This question tests your understanding of how to simplify rational expressions—algebraic fractions that work just like regular fractions but with variables. Rational expressions form a system like rational numbers: they're closed under addition, subtraction, multiplication, and division (as long as we don't divide by zero). This means doing any of these operations on rational expressions gives you another rational expression—the result is still a fraction of polynomials! To simplify (x²-5x+6)/(x²-9), we need to factor both numerator and denominator. The numerator x²-5x+6 factors as (x-2)(x-3), and the denominator x²-9 is a difference of squares: (x+3)(x-3). So we have [(x-2)(x-3)]/[(x+3)(x-3)]. Now we can cancel the common factor (x-3) from numerator and denominator, leaving us with (x-2)/(x+3). Choice A correctly simplifies to (x-2)/(x+3) by factoring and canceling the common factor (x-3). Great work! Choice D shows the original expression without canceling—remember, always simplify by canceling 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. Even when just simplifying a single fraction, factoring is your best friend!

This question tests your understanding of how to simplify rational expressions—algebraic fractions that work just like regular fractions but with variables. Rational expressions form a system like rational numbers: they're closed under addition, subtraction, multiplication, and division (as long as we don't divide by zero). This means doing any of these operations on rational expressions gives you another rational expression—the result is still a fraction of polynomials! To simplify $\frac{x^2-9}{x+3}$, factor the numerator as $(x-3)(x+3)$, then cancel the common $(x+3)$ factor to get $x-3$. Choice B correctly simplifies to $x-3$ by factoring and canceling the common factor. For instance, choice A doesn't factor or cancel, leaving it unsimplified—always factor to spot cancellations! 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.

This question tests your understanding of how to simplify rational expressions—algebraic fractions that work just like regular fractions but with variables. Rational expressions form a system like rational numbers: they're closed under addition, subtraction, multiplication, and division (as long as we don't divide by zero). This means doing any of these operations on rational expressions gives you another rational expression—the result is still a fraction of polynomials! To simplify, factor numerator and denominator: $x^2 - 16 = (x-4)(x+4)$, so $\frac{(x-4)(x+4)}{x+4}$, then cancel the common factor $(x+4)$, leaving $x-4$ (assuming $x \neq -4$ to avoid division by zero). Choice B correctly simplifies to $x-4$ by factoring and canceling the common factor. Great work! For example, choice C is the original expression without simplifying, but always check for factoring opportunities—it's like reducing 4/6 to 2/3 by canceling 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. Example: $[(x^2-4)/x] \cdot[x/(x+2)]$ looks messy, but factor $x^2-4$ to get $[(x+2)(x-2)/x] \cdot[x/(x+2)]$, cancel the $(x+2)$ and $x$, leaving just $(x-2)$. So much cleaner!

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 $ \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 $$ \frac{x+3}{x} \cdot \frac{x}{x-2} $$, the x in the numerator of the second fraction cancels with the x in the denominator of the first, leaving $ \frac{x+3}{x-2} $. Choice B correctly performs the multiplication and simplifies to $ \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.

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 work with x/(x+3) · (x²-9)/x. First, factor x²-9 = (x+3)(x-3). So we have: x/(x+3) · [(x+3)(x-3)]/x. Now we can see that x appears in both a numerator and denominator, so it cancels. Also, (x+3) appears in both a numerator and denominator, so it cancels too. After canceling: we're left with just (x-3), which equals x-3. Choice B correctly performs the multiplication and simplifies to x-3 by factoring x²-9 as (x+3)(x-3) and canceling the common factors x and (x+3). Great work! Choice A leaves it as a fraction, C gives the wrong factor, and D has the wrong denominator. The golden rule for multiplying and dividing rationals: factor everything you can BEFORE you multiply or cancel. Notice how factoring x²-9 as a difference of squares revealed both factors we needed to cancel—this is why factoring first is so powerful!

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 $\frac{1}{2} + \frac{1}{3}$: 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 $(x-1)$ and $(x+1)$, so LCD is $(x-1)(x+1) = x^2 - 1$; rewrite as $[3(x+1) - 1(x-1)] / (x^2 - 1) = (3x + 3 - x + 1) / (x^2 - 1) = (2x + 4) / (x^2 - 1)$, which can be left as is or factored but doesn't simplify further. Choice B correctly performs the subtraction and simplifies to $(2x+4)/(x^2-1)$ 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 $\frac{3}{4} - \frac{1}{5} = (15-4)/20 = 11/20$, 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 $\frac{1}{6} + \frac{1}{4}$: 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: $(2+3)/x = 5/x$.

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!

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!

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!

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-1) - 2/(x-1), we're lucky—the denominators are already the same! When denominators match, we simply subtract the numerators: (x-2)/(x-1). No need to find an LCD or rewrite anything. Choice A correctly subtracts the fractions to get (x-2)/(x-1) by subtracting numerators while keeping the common denominator. Great work! Choice B incorrectly adds instead of subtracts, choice C reverses the subtraction order giving (2-x)/(x-1), and choice D incorrectly cancels the denominator. 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.