Manipulate Algebraic Expressions - GMAT Quantitative Reasoning

1. For all $x \neq \pm 2$, which of the following is equivalent to the expression $\dfrac{2x^{2}-8}{x^{2}-4}$?
   Answer: 2 When you encounter a rational expression like this, your first instinct should be to factor both the numerator and denominator to look for common factors that can be canceled. Let's factor each part systematically. The numerator $2x^2 - 8$ has a common factor of 2, so we can write it as $2(x^2 - 4)$. The denominator $x^2 - 4$ is already in its simplest form, but notice it's a difference of squares: $x^2 - 4 = (x+2)(x-2)$. So our expression becomes: $\dfrac{2(x^2-4)}{x^2-4}$ Since $x^2 - 4$ appears in both numerator and denominator, and we're told $x \neq \pm 2$ (which ensures $x^2 - 4 \neq 0$), we can cancel these terms: $\dfrac{2(x^2-4)}{x^2-4} = 2$ Choice A ($2x-2$) represents what you might get if you incorrectly tried to distribute or factor incorrectly. Choice B ($2x$) could result from mistakenly canceling just the constant terms. Choice D ($2x+2$) might come from confusing the signs during factoring or attempting polynomial long division incorrectly. The key insight is recognizing that after proper factoring, the variable terms completely cancel out, leaving only the constant 2. Strategy tip: When simplifying rational expressions, always factor completely before attempting to cancel. Look especially for difference of squares patterns like $a^2 - b^2 = (a+b)(a-b)$, and remember that domain restrictions (like $x \neq \pm 2$) are given precisely because those values would make denominators zero.
2. If $x\neq 3,-3$, the product $\dfrac{x^{2}-9}{x^{2}-6x+9}\cdot\dfrac{x-3}{x+3}$ is equal to
   Answer: 1 When you encounter algebraic fractions like this, your first instinct should be to factor everything completely before attempting to multiply. This allows you to identify and cancel common factors. Let's factor each expression systematically. The numerator $x^2-9$ is a difference of squares: $(x-3)(x+3)$. The denominator $x^2-6x+9$ is a perfect square trinomial: $(x-3)^2$. The second fraction is already factored. Now substitute the factored forms: $\frac{(x-3)(x+3)}{(x-3)^2} \cdot \frac{x-3}{x+3}$ Multiply the fractions by combining numerators and denominators: $\frac{(x-3)(x+3)(x-3)}{(x-3)^2(x+3)}$ This simplifies to: $\frac{(x-3)^2(x+3)}{(x-3)^2(x+3)}$ Since $x \neq 3$ and $x \neq -3$, we can cancel the common factors $(x-3)^2$ and $(x+3)$, leaving us with $\frac{1}{1} = 1$. The answer is B. Looking at the wrong choices: A) suggests the product equals zero, which would only happen if a numerator factor couldn't be canceled. C) $x-3$ would result from incomplete cancellation. D) $\frac{x-3}{x+3}$ represents what you'd get if you only canceled one factor of $(x-3)$ instead of recognizing the complete cancellation. Strategy tip: Always factor completely before multiplying algebraic fractions. The GMAT often tests whether you can spot perfect squares and differences of squares, then cancel appropriately while respecting domain restrictions.
3. If $\frac{x}{y} + \frac{y}{x} = 5$ and $xy \neq 0$, then $\frac{x^2 + y^2}{xy}$ equals:
   Answer: 5 When you encounter algebraic expressions involving fractions with variables in both numerator and denominator, look for opportunities to manipulate the given equation to match the target expression's structure. Given $\frac{x}{y} + \frac{y}{x} = 5$, you want to find $\frac{x^2 + y^2}{xy}$. The key insight is recognizing that these expressions are closely related. To connect them, multiply both sides of the given equation by $xy$: $xy \cdot \left(\frac{x}{y} + \frac{y}{x}\right) = 5xy$ Distributing on the left side: $xy \cdot \frac{x}{y} + xy \cdot \frac{y}{x} = 5xy$ This simplifies to: $x^2 + y^2 = 5xy$ Now divide both sides by $xy$: $\frac{x^2 + y^2}{xy} = 5$ The answer is (D) 5. Let's examine why the other choices are incorrect. Choice (A) 3 might tempt students who mistakenly subtract terms instead of properly manipulating the equation. Choice (B) 25 could result from incorrectly squaring the entire given equation rather than multiplying by $xy$. Choice (C) 10 might come from doubling the given value of 5, perhaps from confusion about the relationship between the expressions. Strategy tip: When dealing with rational expressions involving sums or differences of fractions, try multiplying through by a common denominator (in this case $xy$) to eliminate fractions and reveal the underlying algebraic relationship. This technique frequently appears on GMAT quantitative questions.
4. For $x\neq -2$, simplify $\dfrac{x^{3}+8}{x^{2}-2x+4}.$
   Answer: $x+2$ When you encounter a rational expression like this, look for opportunities to factor and simplify. The key insight is recognizing that the numerator $x^3 + 8$ is a sum of cubes, which follows the pattern $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$. Since $x^3 + 8 = x^3 + 2^3$, we can factor this as: $x^3 + 8 = (x + 2)(x^2 - 2x + 4)$ Now our expression becomes: $\frac{x^3 + 8}{x^2 - 2x + 4} = \frac{(x + 2)(x^2 - 2x + 4)}{x^2 - 2x + 4}$ Since the factor $(x^2 - 2x + 4)$ appears in both numerator and denominator, we can cancel it out, leaving us with $x + 2$. This confirms answer choice D. Choice A $\frac{x+2}{x^2-2x+4}$ represents what you'd get if you factored the numerator correctly but forgot to simplify by canceling the common factor. Choice B $x-2$ likely comes from incorrectly factoring $x^3 + 8$ as $(x - 2)$ times something, confusing the sum of cubes with a difference. Choice C $x^2 - 4$ might result from misremembering factoring formulas or incorrectly applying difference of squares patterns. Strategy tip: Memorize the sum and difference of cubes formulas: $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ and $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. These appear frequently on the GMAT, especially in rational expression problems.
5. Combine and simplify: $\dfrac{3}{x}+\dfrac{2}{x+1}$ for $$x\neq 0,-1.$
   Answer: $\dfrac{5x+3}{x(x+1)}$ When you see rational expressions that need to be combined, you're working with fraction addition. The key principle is that you can only add fractions when they have a common denominator. To combine $\frac{3}{x}+\frac{2}{x+1}$, you need to find a common denominator. Since the denominators $x$ and $(x+1)$ share no common factors, the least common denominator is their product: $x(x+1)$. Now rewrite each fraction with this common denominator: - $\frac{3}{x} = \frac{3(x+1)}{x(x+1)} = \frac{3x+3}{x(x+1)}$ - $\frac{2}{x+1} = \frac{2x}{x(x+1)}$ Adding these fractions: $\frac{3x+3}{x(x+1)} + \frac{2x}{x(x+1)} = \frac{3x+3+2x}{x(x+1)} = \frac{5x+3}{x(x+1)}$ This confirms answer choice C is correct. Looking at the wrong answers: Choice A gives $\frac{5x-3}{x(x+1)}$, which suggests incorrectly subtracting 3 instead of adding it. Choice B gives $\frac{5x+5}{x(x+1)}$, which likely comes from the error $2 \cdot (x+1) = 2x+2$ instead of $2 \cdot x = 2x$. Choice D gives $\frac{6x+3}{x(x+1)}$, which results from incorrectly combining the $x$-terms as $3x+2x+x = 6x$. Strategy tip: When adding rational expressions, always multiply each fraction by the "missing factors" to create the common denominator, then carefully distribute and combine like terms in the numerator.
6. For $x\neq 0,-7$, simplify $$\dfrac{(x^{2}+3x)^{2}-16x^{2}}{x(x+3)+4x}.$
   Answer: $x^{2}-x$ When you encounter complex rational expressions like this, your goal is to simplify by factoring and canceling common terms. The key is recognizing algebraic patterns, particularly difference of squares. Let's start with the numerator: $(x^2+3x)^2-16x^2$. Notice this follows the difference of squares pattern $a^2-b^2 = (a+b)(a-b)$, where $a = x^2+3x$ and $b = 4x$. So: $(x^2+3x)^2-16x^2 = (x^2+3x+4x)(x^2+3x-4x) = (x^2+7x)(x^2-x)$ The denominator is: $x(x+3)+4x = x^2+3x+4x = x^2+7x$ Now our expression becomes: $\frac{(x^2+7x)(x^2-x)}{x^2+7x}$ Since $x^2+7x$ appears in both numerator and denominator, we can cancel it (noting the restriction $x \neq 0, -7$ ensures this term isn't zero). This leaves us with $x^2-x$. Answer choice A ($x^2+7x$) represents what you'd get if you incorrectly canceled $(x^2-x)$ instead of $(x^2+7x)$. Choice B ($x^2+x$) comes from a sign error when factoring the difference of squares. Choice C ($x^2-7x$) results from incorrectly distributing the negative sign in the factorization. The correct answer is D: $x^2-x$. Strategy tip: Always look for difference of squares patterns ($a^2-b^2$) in GMAT algebra problems. Factor completely before attempting to cancel terms, and double-check your signs throughout the process.
7. If $\frac{1}{1+x} + \frac{1}{1-x} = \frac{k}{1-x^2}$ for all valid values of $x$, then $k$ equals:
   Answer: 2 To find $k$, combine the fractions on the left side: $\frac{1}{1+x} + \frac{1}{1-x} = \frac{(1-x) + (1+x)}{(1+x)(1-x)} = \frac{2}{1-x^2}$. Comparing with $\frac{k}{1-x^2}$, we see that $k = 2$. Choice A results from incorrectly combining the numerators as $(1-x) + (1+x) = 1$. Choice C suggests $k$ depends on $x$, which would mean the equation isn't an identity. Choice D confuses the numerator with the denominator.
8. Let $k=\dfrac{(x+1)^{2}-(x-1)^{2}}{x}$ for all nonzero $x$. What is the value of $k$?
   Answer: 4 When you encounter algebraic expressions with squared binomials like $(x+1)^2$ and $(x-1)^2$, look for opportunities to use the difference of squares pattern or expand strategically to simplify. Let's work through this step-by-step by expanding the numerator. First, expand each squared term: - $(x+1)^2 = x^2 + 2x + 1$ - $(x-1)^2 = x^2 - 2x + 1$ Now substitute these into the original expression: $k = \frac{(x^2 + 2x + 1) - (x^2 - 2x + 1)}{x}$ Simplify the numerator by distributing the negative sign and combining like terms: $k = \frac{x^2 + 2x + 1 - x^2 + 2x - 1}{x} = \frac{4x}{x}$ Since $x \neq 0$, we can cancel the $x$ terms: $k = 4$ The answer is (B) 4. Now let's examine why the other choices are wrong: - (A) 2: This would result from incorrectly calculating the coefficient when the binomials are expanded, possibly forgetting to account for both $+2x$ terms. - **(C) $2x$**: This error occurs if you correctly find $4x$ in the numerator but forget to divide by $x$. - **(D) $4x$**: This is the numerator before simplification—you must complete the division by $x$. Study tip: When you see expressions involving $(a+b)^2 - (a-b)^2$, remember this always simplifies to $4ab$. This pattern can save time on similar problems.
9. If $P = \frac{x^2 - 1}{x + 1}$ and $Q = \frac{x^2 + 2x + 1}{x + 1}$, then $P - Q$ equals: Give your answer in simplest form.
   Answer: $-2$ When you encounter rational expressions that need to be subtracted, the key is to work methodically with common denominators and algebraic simplification. To find $P - Q$, you need to subtract the fractions: $\frac{x^2 - 1}{x + 1} - \frac{x^2 + 2x + 1}{x + 1}$. Since both fractions already have the same denominator $(x + 1)$, you can subtract the numerators directly: $P - Q = \frac{(x^2 - 1) - (x^2 + 2x + 1)}{x + 1}$. Expanding the numerator: $(x^2 - 1) - (x^2 + 2x + 1) = x^2 - 1 - x^2 - 2x - 1 = -2x - 2$. This gives you $\frac{-2x - 2}{x + 1}$. Now factor the numerator: $-2x - 2 = -2(x + 1)$. So $P - Q = \frac{-2(x + 1)}{x + 1} = -2$ (assuming $x \neq -1$). Choice A gives $-2(x + 1)$, which represents the numerator before canceling the common factor. This suggests you stopped one step short of full simplification. Choice B gives $\frac{-2x - 2}{x + 1}$, which is correct before factoring and canceling. This is the intermediate step, not the final simplified form. Choice D gives $2$, indicating a sign error when distributing the negative through the second numerator. The correct answer is C: $-2$. Strategy tip: Always check if you can factor and cancel common terms when working with rational expressions. The GMAT often tests whether you'll recognize when expressions can be simplified beyond their initial form.
10. If $a = \frac{2x}{x-1}$ and $b = \frac{x+1}{x-1}$, then $\frac{a-b}{a+b}$ equals:
    Answer: $\frac{x-1}{3x+1}$ First find $a-b = \frac{2x}{x-1} - \frac{x+1}{x-1} = \frac{2x-(x+1)}{x-1} = \frac{x-1}{x-1} = 1$. Next find $a+b = \frac{2x}{x-1} + \frac{x+1}{x-1} = \frac{2x+(x+1)}{x-1} = \frac{3x+1}{x-1}$. Therefore $\frac{a-b}{a+b} = \frac{1}{\frac{3x+1}{x-1}} = \frac{x-1}{3x+1}$. Choice B results from sign errors in computing $a+b$. Choice C results from incorrectly computing $a+b = 3x-1$ instead of $3x+1$. Choice D combines errors from both the numerator and denominator calculations.
11. Suppose $(p x+4)(x-2)+(2-x)(x+2)$ is written as $ax^{2}+bx+c.\$ If$a=7,$what is the value of$p?$$
    Answer: 8 When you encounter polynomial expansion problems where you need to find an unknown coefficient, your approach should be to expand the given expression and match coefficients with the target form. Let's expand $(px+4)(x-2)+(2-x)(x+2)$ step by step. First, expand $(px+4)(x-2)$: $(px+4)(x-2) = px^2 - 2px + 4x - 8$ Next, expand $(2-x)(x+2)$: $(2-x)(x+2) = 2x + 4 - x^2 - 2x = 4 - x^2$ Combining both parts: $px^2 - 2px + 4x - 8 + 4 - x^2$ $= (p-1)x^2 + (-2p+4)x + (-4)$ Since this equals $ax^2 + bx + c$ and we're told $a = 7$, we need the coefficient of $x^2$ to equal 7: $p - 1 = 7$ $p = 8$ Choice (A) 5 would give us $a = 5 - 1 = 4$, not 7. Choice (B) 6 would give us $a = 6 - 1 = 5$. Choice (C) 7 would give us $a = 7 - 1 = 6$. Only choice (D) 8 gives us the required $a = 8 - 1 = 7$. Study tip: In coefficient-matching problems, expand systematically and collect like terms carefully. Always double-check your algebra by substituting your answer back into the original constraint—here, verify that $p = 8$ indeed produces $a = 7$.
12. For $x\neq 3$, the expression $\dfrac{x^{3}-27}{x^{2}+3x+9}$ simplifies to
    Answer: $x-3$ When you encounter a rational expression that needs simplification, look for opportunities to factor and cancel common terms. This question tests your ability to recognize and apply the difference of cubes formula. The key insight is recognizing that $x^3 - 27$ is a difference of cubes, since $27 = 3^3$. The difference of cubes formula states that $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$. Applying this with $a = x$ and $b = 3$: $x^3 - 27 = x^3 - 3^3 = (x-3)(x^2 + 3x + 9)$ Now you can substitute this factorization into the original expression: $\frac{x^3-27}{x^2+3x+9} = \frac{(x-3)(x^2+3x+9)}{x^2+3x+9}$ Since $x \neq 3$, the factor $(x^2+3x+9)$ is never zero and can be canceled from numerator and denominator, leaving $x-3$. Choice A, $\frac{x-3}{x^2+3x+9}$, represents the expression before canceling the common factor. Choice B, $x+3$, might result from incorrectly applying the difference of squares formula instead of difference of cubes. Choice C, $x^2-9$, could come from confusing this with factoring $x^2-9 = (x-3)(x+3)$ and forgetting about the cubic terms. Remember that difference of cubes problems appear frequently on the GMAT. Memorize the formula $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ and always check whether rational expressions can be simplified by factoring and canceling common terms.
13. If $\frac{1}{x-2} + \frac{1}{x+2} = \frac{2x}{x^2-4}$, then which statement about $x$ is correct?
    Answer: The equation is true for all real $x$ except $x = \pm 2$ Start with the left side: $\frac{1}{x-2} + \frac{1}{x+2} = \frac{(x+2) + (x-2)}{(x-2)(x+2)} = \frac{2x}{x^2-4}$. This matches the right side exactly, making it an identity. However, both sides are undefined when $x = \pm 2$ (since $x^2-4 = 0$), so the equation holds for all real $x$ except $x = \pm 2$. Choice B incorrectly thinks we need to solve $2x = 2x$, which would give $0 = 0$ (true for all $x$), not $x = 0$. Choice C incorrectly claims the equation holds at $x = \pm 2$ where it's undefined. Choice D incorrectly suggests we get $0 = 2x$, which would come from a calculation error.
14. Which of the following is equivalent to $(x+2)(x-5)-(x-3)(x+4)?$
    Answer: $2-4x$ When you encounter algebraic expressions with multiple binomial products, your goal is to expand each product using FOIL (First, Outer, Inner, Last), then combine like terms carefully. Let's expand each product systematically. For $(x+2)(x-5)$: multiply $x \cdot x = x^2$, then $x \cdot (-5) = -5x$, then $2 \cdot x = 2x$, and finally $2 \cdot (-5) = -10$. This gives us $x^2 - 5x + 2x - 10 = x^2 - 3x - 10$. For $(x-3)(x+4)$: multiply $x \cdot x = x^2$, then $x \cdot 4 = 4x$, then $(-3) \cdot x = -3x$, and finally $(-3) \cdot 4 = -12$. This gives us $x^2 + 4x - 3x - 12 = x^2 + x - 12$. Now substitute back into the original expression: $(x^2 - 3x - 10) - (x^2 + x - 12)$. Distribute the negative sign: $x^2 - 3x - 10 - x^2 - x + 12$. Combining like terms: the $x^2$ terms cancel, $-3x - x = -4x$, and $-10 + 12 = 2$. The result is $2 - 4x$. Choice A gives $4x + 2$ (wrong sign on the $x$ term), choice B gives $4x - 2$ (wrong signs on both terms), and choice C gives $2 + 4x$ (correct constant but wrong sign on $x$). These errors typically occur from sign mistakes during distribution or combining like terms. Remember: when subtracting polynomials, distribute the negative sign to every term in the second polynomial, then combine like terms methodically to avoid sign errors.
15. For all nonzero $x$, the expression $\dfrac{1}{1-\dfrac{1}{1-\dfrac{1}{x}}}$ is equivalent to
    Answer: $1-x$ When you encounter complex fractions with multiple layers, the key is to simplify from the inside out, working systematically through each nested fraction. Let's start with the innermost fraction and build outward. First, simplify $1-\frac{1}{x}$: $1-\frac{1}{x} = \frac{x}{x} - \frac{1}{x} = \frac{x-1}{x}$ Now we can rewrite the middle fraction: $1-\frac{1}{1-\frac{1}{x}} = 1-\frac{1}{\frac{x-1}{x}}$ When dividing by a fraction, multiply by its reciprocal: $1-\frac{1}{\frac{x-1}{x}} = 1-\frac{x}{x-1}$ Converting to a common denominator: $1-\frac{x}{x-1} = \frac{x-1}{x-1} - \frac{x}{x-1} = \frac{x-1-x}{x-1} = \frac{-1}{x-1}$ Finally, the entire expression becomes: $\frac{1}{1-\frac{1}{1-\frac{1}{x}}} = \frac{1}{\frac{-1}{x-1}} = \frac{1 \cdot (x-1)}{-1} = -(x-1) = 1-x$ The answer is (B) $1-x$. Looking at the wrong choices: (A) $x-1$ is the negative of our answer, likely resulting from a sign error. (C) $\frac{1}{x}-1$ might come from incorrectly simplifying the innermost fraction. (D) $1-\frac{1}{x}$ is just the value of the innermost expression, not the full simplification. Strategy tip: With nested fractions, always work inside-out and double-check your algebra at each step. Sign errors are particularly common in these problems.
16. When $(2x-3)^{3}+(3x-2)^{3}$ is expanded and like terms are combined, what is the coefficient of $x^{2}?$
    Answer: -90 When you encounter polynomial expansion problems involving cubes, the key is recognizing that you don't need to fully expand everything—focus only on the terms that will contribute to the coefficient you're seeking. To find the coefficient of $x^2$ in $(2x-3)^3+(3x-2)^3$, use the binomial theorem. For $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$, you only need the terms containing $x^2$. For $(2x-3)^3$: The $x^2$ term comes from $3(2x)^2(-3) = 3 \cdot 4x^2 \cdot (-3) = -36x^2$ For $(3x-2)^3$: The $x^2$ term comes from $3(3x)^2(-2) = 3 \cdot 9x^2 \cdot (-2) = -54x^2$ Adding these: $-36x^2 + (-54x^2) = -90x^2$ The coefficient of $x^2$ is $-90$. Answer choice (A) $-36$ represents only the $x^2$ coefficient from the first cubic term, ignoring the second entirely. Answer choice (B) $-54$ captures only the $x^2$ coefficient from the second cubic term, missing the first. Answer choice (D) $90$ gets the magnitude correct but has the wrong sign—a common error when handling multiple negative terms. Strategy tip: When expanding polynomials for specific coefficients, use the binomial theorem selectively. Calculate only the terms you need rather than expanding everything, which saves time and reduces arithmetic errors. Always double-check your signs when combining terms with negatives.
17. Simplify $\dfrac{x^{2}-1}{x-1}-\dfrac{x^{2}-1}{x+1}$ for all $x\neq\pm1.$
    Answer: 2 When you encounter rational expressions with factorizable polynomials, always look for opportunities to factor and simplify before performing operations. This problem tests your ability to factor differences of squares and find common denominators. Start by factoring the numerator $x^2-1$, which is a difference of squares: $x^2-1 = (x-1)(x+1)$. Now rewrite the expression: $\frac{(x-1)(x+1)}{x-1} - \frac{(x-1)(x+1)}{x+1}$ Since $x \neq \pm 1$, you can cancel common factors in each fraction: - First fraction: $\frac{(x-1)(x+1)}{x-1} = x+1$ - Second fraction: $\frac{(x-1)(x+1)}{x+1} = x-1$ The expression becomes: $(x+1) - (x-1) = x + 1 - x + 1 = 2$ Choice A (0) is incorrect because the two simplified fractions don't cancel each other out completely—they differ by 2. Choice B (1) represents a common computational error where students might think the constant terms cancel instead of adding. Choice D ($2x$) occurs when students mistakenly combine the variable terms instead of recognizing they cancel out, leaving only the constants. The correct answer is C. Study tip: When working with rational expressions containing factorable polynomials, always factor first, then simplify each fraction separately before combining. Watch for difference of squares patterns like $x^2-1$, and remember that subtraction affects all terms in parentheses, not just the first one.

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