Simplify Algebraic Expressions - International Brotherhood of Electrical Workers Exam IBEW

1. Simplify the expression by combining the fractional terms: $\frac{x}{3} + \frac{x}{6}$
   Answer: $\frac{x}{2}$ When you encounter fractions with the same variable in the numerator but different denominators, you need to find a common denominator to combine them effectively. To add $\frac{x}{3} + \frac{x}{6}$, first identify the least common denominator (LCD) of 3 and 6. Since 6 is divisible by 3, the LCD is 6. Convert the first fraction: $\frac{x}{3} = \frac{2x}{6}$ by multiplying both numerator and denominator by 2. The second fraction already has denominator 6. Now you can add: $\frac{2x}{6} + \frac{x}{6} = \frac{2x + x}{6} = \frac{3x}{6}$. Simplify by dividing both numerator and denominator by their greatest common factor of 3: $\frac{3x}{6} = \frac{x}{2}$. Choice A ($\frac{x}{9}$) incorrectly multiplies the denominators (3 × 6 = 18, then somehow becomes 9), which is a common mistake when students confuse addition with multiplication of fractions. Choice C ($\frac{x^2}{18}$) incorrectly squares the variable and multiplies denominators instead of finding the LCD. Choice D ($\frac{2x}{9}$) appears to result from adding denominators in the final step (3 + 6 = 9) rather than properly combining fractions. For IBEW electrical calculations, you'll frequently work with fractions when dealing with parallel resistances, voltage dividers, and current calculations. Always remember: to add fractions, find the LCD first, convert both fractions, then add the numerators while keeping the common denominator. This systematic approach prevents the arithmetic errors that create these distractors.
2. Simplify the expression: $c(c + 4) - 2(c^2 - 3)$
   Answer: $-c^2 + 4c + 6$ This question tests your ability to distribute terms and combine like terms—fundamental algebraic skills you'll use regularly in electrical calculations involving power, resistance, and circuit analysis. To solve $c(c + 4) - 2(c^2 - 3)$, you need to distribute each term, then combine like terms. First, distribute $c$ through $(c + 4)$: $c \cdot c + c \cdot 4 = c^2 + 4c$. Next, distribute $-2$ through $(c^2 - 3)$: $-2 \cdot c^2 - 2 \cdot (-3) = -2c^2 + 6$. Notice that multiplying by the negative 3 gives you positive 6. Now combine: $c^2 + 4c - 2c^2 + 6$. Grouping like terms: $(c^2 - 2c^2) + 4c + 6 = -c^2 + 4c + 6$. This matches answer choice A. Answer B ($-c^2 + 4c - 6$) represents the common error of incorrectly distributing the $-2$ through $(c^2 - 3)$, getting $-6$ instead of $+6$. Answer C ($3c^2 + 4c - 6$) shows two mistakes: adding the $c^2$ terms incorrectly ($1 + 2 = 3$ instead of $1 - 2 = -1$) and the sign error with the constant term. Answer D ($-c^2 + 2c - 6$) combines both the constant term sign error and incorrectly handles the $4c$ term. When distributing negative numbers, pay extra attention to sign changes—this is where most algebra mistakes occur. Always double-check that $-2 \times (-3) = +6$, not $-6$.
3. Simplify the expression $-3(x-y) + 2(y-x)$ completely.
   Answer: $-5x + 5y$ When you encounter algebraic expressions with parentheses and like terms, your goal is to distribute carefully and then combine terms systematically. Start by distributing each coefficient through its parentheses. For $-3(x-y)$, multiply $-3$ by both $x$ and $-y$: this gives you $-3x + 3y$. For $2(y-x)$, multiply $2$ by both $y$ and $-x$: this gives you $2y - 2x$. Now rewrite the entire expression: $-3x + 3y + 2y - 2x$. Rearrange to group like terms: $(-3x - 2x) + (3y + 2y) = -5x + 5y$. Answer choice A ($-x + y$) results from incorrectly thinking the coefficients somehow cancel out completely. Answer choice B ($-5x + y$) comes from properly combining the $x$ terms but failing to add the $y$ terms correctly—you'd get this if you mistakenly thought $3y + 2y = y$. Answer choice C ($-x + 5y$) represents the opposite error: correctly combining $y$ terms but botching the $x$ terms, perhaps by thinking $-3x - 2x = -x$. The correct answer is D: $-5x + 5y$. Study tip: When distributing negative signs, be extra careful with the second term in each parentheses. Write out every step rather than trying to do multiple operations mentally—this prevents sign errors that commonly appear on electrical apprenticeship exams where algebraic manipulation is tested.
4. Which simplified expression represents the perimeter of the panel?
   Answer: $8w - 4$ This problem tests your ability to substitute algebraic expressions into formulas and simplify the result—a skill you'll use regularly when calculating electrical loads and panel specifications. Given that the width is $w$ and the length is $3w - 2$, you substitute these values into the perimeter formula $P = 2l + 2w$. This becomes: $P = 2(3w - 2) + 2w$. First, distribute the 2 through the parentheses: $2(3w - 2) = 6w - 4$. Then add the remaining term: $P = 6w - 4 + 2w = 8w - 4$. Answer A ($4w - 2$) represents a common error where students only use one length and one width instead of doubling both dimensions. This misses the fundamental concept that a perimeter includes all four sides of a rectangle. Answer B ($6w - 4$) occurs when you correctly distribute $2(3w - 2)$ but forget to add the $2w$ term for the second width. This shows incomplete application of the perimeter formula. Answer D ($8w - 2$) results from incorrectly distributing: calculating $2(3w - 2)$ as $6w - 2$ instead of $6w - 4$. This is a fundamental algebra mistake where you don't multiply the constant term by 2. The correct answer is C: $8w - 4$. Study tip: When working with perimeter problems involving expressions, always write out the complete formula substitution first, then work through the algebra step-by-step. Double-check your distribution—multiply coefficients by every term inside parentheses.
5. Simplify the following expression completely: $7p + 2(p - 4)$
   Answer: $9p - 8$ When you encounter algebraic expressions with parentheses, your first step is to distribute (multiply) any coefficients outside the parentheses to all terms inside. This follows the distributive property: $a(b + c) = ab + ac$. Let's work through $7p + 2(p - 4)$ step by step. First, distribute the 2 to both terms inside the parentheses: $2(p - 4) = 2p - 8$. Now your expression becomes $7p + 2p - 8$. Next, combine like terms by adding the coefficients of the $p$ terms: $7p + 2p = 9p$. Your final simplified expression is $9p - 8$. Looking at the wrong answers: Choice B ($9p - 4$) correctly distributes to get $9p$, but makes an error with the constant term. This happens when you forget that $2 \times (-4) = -8$, not $-4$. Choice C ($7p - 8$) gets the constant term right but fails to combine the $p$ terms properly—you can't ignore the $2p$ that comes from distribution. Choice D ($1p$) appears to result from incorrectly thinking you subtract coefficients ($7 - 2 = 5$, then somehow getting to 1) and completely losing track of the constant term. Remember this key pattern: distribution comes first, then combine like terms. Many algebra mistakes on technical exams happen when students rush through the order of operations. Always distribute completely before attempting to combine terms, and double-check your arithmetic with negative numbers.
6. Simplify the expression: $\frac{2}{3}(9a - 12) + 5$
   Answer: $6a - 3$ When you encounter algebraic expressions with parentheses and mixed operations, you need to follow the order of operations carefully, starting with the distributive property to eliminate parentheses. To simplify $\frac{2}{3}(9a - 12) + 5$, first distribute $\frac{2}{3}$ to both terms inside the parentheses. Multiply $\frac{2}{3} \times 9a = \frac{18a}{3} = 6a$, and $\frac{2}{3} \times (-12) = \frac{-24}{3} = -8$. This gives you $6a - 8 + 5$. Finally, combine the constant terms: $-8 + 5 = -3$, resulting in $6a - 3$. Looking at the incorrect choices: Choice B ($6a - 7$) suggests you correctly distributed to get $6a$ but made an error combining the constants, possibly calculating $-12 + 5 = -7$ instead of first applying the distributive property to get $-8$. Choice C ($6a + 3$) indicates you distributed correctly but incorrectly combined $-8 + 5$ as $+3$, likely from a sign error. Choice D ($9a - 7$) shows you failed to distribute properly, keeping the $9a$ unchanged and somehow arriving at $-7$ for the constant term. For IBEW algebraic problems, always work methodically through the order of operations. Write out each step clearly, especially when distributing fractions, and double-check your arithmetic with constants. These systematic approaches prevent the small calculation errors that create wrong answer choices.
7. For a circuit sketch, simplify: $R_T=R+2(R+4)-3R$.
   Answer: $R_T=8$ This question tests the ability to simplify algebraic expressions involving variables, a key skill in algebra and functions. Simplifying algebraic expressions involves combining like terms, applying the distributive property, and using correct operations to reduce expressions to their simplest form. In the given scenario, simplifying the expression ($R_T$ = R + 2(R + 4) - 3R) involves distributing 2 to get 2R + 8, then combining R + 2R - 3R + 8 to 8. Choice A is correct because it accurately simplifies the expression by canceling out the R terms to zero and leaving the constant. Choice D is incorrect because it multiplies constants by R unnecessarily, demonstrating a misunderstanding of combining terms. Teaching strategies include practicing the order of operations, reinforcing the combination of like terms, and using real-life applications to contextualize algebra. Encourage students to check their work by substituting values into the original and simplified expressions.
8. Simplify the expression: $-4(c - 2d) + 2(3c + d)$
   Answer: $2c + 10d$ This question tests your ability to distribute multiplication across parentheses and combine like terms - fundamental algebra skills you'll use regularly in electrical calculations involving circuits and formulas. To solve $-4(c - 2d) + 2(3c + d)$, you need to distribute each coefficient to every term inside its parentheses, then combine like terms. First, distribute $-4$ to both terms in the first parentheses: $-4(c - 2d) = -4c + 8d$. Notice that $-4 \times (-2d) = +8d$. Next, distribute $2$ to both terms in the second parentheses: $2(3c + d) = 6c + 2d$. Now you have: $-4c + 8d + 6c + 2d$ Combine like terms by grouping $c$ terms and $d$ terms separately: $(-4c + 6c) + (8d + 2d) = 2c + 10d$ This confirms answer B is correct. Let's examine why the other options are wrong. Answer A ($2c + 6d$) incorrectly adds the $d$ terms as $8d + 2d = 6d$ instead of $10d$. Answer C ($2c - 6d$) makes two errors: it gets the wrong coefficient for the $d$ terms and the wrong sign. Answer D ($-10c + 10d$) correctly finds $10d$ but incorrectly combines the $c$ terms as $-4c + 6c = -10c$ instead of $+2c$. Remember: when distributing negative numbers, pay careful attention to sign changes, and always double-check your arithmetic when combining like terms.
9. Simplify the expression: $8x - (x - [2x - 4])$
   Answer: $9x - 4$ When you encounter expressions with nested brackets and parentheses, the key is working from the inside out, carefully tracking positive and negative signs as you remove each layer of grouping symbols. Start with the innermost brackets: $[2x - 4]$. This stays as is since there's nothing inside to simplify further. Next, work on the parentheses: $(x - [2x - 4])$. Distribute the negative sign in front of the brackets: $x - 2x + 4 = -x + 4$. Now substitute this back into the original expression: $8x - (-x + 4)$. Distribute the negative sign: $8x + x - 4 = 9x - 4$. Looking at the wrong answers: Choice A ($5x + 4$) likely comes from incorrectly combining $8x - x - 2x$ and getting the sign wrong on the constant term. Choice B ($7x + 4$) results from the common error of computing $8x - x = 7x$ but forgetting about the $2x$ term entirely, plus sign errors. Choice C ($9x + 4$) gets the coefficient right but makes the final sign error, treating $8x - (-x + 4)$ as $8x + x + 4$ instead of $8x + x - 4$. The correct answer is D: $9x - 4$. Study tip: For nested grouping symbols, always work inside-out and be extra careful with negative signs. Write out each step clearly rather than trying to do multiple operations mentally—sign errors are the most common mistake on algebraic simplification problems.
10. In a junction box, simplify: $R_T=7+2R-3-R$.
    Answer: $R_T=4+R$ This question tests the ability to simplify algebraic expressions involving variables, a key skill in algebra and functions. Simplifying algebraic expressions involves combining like terms, applying the distributive property, and using correct operations to reduce expressions to their simplest form. In the given scenario, simplifying the expression ($R_T$ = 7 + 2R - 3 - R) involves combining constants 7 - 3 to 4 and terms 2R - R to R, leading to ($R_T$ = 4 + R). Choice A is correct because it accurately simplifies the expression by grouping like terms properly. Choice B is incorrect because it adds extra constants, demonstrating a misunderstanding of combining constants. Teaching strategies include practicing the order of operations, reinforcing the combination of like terms, and using real-life applications to contextualize algebra. Encourage students to check their work by substituting values into the original and simplified expressions.
11. During load calc, simplify: $V=IR+2IR-IR+8I$.
    Answer: $V=2IR+8I$ This question tests the ability to simplify algebraic expressions involving variables, a key skill in algebra and functions. Simplifying algebraic expressions involves combining like terms, applying the distributive property, and using correct operations to reduce expressions to their simplest form. In the given scenario, simplifying the expression (V = IR + 2IR - IR + 8I) involves combining the IR terms to 2IR and adding 8I, leading to (V = 2IR + 8I). Choice A is correct because it accurately simplifies the expression by combining coefficients of IR properly. Choice C is incorrect because it subtracts the constant term, demonstrating a misunderstanding of adding terms. Teaching strategies include practicing the order of operations, reinforcing the combination of like terms, and using real-life applications to contextualize algebra. Encourage students to check their work by substituting values into the original and simplified expressions.
12. In a series circuit, simplify: $R_T=5R-(2R+4)+R$.
    Answer: $R_T=4R-4$ This question tests the ability to simplify algebraic expressions involving variables, a key skill in algebra and functions. Simplifying algebraic expressions involves combining like terms, applying the distributive property, and using correct operations to reduce expressions to their simplest form. In the given scenario, simplifying the expression ($R_T$ = 5R - (2R + 4) + R) involves distributing the negative to 5R - 2R - 4 + R, combining to 4R - 4. Choice A is correct because it accurately simplifies the expression by handling the parentheses and combining like terms. Choice B is incorrect because it undercounts the R terms, demonstrating a misunderstanding of distribution. Teaching strategies include practicing the order of operations, reinforcing the combination of like terms, and using real-life applications to contextualize algebra. Encourage students to check their work by substituting values into the original and simplified expressions.
13. Simplify the expression: $3x^2 + 5x - 8 + 2x^2 - 3x - 1$
    Answer: $5x^2 + 2x - 9$ This question tests your ability to combine like terms in polynomial expressions, a fundamental algebraic skill you'll use frequently in electrical calculations involving power, resistance, and circuit analysis. To simplify $3x^2 + 5x - 8 + 2x^2 - 3x - 1$, you need to identify and combine terms with the same variable and exponent. Group the like terms: $(3x^2 + 2x^2) + (5x - 3x) + (-8 - 1)$. The $x^2$ terms combine to give $5x^2$, the $x$ terms combine to give $2x$, and the constants combine to give $-9$. This yields $5x^2 + 2x - 9$, which is answer choice C. Let's examine why the other answers are incorrect. Choice A ($5x^4 + 2x - 9$) incorrectly shows $x^4$ instead of $x^2$—this suggests confusion about how exponents work when adding terms. You don't add exponents when combining like terms; the exponent stays the same. Choice B ($5x^2 + 8x - 9$) correctly handles the $x^2$ and constant terms but adds the $x$ terms incorrectly: $5x + (-3x) = 2x$, not $8x$. Choice D ($7x - 9$) completely ignores the $x^2$ terms and only considers the linear and constant terms. Remember this key principle: when combining like terms, only add or subtract the coefficients—the variable part stays unchanged. Always double-check by grouping terms with identical variables and exponents before combining. This skill is essential for solving electrical formulas involving multiple variables.
14. Simplify the following expression: $5x - [2y - (3x + y)]$
    Answer: $8x - y$ When you encounter algebraic expressions with nested brackets and parentheses, the key is working systematically from the inside out, carefully tracking positive and negative signs at each step. Start with the innermost parentheses: $(3x + y)$. Now work outward to the brackets. Inside the brackets you have $2y - (3x + y)$. When you subtract the entire expression $(3x + y)$, you must distribute the negative sign to each term: $2y - 3x - y$. Combining like terms gives you $-3x + y$. Now substitute this back into the original expression: $5x - (-3x + y)$. Again, distribute the negative sign: $5x + 3x - y$. Combining like terms yields $8x - y$. Looking at the wrong answers: Choice A ($2x - 3y$) likely results from sign errors when distributing the negative signs and possibly combining terms incorrectly. Choice B ($2x - y$) suggests you correctly handled the $y$ terms but made errors with the $x$ terms, possibly forgetting to distribute one of the negative signs to $3x$. Choice C ($8x - 3y$) shows correct handling of the $x$ terms but incorrect management of the $y$ terms—you might have incorrectly combined $2y$ and $y$ or mishandled signs. For bracket and parentheses problems, always work inside-out and write each step clearly. Pay special attention to negative signs—they're the most common source of errors. Double-check by expanding your final answer back to verify it matches the original expression.
15. For current flow, simplify: $I=V/(4R) + V/(4R) + V/(2R)$.
    Answer: $I=V/R$ This question tests the ability to simplify algebraic expressions involving variables, a key skill in algebra and functions. Simplifying algebraic expressions involves combining like terms, applying the distributive property, and using correct operations to reduce expressions to their simplest form. In the given scenario, simplifying the expression ($I = \frac{V}{4R} + \frac{V}{4R} + \frac{V}{2R}$) involves finding a common denominator of 4R and combining the terms, leading to ($\frac{4V}{4R} = \frac{V}{R}$). Choice A is correct because it accurately simplifies the expression by rewriting ($\frac{V}{2R}$) as ($\frac{2V}{4R}$) and adding ($\frac{V}{4R} + \frac{V}{4R} + \frac{2V}{4R} = \frac{4V}{4R} = \frac{V}{R}$). Choice C is incorrect because it results from adding the numerators without properly adjusting the third term to the common denominator, demonstrating a misunderstanding of combining fractions with different denominators. Teaching strategies include practicing the order of operations, reinforcing the combination of like terms, and using real-life applications to contextualize algebra. Encourage students to check their work by substituting values into the original and simplified expressions.
16. Simplify the following expression: $8a - 4 - 6a - 3$
    Answer: $2a - 7$ When you encounter algebraic expressions that need simplifying, you're working with combining like terms—a fundamental skill that appears frequently on electrical calculations where you'll manipulate formulas involving voltage, current, and resistance. To simplify $8a - 4 - 6a - 3$, you need to group and combine like terms. Like terms have the same variable with the same exponent. Here you have terms with $a$ and constant terms without variables. First, combine the $a$ terms: $8a - 6a = 2a$. Then combine the constants: $-4 - 3 = -7$. This gives you $2a - 7$. Looking at the wrong answers: Choice A ($2a - 1$) correctly combines the $a$ terms but makes an error with the constants, likely adding $-4 + (-3)$ incorrectly as $-1$ instead of $-7$. Choice C ($-5a$) incorrectly combines $8a - 6a$ as $-5a$ (possibly confusing subtraction direction) and completely ignores the constant terms. Choice D ($14a - 7$) adds the $a$ coefficients instead of subtracting: $8 + 6 = 14$, which would be wrong even if the expression were $8a + 6a$. The correct answer is B: $2a - 7$. Study tip: Always separate like terms first, then perform operations within each group. Double-check your arithmetic with constants—negative signs are easy to mishandle. This systematic approach prevents the mixing errors that create these common wrong answers.
17. Simplify the expression: $\frac{w^2}{2} + 3 - \frac{w^2}{4} - 1$
    Answer: $\frac{w^2}{4} + 2$ This question tests your ability to combine like terms and work with fractions in algebraic expressions. When simplifying expressions with multiple terms, you need to group similar terms together and perform operations carefully. Start by identifying like terms: the $w^2$ terms and the constant terms. For the $w^2$ terms, you have $\frac{w^2}{2}$ and $-\frac{w^2}{4}$. To subtract these fractions, find a common denominator of 4: $\frac{w^2}{2} = \frac{2w^2}{4}$. So $\frac{2w^2}{4} - \frac{w^2}{4} = \frac{w^2}{4}$. For the constants: $3 - 1 = 2$. Therefore, the simplified expression is $\frac{w^2}{4} + 2$, which is answer choice B. Looking at the wrong answers: Choice A keeps $\frac{w^2}{2}$ unchanged, suggesting you forgot to combine the $w^2$ terms entirely. Choice C incorrectly adds the constants as $3 + 1 = 4$ instead of subtracting to get $3 - 1 = 2$. Choice D shows $\frac{3w^2}{4}$, which would result from incorrectly adding the fractions: $\frac{2w^2}{4} + \frac{w^2}{4} = \frac{3w^2}{4}$ instead of subtracting them. When simplifying algebraic expressions, always identify like terms first, then handle fraction operations by finding common denominators. Pay close attention to positive and negative signs—they determine whether you add or subtract terms. Double-check your arithmetic, especially when working with fractions.
18. Three resistors in a series have resistances of $(R+50)$ ohms, $(2R-10)$ ohms, and $(4R+25)$ ohms. Total resistance in a series is the sum of individual resistances. What is the simplified expression for the total resistance?
    Answer: $7R + 65$ When you encounter series resistance problems with algebraic expressions, remember that series circuits follow one fundamental rule: total resistance equals the sum of all individual resistances. This is true whether you're working with numbers or variables. To find the total resistance, you simply add the three given resistances: $(R+50) + (2R-10) + (4R+25)$. When combining like terms, collect all the R terms together and all the constant terms together. The R terms give you: $R + 2R + 4R = 7R$. The constant terms give you: $50 + (-10) + 25 = 50 - 10 + 25 = 65$. Therefore, the total resistance is $7R + 65$. Looking at the wrong answers: Choice A ($6R + 65$) correctly adds the constants but miscounts the R terms—you'd get this if you missed one of the R coefficients. Choice B ($7R + 85$) correctly identifies the R coefficient but makes an arithmetic error with the constants, possibly adding $50 + 10 + 25 = 85$ instead of properly handling the negative sign. Choice D ($7R - 35$) gets the R terms right but makes a significant error with the constants, perhaps subtracting instead of adding or mishandling multiple signs. The key strategy here is careful algebraic manipulation: always distribute any negative signs properly, group like terms systematically, and double-check your arithmetic with the constants. Series resistance problems become straightforward when you methodically combine terms rather than rushing through the algebra.
19. Simplify the expression with nested parentheses: $a - [3a - (2a + 1)]$
    Answer: $1$ When you encounter expressions with nested parentheses like this one, you must work from the inside out, carefully tracking positive and negative signs as you remove each set of grouping symbols. Start with the innermost parentheses: $a - [3a - (2a + 1)]$. First, distribute the negative sign in front of $(2a + 1)$: $a - [3a - 2a - 1]$. Now simplify inside the brackets by combining like terms: $3a - 2a = a$, so you have $a - [a - 1]$. Next, distribute the negative sign in front of the brackets: $a - a + 1$. The $a$ terms cancel out, leaving you with $1$. Let's examine why the other answers are incorrect. Choice A ($0$) would result if you mistakenly thought all terms canceled out, perhaps by incorrectly handling the constant term. Choice C ($2a - 1$) suggests an error in combining like terms or distributing negative signs—you might get this if you incorrectly simplified $3a - 2a$ or mishandled the final distribution. Choice D ($-4a - 1$) indicates a major error in sign distribution, possibly adding coefficients instead of subtracting or making multiple sign errors throughout the problem. The correct answer is B. For nested parentheses problems on the IBEW exam, always work systematically from inside to outside, and pay extra attention to negative signs—they're the most common source of errors. Double-check your work by substituting a simple number like $a = 1$ into both the original expression and your simplified result.
20. Combine the like terms in the expression: $6xy + 3x - 4y - 2xy + 5y$
    Answer: $4xy + 3x + y$ When you encounter algebraic expressions with multiple terms, your goal is to combine "like terms" - terms that have identical variable parts. This is a fundamental skill you'll use throughout electrical calculations involving formulas and equations. To combine like terms in $6xy + 3x - 4y - 2xy + 5y$, group terms with matching variables. The $xy$ terms are $6xy$ and $-2xy$, which combine to give $4xy$. The $x$ term stands alone as $3x$. The $y$ terms are $-4y$ and $+5y$, which combine to give $+y$. Your final expression is $4xy + 3x + y$. Looking at the wrong answers: Choice A ($4xy - x - y$) incorrectly shows the $x$ and $y$ terms as negative, suggesting errors in sign handling when combining terms. Choice C ($8xy + 3x + y$) adds the $xy$ terms incorrectly - $6xy + (-2xy) = 4xy$, not $8xy$. This likely comes from adding $6 + 2 = 8$ while ignoring the negative sign. Choice D ($8xy$) makes the same error with the $xy$ terms but also completely drops the $x$ and $y$ terms, which is mathematically invalid since they cannot be combined with $xy$ terms. Remember this key strategy: like terms must have identical variable parts to be combined. $xy$, $x$, and $y$ are all different types of terms and stay separate. Always double-check your signs when adding and subtracting terms.

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