−x+1)/(x2−1)=(2x+4)/(x2−1)[3(x+1) - 1(x-1)] / (x^2 - 1) = (3x + 3 - x + 1) / (x^2 - 1) = (2x + 4) / (x^2 - 1), which can be left as is or factored but doesn't simplify further. Choice B correctly performs the subtraction and simplifies to (2x+4)/(x2−1)(2x+4)/(x^2-1) by finding the LCD and combining numerators. Great work! For example, choice A might come from incorrectly adding instead of subtracting or mishandling numerators, but double-check the signs when subtracting—it's like 34−15=(15−4)/20=11/20\frac{3}{4} - \frac{1}{5} = (15-4)/20 = 11/20, and practice makes it easier! For adding/subtracting: (1) Factor all denominators to see what you're working with, (2) Find LCD by taking each factor to its highest power, (3) Multiply numerator and denominator of each fraction by what's needed to get LCD, (4) Add/subtract numerators, (5) Simplify if possible. It's exactly like 16+14\frac{1}{6} + \frac{1}{4}: LCD = 12, rewrite as 2/12+3/12=5/122/12 + 3/12 = 5/12, just with variables! Common mistake: trying to cancel before getting common denominator in addition. You can only cancel FACTORS (things being multiplied), not TERMS (things being added). So in [2/x]+[3/x][2/x] + [3/x], you cannot cancel the x's—you can only add numerators because denominators are already the same: (2+3)/x=5/x(2+3)/x = 5/x.

Question 7

Perform the indicated operation and simplify: 3x⋅x2x+2\frac{3}{x}\cdot\frac{x^2}{x+2}

  1. 3xx+2\frac{3x}{x+2} (correct answer)
  2. 3x2x+2\frac{3x^2}{x+2}
  3. 3x+2\frac{3}{x+2}
  4. 3xx(x+2)\frac{3x}{x(x+2)}
Explanation: This question tests your understanding of how to multiply rational expressions—algebraic fractions that work just like regular fractions but with variables. Multiplying rational expressions works just like multiplying numeric fractions: multiply the numerators together and multiply the denominators together, giving (a/b)·(c/d) = (ac)/(bd). But here's the smart way: factor first, cancel common factors, THEN multiply—it keeps the numbers smaller and the result already simplified! For 3/x · x²/(x+2), we multiply numerators to get 3·x² = 3x² and denominators to get x·(x+2) = x(x+2), giving us 3x²/[x(x+2)]. Notice we can cancel one factor of x from both numerator and denominator: 3x²/[x(x+2)] = 3x/(x+2). Choice A correctly shows 3x/(x+2) as the simplified result. Great work! Choice B shows 3x²/(x+2) without canceling the common factor of x—always simplify by canceling common factors. The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. Example: [(x²-4)/x]·[x/(x+2)] looks messy, but factor x²-4 to get [(x+2)(x-2)/x]·[x/(x+2)], cancel the (x+2) and x, leaving just (x-2). So much cleaner!

Question 8

Add and simplify (remember: rational expressions are closed under addition, so your result should be a rational expression in simplest form): 1x+2x+1\frac{1}{x}+\frac{2}{x+1}

  1. 32x+1\frac{3}{2x+1}
  2. 3x+1\frac{3}{x+1}
  3. 3x(x+1)\frac{3}{x(x+1)}
  4. 3x+1x(x+1)\frac{3x+1}{x(x+1)} (correct answer)
Explanation: This question tests your understanding of how to add rational expressions—algebraic fractions that work just like regular fractions but with variables. Adding or subtracting rational expressions requires a common denominator, just like with 1/2 + 1/3: we find the LCD (least common denominator), rewrite each fraction with that denominator, then add or subtract the numerators while keeping the denominator the same. The tricky part is finding the LCD when denominators have variables! For 1/x + 2/(x+1), the denominators are x and (x+1). Since these share no common factors, the LCD is their product: x(x+1). We rewrite: 1/x = (x+1)/[x(x+1)] and 2/(x+1) = 2x/[x(x+1)]. Now we can add: (x+1)/[x(x+1)] + 2x/[x(x+1)] = (x+1+2x)/[x(x+1)] = (3x+1)/[x(x+1)]. Choice D correctly shows (3x+1)/[x(x+1)] as the final answer. Great work! Choice A shows 3/(2x+1)—remember we need to find a common denominator first, not just add the numerators. For adding/subtracting: (1) Factor all denominators to see what you're working with, (2) Find LCD by taking each factor to its highest power, (3) Multiply numerator and denominator of each fraction by what's needed to get LCD, (4) Add/subtract numerators, (5) Simplify if possible. It's exactly like 1/6 + 1/4: LCD = 12, rewrite as 2/12 + 3/12 = 5/12, just with variables!

Question 9

Subtract and simplify (the difference of two rational expressions is a rational expression): xx−2−3x−2\frac{x}{x-2}-\frac{3}{x-2}

  1. x−3x−2\frac{x-3}{x-2} (correct answer)
  2. x−1x−2\frac{x-1}{x-2}
  3. x−3x\frac{x-3}{x}
  4. xx−2−3x\frac{x}{x-2}-\frac{3}{x}
Explanation: This question tests your understanding of how to subtract rational expressions—algebraic fractions that work just like regular fractions but with variables. Adding or subtracting rational expressions requires a common denominator, just like with 1/2 + 1/3: we find the LCD (least common denominator), rewrite each fraction with that denominator, then add or subtract the numerators while keeping the denominator the same. The tricky part is finding the LCD when denominators have variables! For x/(x-2) - 3/(x-2), we're lucky—the denominators are already the same! Just like 5/7 - 2/7 = 3/7, we subtract the numerators and keep the denominator: x/(x-2) - 3/(x-2) = (x-3)/(x-2). Choice A correctly subtracts the numerators to get (x-3)/(x-2) since the denominators were already the same. Great work! Choice B incorrectly gets x-1 in the numerator, C has the wrong denominator, and D shows the original expression without simplifying. Common mistake: trying to cancel before getting common denominator in addition. You can only cancel FACTORS (things being multiplied), not TERMS (things being added). So in x/(x-2) - 3/(x-2), you cannot cancel anything—you can only subtract numerators because denominators are already the same: (x-3)/(x-2). When denominators match, it's just like regular fraction subtraction!

Question 10

Perform the indicated operation and simplify (the product is a rational expression): 3x⋅x2x+2\frac{3}{x}\cdot\frac{x^2}{x+2}

  1. 3x2x(x+2)\frac{3x^2}{x(x+2)}
  2. 3x+2\frac{3}{x+2}
  3. 3xx+2\frac{3x}{x+2} (correct answer)
  4. 3x2x+2\frac{3x^2}{x+2}
Explanation: This question tests your understanding of how to multiply rational expressions—algebraic fractions that work just like regular fractions but with variables. Multiplying rational expressions works just like multiplying numeric fractions: multiply the numerators together and multiply the denominators together, giving (a/b)·(c/d) = (ac)/(bd). But here's the smart way: factor first, cancel common factors, THEN multiply—it keeps the numbers smaller and the result already simplified! Let's multiply: (3/x) · (x²/(x+2)) = (3·x²)/(x·(x+2)) = 3x²/[x(x+2)]. Now we can simplify by canceling one factor of x from numerator and denominator: 3x²/[x(x+2)] = 3x/(x+2). Choice C correctly multiplies and simplifies to 3x/(x+2) by multiplying numerators and denominators, then canceling the common factor of x. Great work! Choice A shows the unsimplified form, choice B cancels too many x's (you can only cancel one!), and choice D has the wrong denominator. The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. This prevents working with huge expressions and catches opportunities to simplify. Example: [(x²-4)/x]·[x/(x+2)] looks messy, but factor x²-4 to get [(x+2)(x-2)/x]·[x/(x+2)], cancel the (x+2) and x, leaving just (x-2). So much cleaner!

Question 11

Perform the indicated operation and simplify: 3x⋅x2x+2\frac{3}{x} \cdot \frac{x^2}{x+2}

  1. 3xx+2\frac{3x}{x+2} (correct answer)
  2. 3x2x+2\frac{3x^2}{x+2}
  3. 3x+2\frac{3}{x+2}
  4. x+23x\frac{x+2}{3x}
Explanation: This question tests your understanding of how to multiply rational expressions—algebraic fractions that work just like regular fractions but with variables. Multiplying rational expressions works just like multiplying numeric fractions: multiply the numerators together and multiply the denominators together, giving (ab⋅cd=acbd)(\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}). But here's the smart way: factor first, cancel common factors, THEN multiply—it keeps the numbers smaller and the result already simplified! For (3x⋅x2x+2)(\frac{3}{x} \cdot \frac{x^2}{x+2}), multiply to get (3x2x(x+2))(\frac{3x^2}{x(x+2)}), then cancel one x from numerator and denominator to leave (3xx+2)(\frac{3x}{x+2}). Choice A correctly multiplies and simplifies to (3xx+2)(\frac{3x}{x+2}) by canceling the common x factor. For example, choice B doesn't cancel the x, so it's not fully simplified—look for those common factors! The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. This prevents working with huge expressions and catches opportunities to simplify.

Question 12

Multiply and simplify (factor first, then cancel common factors): x2−1x2+3x+2⋅x+2x−1\frac{x^2-1}{x^2+3x+2}\cdot\frac{x+2}{x-1}

  1. x+1x+2\frac{x+1}{x+2}
  2. x2−1x2+3x+2\frac{x^2-1}{x^2+3x+2}
  3. x−1x+1\frac{x-1}{x+1}
  4. x+1x\frac{x+1}{x} (correct answer)
Explanation: This question tests your understanding of how to multiply rational expressions—algebraic fractions that work just like regular fractions but with variables. Multiplying rational expressions works just like multiplying numeric fractions: multiply the numerators together and multiply the denominators together, giving (a/b)·(c/d) = (ac)/(bd). But here's the smart way: factor first, cancel common factors, THEN multiply—it keeps the numbers smaller and the result already simplified! Let's factor everything: (x²-1)/(x²+3x+2) = [(x+1)(x-1)]/[(x+1)(x+2)] and we're multiplying by (x+2)/(x-1). Now we can see common factors: (x+1) and (x-1) each appear once in a numerator and denominator, as does (x+2). After canceling: [(x+1)(x-1)]/[(x+1)(x+2)] · (x+2)/(x-1) = (x+1)/x. Choice B correctly performs the multiplication and simplifies to (x+1)/x by factoring and canceling all common factors. Great work! Choice A has an extra +2 in the denominator, C and D show different incorrect simplifications. The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. This prevents working with huge expressions and catches opportunities to simplify. Notice how factoring x²+3x+2 as (x+1)(x+2) revealed the (x+1) factor that could cancel—without factoring first, you'd miss this!

Question 13

Add the rational expressions and simplify. (Because rational expressions are closed under addition, the sum is also a rational expression.) 2x−3+5x−3\frac{2}{x-3}+\frac{5}{x-3}

  1. 7x\frac{7}{x}
  2. 7x−3\frac{7}{x-3} (correct answer)
  3. 7x−6\frac{7}{x-6}
  4. 10x−3\frac{10}{x-3}
Explanation: This question tests your understanding of how to add rational expressions—algebraic fractions that work just like regular fractions but with variables. Adding or subtracting rational expressions requires a common denominator, just like with 1/2 + 1/3: we find the LCD (least common denominator), rewrite each fraction with that denominator, then add or subtract the numerators while keeping the denominator the same. The tricky part is finding the LCD when denominators have variables! Since the denominators are already the same (x-3), simply add the numerators: (2 + 5)/(x-3) = 7/(x-3), and it doesn't simplify further. Choice A correctly performs the addition and simplifies to 7/(x-3) by combining numerators over the common denominator. Great work! For example, choice C might come from subtracting 3 from the denominator or something similar, but keep the denominator as is when adding—only the numerators combine, just like 1/5 + 2/5 = 3/5, not changing the 5. For adding/subtracting: (1) Factor all denominators to see what you're working with, (2) Find LCD by taking each factor to its highest power, (3) Multiply numerator and denominator of each fraction by what's needed to get LCD, (4) Add/subtract numerators, (5) Simplify if possible. It's exactly like 1/6 + 1/4: LCD = 12, rewrite as 2/12 + 3/12 = 5/12, just with variables! Common mistake: trying to cancel before getting common denominator in addition. You can only cancel FACTORS (things being multiplied), not TERMS (things being added). So in [2/x] + [3/x], you cannot cancel the x's—you can only add numerators because denominators are already the same: (2+3)/x = 5/x.

Question 14

Perform the indicated operation and simplify (the product is a rational expression, showing closure under multiplication): 3xx2−4⋅x−26\frac{3x}{x^2-4}\cdot\frac{x-2}{6}

  1. x2(x−2)\frac{x}{2(x-2)}
  2. x2(x+2)\frac{x}{2(x+2)} (correct answer)
  3. 3x(x−2)6(x2−4)\frac{3x(x-2)}{6(x^2-4)}
  4. x−22(x+2)\frac{x-2}{2(x+2)}
Explanation: This question tests your understanding of how to multiply rational expressions—algebraic fractions that work just like regular fractions but with variables. Multiplying rational expressions works just like multiplying numeric fractions: multiply the numerators together and multiply the denominators together, giving (a/b)·(c/d) = (ac)/(bd). But here's the smart way: factor first, cancel common factors, THEN multiply—it keeps the numbers smaller and the result already simplified! Let's factor first: x²-4 = (x+2)(x-2). So we have [3x/(x+2)(x-2)]·[(x-2)/6]. Before multiplying, we can cancel: (x-2) appears in both a numerator and denominator, so it cancels. We can also simplify 3/6 = 1/2. This leaves us with [x/(x+2)]·[1/2] = x/[2(x+2)]. Choice B correctly performs the multiplication and simplifies to x/[2(x+2)] by factoring and canceling the (x-2) term. Great work! Choice A incorrectly has (x-2) in the denominator instead of (x+2)—remember x²-4 = (x+2)(x-2), not 2(x-2)! The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. This prevents working with huge expressions and catches opportunities to simplify.

Question 15

Multiply and simplify the rational expressions (the result is also a rational expression): x+3x⋅xx−2\frac{x+3}{x} \cdot \frac{x}{x-2}

  1. x(x+3)x−2\frac{x(x+3)}{x-2}
  2. x+3x−2\frac{x+3}{x-2} (correct answer)
  3. x+3x(x−2)\frac{x+3}{x(x-2)}
  4. x−2x+3\frac{x-2}{x+3}
Explanation: This question tests your understanding of how to multiply rational expressions—algebraic fractions that work just like regular fractions but with variables. Multiplying rational expressions works just like multiplying numeric fractions: multiply the numerators together and multiply the denominators together, giving ab⋅cd=acbd\frac{a}{b} \cdot \frac{c}{d} = \frac{a c}{b d}. But here's the smart way: factor first, cancel common factors, THEN multiply—it keeps the numbers smaller and the result already simplified! For x+3x⋅xx−2\frac{x+3}{x} \cdot \frac{x}{x-2}, the x in the numerator of the second fraction cancels with the x in the denominator of the first, leaving x+3x−2\frac{x+3}{x-2}. Choice B correctly performs the multiplication and simplifies to x+3x−2\frac{x+3}{x-2} by canceling the common x factors. For example, choice A forgets to cancel the common x, resulting in an unsimplified expression—remember to always look for cancellations after multiplying! The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. This prevents working with huge expressions and catches opportunities to simplify.

Question 16

Multiply and simplify the rational expressions (the result is also a rational expression): x+3x⋅xx−2\frac{x+3}{x}\cdot\frac{x}{x-2}

  1. x+3x(x−2)\frac{x+3}{x(x-2)}
  2. x−2x+3\frac{x-2}{x+3}
  3. x+3x−2\frac{x+3}{x-2} (correct answer)
  4. x(x+3)x−2\frac{x(x+3)}{x-2}
Explanation: This question tests your understanding of how to multiply rational expressions—algebraic fractions that work just like regular fractions but with variables. Multiplying rational expressions works just like multiplying numeric fractions: multiply the numerators together and multiply the denominators together, giving (a/b)·(c/d) = (ac)/(bd). But here's the smart way: factor first, cancel common factors, THEN multiply—it keeps the numbers smaller and the result already simplified! For (x+3)/x · x/(x-2), we multiply numerators to get x(x+3) and denominators to get x(x-2), giving us x(x+3)/[x(x-2)]. Notice we can cancel the common factor x from both numerator and denominator, leaving (x+3)/(x-2). Choice C correctly shows this simplified result of (x+3)/(x-2). Great work! Choice A shows the unsimplified form with x still in the denominator—remember to always cancel common factors. The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. This prevents working with huge expressions and catches opportunities to simplify.

Question 17

Subtract and simplify (the result should be a simplified rational expression): 4x+2−1x+2\frac{4}{x+2}-\frac{1}{x+2}

  1. 5x+2\frac{5}{x+2}
  2. 32x+2\frac{3}{2x+2}
  3. 3x+2\frac{3}{x+2} (correct answer)
  4. 3x\frac{3}{x}
Explanation: This question tests your understanding of how to subtract rational expressions—algebraic fractions that work just like regular fractions but with variables. Adding or subtracting rational expressions requires a common denominator, just like with 1/2 + 1/3: we find the LCD (least common denominator), rewrite each fraction with that denominator, then add or subtract the numerators while keeping the denominator the same. The tricky part is finding the LCD when denominators have variables! For 4/(x+2) - 1/(x+2), the denominators are already the same—how convenient! Just like 7/9 - 4/9 = 3/9, we subtract the numerators and keep the denominator: 4/(x+2) - 1/(x+2) = (4-1)/(x+2) = 3/(x+2). Choice A correctly subtracts the numerators to get 3/(x+2) since the denominators were already identical. Great work! Choice B has the wrong denominator (just x), C incorrectly adds instead of subtracts to get 5, and D changes the denominator unnecessarily. Common mistake: trying to cancel before getting common denominator in addition. You can only cancel FACTORS (things being multiplied), not TERMS (things being added). When denominators already match, subtraction is super simple—just subtract the numerators like regular fractions!

Question 18

Multiply and simplify the rational expressions (the product of two rational expressions is also a rational expression): x2−9x2−4⋅x+2x−3\frac{x^2-9}{x^2-4}\cdot\frac{x+2}{x-3}

  1. x+3x−2\frac{x+3}{x-2} (correct answer)
  2. x2−9x2−4\frac{x^2-9}{x^2-4}
  3. x2−9(x−2)(x−3)\frac{x^2-9}{(x-2)(x-3)}
  4. x+3x+2\frac{x+3}{x+2}
Explanation: This question tests your understanding of how to multiply rational expressions—algebraic fractions that work just like regular fractions but with variables. Multiplying rational expressions works just like multiplying numeric fractions: multiply the numerators together and multiply the denominators together, giving (a/b)·(c/d) = (ac)/(bd). But here's the smart way: factor first, cancel common factors, THEN multiply—it keeps the numbers smaller and the result already simplified! Let's factor first: (x²-9)/(x²-4) = [(x+3)(x-3)]/[(x+2)(x-2)] and we're multiplying by (x+2)/(x-3). Now we can see that (x-3) appears in both a numerator and denominator, as does (x+2), so they cancel: [(x+3)(x-3)]/[(x+2)(x-2)] · (x+2)/(x-3) = (x+3)/(x-2). Choice A correctly performs the multiplication and simplifies to (x+3)/(x-2) by factoring and canceling the common factors (x-3) and (x+2). Great work! Choice B shows the original first fraction without any multiplication, while C and D have errors in their simplification process. The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. This prevents working with huge expressions and catches opportunities to simplify. Remember, when multiplying fractions, any factor that appears in both a numerator and denominator (anywhere in the problem) can be canceled—it's like reducing 6/8 · 4/3 by noticing the 3 and 4 can cancel across fractions!

Question 19

Multiply and simplify: x2−1x2−4⋅x+2x+1\frac{x^2-1}{x^2-4}\cdot\frac{x+2}{x+1}

  1. x−1x−2\frac{x-1}{x-2} (correct answer)
  2. x+1x+2\frac{x+1}{x+2}
  3. x2−1x2−4\frac{x^2-1}{x^2-4}
  4. x+2x−2\frac{x+2}{x-2}
Explanation: This question tests your understanding of how to multiply rational expressions—algebraic fractions that work just like regular fractions but with variables. Multiplying rational expressions works just like multiplying numeric fractions: multiply the numerators together and multiply the denominators together, giving (ab)⋅(cd)=acbd(\frac{a}{b}) \cdot (\frac{c}{d}) = \frac{ac}{bd}. But here's the smart way: factor first, cancel common factors, THEN multiply—it keeps the numbers smaller and the result already simplified! For (x2−1x2−4⋅x+2x+1)(\frac{x^2-1}{x^2-4} \cdot \frac{x+2}{x+1}), factor to ((x−1)(x+1)(x−2)(x+2)⋅x+2x+1)(\frac{(x-1)(x+1)}{(x-2)(x+2)} \cdot \frac{x+2}{x+1}), then cancel (x+1)(x+1) and (x+2)(x+2) to leave (x−1x−2)(\frac{x-1}{x-2}). Choice A correctly multiplies and simplifies to (x−1x−2)(\frac{x-1}{x-2}) by factoring and canceling common factors. For example, choice C doesn't factor or cancel fully—always factor differences of squares! The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. This prevents working with huge expressions and catches opportunities to simplify.

Question 20

Add the rational expressions (polynomials divided by polynomials) and simplify. (The sum is a rational expression because the set is closed under addition.) 1x+2x+1\frac{1}{x}+\frac{2}{x+1}

  1. 3x+1x(x+1)\frac{3x+1}{x(x+1)} (correct answer)
  2. x+2x(x+1)\frac{x+2}{x(x+1)}
  3. 32x+1\frac{3}{2x+1}
  4. 3x(x+1)\frac{3}{x(x+1)}
Explanation: This question tests your understanding of how to add rational expressions—algebraic fractions that work just like regular fractions but with variables. Adding or subtracting rational expressions requires a common denominator, just like with 1/2+1/31/2 + 1/3: we find the LCD (least common denominator), rewrite each fraction with that denominator, then add or subtract the numerators while keeping the denominator the same. The tricky part is finding the LCD when denominators have variables! The denominators are xx and (x+1)(x+1), so LCD is x(x+1)x(x+1); rewrite as 1⋅(x+1)+2⋅xx(x+1)=x+1+2xx(x+1)=3x+1x(x+1)\frac{1 \cdot (x+1) + 2 \cdot x}{x(x+1)} = \frac{x+1 + 2x}{x(x+1)} = \frac{3x+1}{x(x+1)}, which doesn't simplify further. Choice B correctly performs the addition and simplifies to 3x+1x(x+1)\frac{3x+1}{x(x+1)} by finding the LCD and combining numerators. Great work! For example, choice C might result from adding numerators without the common denominator properly, like just 1+2=3 over something, but always build up to the LCD first—it's a key step, and you're doing great by practicing it. For adding/subtracting: (1) Factor all denominators to see what you're working with, (2) Find LCD by taking each factor to its highest power, (3) Multiply numerator and denominator of each fraction by what's needed to get LCD, (4) Add/subtract numerators, (5) Simplify if possible. It's exactly like 1/6+1/41/6 + 1/4: LCD = 12, rewrite as 2/12+3/12=5/122/12 + 3/12 = 5/12, just with variables! Common mistake: trying to cancel before getting common denominator in addition. You can only cancel FACTORS (things being multiplied), not TERMS (things being added). So in [2/x]+[3/x][2/x] + [3/x], you cannot cancel the x's—you can only add numerators because denominators are already the same: (2+3)/x=5/x(2+3)/x = 5/x.