IBEW: Electrical Training Alliance Aptitude Test Quiz: Simplify Algebraic Expressions

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Simplify Algebraic ExpressionsQuestion 1 of 20

For current flow, simplify: $I=V/(4R) + V/(4R) + V/(2R)$ .

I=V/R

I=V/(2R)

I=3V/(4R)

I=2V/R

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IBEW: Electrical Training Alliance Aptitude Test Quiz

IBEW: Electrical Training Alliance Aptitude Test Quiz: Simplify Algebraic Expressions

Practice Simplify Algebraic Expressions in IBEW: Electrical Training Alliance Aptitude Test 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 Simplify Algebraic Expressions, giving you a quick way to practice the rules, question types, and explanations that matter most for IBEW: Electrical Training Alliance Aptitude Test.

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.

All questions

Question 1

For current flow, simplify: $I=V/(4R) + V/(4R) + V/(2R)$ .

$I=V/R$ (correct answer)
$I=V/(2R)$
$I=3V/(4R)$
$I=2V/R$

\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.

Question 2

Simplify the expression $-3(x-y) + 2(y-x)$ completely.

$-x + y$
$-5x + y$
$-x + 5y$
$-5x + 5y$ (correct answer)

Explanation: 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.

Question 3

Simplify the following expression: $8a - 4 - 6a - 3$

$2a - 1$
$2a - 7$ (correct answer)
$-5a$
$14a - 7$

Explanation: 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

-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.

Question 4

Simplify the expression: $c(c + 4) - 2(c^2 - 3)$

$-c^2 + 4c + 6$ (correct answer)
$-c^2 + 4c - 6$
$3c^2 + 4c - 6$
$-c^2 + 2c - 6$

Explanation: 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

Question 5

Simplify the expression: $15 - (4y - 6) + 2y$

$9 - 2y$
$15 - 6y$
$21 - 2y$ (correct answer)
$21 - 6y$

Explanation: When you encounter algebraic expressions with parentheses, your success depends on carefully applying the distributive property and combining like terms in the correct order. Let's work through

15 - (4y - 6) + 2y

step by step. First, distribute the negative sign to everything inside the parentheses:

15 - 4y + 6 + 2y

. The negative sign changes

(4y - 6)

-4y + 6

. Next, rearrange to group like terms:

15 + 6 - 4y + 2y

. Combine the constants:

15 + 6 = 21

. Combine the variable terms:

-4y + 2y = -2y

. This gives you

21 - 2y

, which is answer C. Answer A (

9 - 2y

) correctly handles the variable terms but makes an error with the constants—likely subtracting 6 from 15 instead of adding it after distribution. Answer B (

15 - 6y

) incorrectly treats the constant 6 as if it stays negative and fails to combine it with 15, while also getting the wrong coefficient for y. Answer D (

21 - 6y

) gets the constant term right but forgets to add the

+2y

term at the end, keeping only the

-4y

from the parentheses. The key strategy for IBEW algebra problems: always distribute first, then immediately rewrite your expression to group like terms together. This prevents sign errors and ensures you don't forget any terms. Double-check by substituting a test value if time permits.

Question 6

Use the distributive property to simplify the expression: $4(x + 3)$

$4x + 3$
$7x$
$x + 12$
$4x + 12$ (correct answer)

Explanation: The distributive property is a fundamental algebraic concept that electrical workers use regularly when calculating circuit values and performing load calculations. When you see an expression like

4(x + 3)

, you need to "distribute" the number outside the parentheses to each term inside. To apply the distributive property, multiply the term outside the parentheses by each term inside:

4(x + 3) = 4 \cdot x + 4 \cdot 3 = 4x + 12

. This gives us answer choice D. Let's examine why the other options are incorrect. Choice A (

4x + 3

) represents a common mistake where you only multiply the first term by 4, forgetting to distribute to the constant 3. Choice B (

7x

) incorrectly treats the expression as if it were

4x + 3x

, combining unlike terms that can't actually be combined. Choice C (

x + 12

) shows another partial error where you've correctly multiplied

4 \times 3 = 12

but failed to multiply the 4 by the

x

term. Remember this key pattern: when you see a number (or variable) followed by parentheses, that signals multiplication and requires distribution. In electrical work, you'll encounter similar situations when calculating total power across multiple loads or determining voltage drops across series components. Always distribute to every term inside the parentheses—missing even one term will lead to incorrect calculations that could affect circuit safety and performance.

Question 7

Simplify the expression: $-(3m - 5n)$

$-3m - 5n$
$3m + 5n$
$-3m + 5n$ (correct answer)
$2mn$

Explanation: When you encounter a negative sign in front of parentheses, you're dealing with the distributive property. The negative sign acts like multiplying by -1, which means you distribute it to every term inside the parentheses. To simplify

-(3m - 5n)

, multiply each term inside the parentheses by -1:

The first term: $(-1) \times 3m = -3m$
The second term: $(-1) \times (-5n) = +5n$

This gives you

-3m + 5n

, which is answer C. Let's examine why the other options are incorrect. Answer A (

-3m - 5n

) represents a common mistake where you only distribute the negative sign to the first term, leaving

-5n

instead of recognizing that

(-1) \times (-5n) = +5n

. Answer B (

3m + 5n

) suggests you somehow made both terms positive, which would only happen if you had

+(3m - 5n)

and incorrectly handled the signs. Answer D (

2mn

) makes no mathematical sense here—there's no operation that would combine unlike terms

3m

and

5n

into a product. The key strategy for these problems is to remember that a negative sign in front of parentheses changes the sign of every term inside. When distributing a negative, pay special attention to terms that are already negative—two negatives make a positive. Always double-check by distributing systematically to each term rather than trying to do it all at once.

Question 8

Simplify the expression: $3x^2 + 5x - 8 + 2x^2 - 3x - 1$

$5x^4 + 2x - 9$
$5x^2 + 8x - 9$
$5x^2 + 2x - 9$ (correct answer)
$7x - 9$

Explanation: 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.

Question 9

Simplify the following expression: $5x - [2y - (3x + y)]$

$2x - 3y$
$2x - y$
$8x - 3y$
$8x - y$ (correct answer)

Explanation: 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.

Question 10

Simplify the following expression: $5(2a - b) - 3(a - 2b)$

$7a - 11b$
$7a + b$ (correct answer)
$7a - 7b$
$13a + b$

Explanation: This problem tests your ability to distribute terms and combine like terms—fundamental algebraic skills you'll use constantly in electrical calculations involving voltage drops, power formulas, and circuit analysis. To solve

5(2a - b) - 3(a - 2b)

, start by distributing each coefficient to the terms inside the parentheses. For the first part:

5(2a - b) = 5 \cdot 2a + 5 \cdot (-b) = 10a - 5b

. For the second part:

-3(a - 2b) = -3 \cdot a + (-3) \cdot (-2b) = -3a + 6b

. Notice that the negative sign in front of the 3 affects both terms inside the parentheses. Now combine:

10a - 5b - 3a + 6b

. Group like terms:

(10a - 3a) + (-5b + 6b) = 7a + 1b = 7a + b

. This matches answer choice B. Let's examine why the other answers are wrong. Choice A (

7a - 11b

) suggests you incorrectly subtracted the

b

terms:

-5b - 6b = -11b

, but you should have calculated

-5b + 6b = +b

. Choice C (

7a - 7b

) indicates you might have distributed incorrectly, perhaps treating

-3(-2b)

-6b

instead of

+6b

. Choice D (

13a + b

) shows an error in combining the

a

terms—adding

10a + 3a

instead of

10a - 3a

. Remember: when distributing negative signs, every term inside the parentheses changes sign. Double-check your work by carefully tracking positive and negative signs—this attention to detail is crucial for electrical calculations where sign errors can indicate dangerous wiring mistakes.

Question 11

Simplify the expression: $\frac{w^2}{2} + 3 - \frac{w^2}{4} - 1$

$\frac{w^2}{2} + 2$
$\frac{w^2}{4} + 2$ (correct answer)
$\frac{w^2}{4} + 4$
$\frac{3w^2}{4} + 2$

Explanation: 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.

Question 12

Simplify the algebraic expression: $5x + 3y - 2x + 7y$

$3x + 10y$ (correct answer)
$3x + 4y$
$7x + 10y$
$13xy$

Explanation: This question tests your ability to combine like terms in algebraic expressions, a fundamental skill you'll need when working with electrical formulas that involve multiple variables. To simplify

5x + 3y - 2x + 7y

, you need to group and combine terms that have the same variable. Start by identifying like terms: the x-terms are

5x

and

-2x

, while the y-terms are

3y

and

7y

. For the x-terms:

5x - 2x = 3x

For the y-terms:

3y + 7y = 10y

Therefore, the simplified expression is

3x + 10y

, which is answer choice A. Looking at the wrong answers: Choice B (

3x + 4y

) correctly finds

3x

but incorrectly calculates the y-terms, likely by subtracting

3y - 7y = -4y

instead of adding

3y + 7y = 10y

. Choice C (

7x + 10y

) gets the y-terms right but adds the x-terms incorrectly as

5x + 2x = 7x

, forgetting that the second term is negative. Choice D (

13xy

) represents a fundamental misunderstanding—you cannot combine unlike terms (x and y terms) into a single term, and this answer incorrectly adds all coefficients (5 + 3 + 2 + 7) while changing the variables to

xy

. When combining like terms, always pay careful attention to positive and negative signs, and remember that you can only combine terms with identical variables and exponents.

Question 13

Simplify the expression: $7 - 3(2x - 5) + 4x$

$-2x + 22$ (correct answer)
$-2x - 8$
$10x + 22$
$12x - 20$

Explanation: This question tests your ability to simplify algebraic expressions using the distributive property and combining like terms—fundamental skills you'll need for electrical calculations involving formulas and variables. To simplify

7 - 3(2x - 5) + 4x

, start by distributing the

-3

through the parentheses. Remember that

-3

multiplies both terms inside:

-3(2x) = -6x

and

-3(-5) = +15

. This gives you:

7 - 6x + 15 + 4x

. Next, combine like terms by grouping constants together and variable terms together:

(7 + 15) + (-6x + 4x) = 22 + (-2x) = 22 - 2x

. Written in standard form, this is

-2x + 22

. Looking at the wrong answers: Choice B (

-2x - 8

) correctly handles the variable terms but makes an error with the constants—likely forgetting that

-3 \times (-5) = +15

, not

-15

. Choice C (

10x + 22

) gets the constant term right but incorrectly adds the variable coefficients as

-6x + 4x = 10x

instead of

-2x

. Choice D (

12x - 20

) appears to result from incorrectly distributing and then making sign errors throughout the simplification process. The correct answer is A:

-2x + 22

. Study tip: When distributing negative numbers, carefully track your signs at each step. Write out

-3(2x - 5)

(-3)(2x) + (-3)(-5)

to avoid the common trap of forgetting that negative times negative equals positive.

Question 14

What is the simplified form of $\frac{1}{2}(4x + 6) - 3x$ ?

$-x + 6$
$2x + 3$
$-x + 3$ (correct answer)
$5x + 3$

\frac{1}{2}(4x + 6)

. Multiply

\frac{1}{2}

by each term inside the parentheses:

\frac{1}{2} \times 4x = 2x

and

\frac{1}{2} \times 6 = 3

. So

\frac{1}{2}(4x + 6) = 2x + 3

. Now your expression becomes:

2x + 3 - 3x

. Next, combine like terms by grouping the

x

terms together:

2x - 3x + 3 = -x + 3

. The answer is C)

-x + 3

. Looking at the wrong answers: A)

-x + 6

correctly finds

-x

but fails to apply the distributive property, keeping the 6 instead of recognizing that

\frac{1}{2} \times 6 = 3

. B)

2x + 3

represents only the first step—distributing

\frac{1}{2}(4x + 6)

—but forgetting to subtract the

3x

term entirely. D)

5x + 3

makes the error of adding

2x + 3x = 5x

instead of subtracting

2x - 3x = -x

. When simplifying algebraic expressions, always work in order: first handle parentheses using the distributive property, then combine like terms. Pay careful attention to positive and negative signs—they're easy to mix up but completely change your final answer.

Question 15

In a control circuit, simplify: $I=(2V+6)/(2R)$ .

$I=(V+3)/R$ (correct answer)
$I=(2V+3)/R$
$I=(V+3)/(2R)$
$I=(V+6)/R$

Explanation: 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 = (2V + 6)/(2R)) involves factoring 2 from the numerator to get 2(V + 3)/(2R), which cancels to (V + 3)/R. Choice A is correct because it accurately simplifies the expression by dividing numerator and denominator by 2. Choice B is incorrect because it fails to simplify the fraction, demonstrating a misunderstanding of factoring. 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.

Question 16

For a branch circuit, simplify: $I=(3V-9)/(3R)$ .

$I=(V-3)/R$ (correct answer)
$I=(V-9)/R$
$I=(3V-3)/R$
$I=(V-3)/(3R)$

Explanation: 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 = (3V - 9)/(3R)) involves factoring 3 from the numerator to 3(V - 3)/(3R), which simplifies to (V - 3)/R. Choice A is correct because it accurately simplifies the expression by canceling the 3 in numerator and denominator. Choice B is incorrect because it divides incorrectly, demonstrating a misunderstanding of factoring. 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.

Question 17

An electrician has a length of conduit measuring $(5x + 7)$ meters. From this, two pieces are used, one of length $x$ meters and another of length $(x + 2)$ meters.

What expression represents the remaining length of the conduit?

$3x + 5$ (correct answer)
$3x + 9$
$4x + 5$
$5x + 5$

Explanation: This is a classic algebraic word problem about subtracting lengths, a skill you'll use frequently when calculating conduit runs and material usage on electrical jobs. To find the remaining length, you need to subtract the two pieces used from the original length. Start with the original conduit length of

(5x + 7)

meters. The electrician uses two pieces: one measuring

x

meters and another measuring

(x + 2)

meters. Set up the subtraction: Remaining length = Original length - First piece - Second piece

\text{Remaining} = (5x + 7) - x - (x + 2)

Distribute the negative signs and combine like terms:

= 5x + 7 - x - x - 2

= 5x - 2x + 7 - 2

= 3x + 5

The correct answer is A)

3x + 5

. Now let's see where the other answers go wrong. Choice B)

3x + 9

likely comes from adding instead of subtracting the constant terms:

7 + 2 = 9

. Choice C)

4x + 5

results from incorrectly combining the

x

terms—perhaps only subtracting one

x

instead of both

x

and

(x + 2)

. Choice D)

5x + 5

suggests the student subtracted only the constant terms without touching the

x

terms at all. Study tip: When working with algebraic expressions involving subtraction, always distribute negative signs carefully and double-check your like-term combinations. Set up the problem systematically: original minus what's removed equals what remains.

Question 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?

$6R + 65$
$7R + 85$
$7R + 65$ (correct answer)
$7R - 35$

Explanation: 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.

Question 19

A wire has a length represented by $(12L - 5)$ feet. If a piece of length $(4L + 3)$ feet is cut off, what expression represents the length of the remaining wire?

Find the simplified expression for the remaining length of the wire.

$8L - 2$
$8L - 8$ (correct answer)
$16L - 2$
$16L + 8$

Explanation: This question tests your ability to work with algebraic expressions involving subtraction, a skill you'll use frequently when calculating wire lengths and material requirements in electrical work. To find the remaining wire length, you need to subtract the cut piece from the original length. Set up the expression as: original length minus cut length =

(12L - 5) - (4L + 3)

. When subtracting expressions, distribute the negative sign to every term in the second expression:

(12L - 5) - (4L + 3) = 12L - 5 - 4L - 3

. Combine like terms:

12L - 4L = 8L

and

-5 - 3 = -8

. This gives you

8L - 8

. Looking at the wrong answers: Choice A (

8L - 2

) correctly finds

8L

but makes an error in the constant term, likely from calculating

-5 - 3 = -2

instead of

-8

. Choice C (

16L - 2

) incorrectly adds the

L

terms (

12L + 4L = 16L

) instead of subtracting, and also miscalculates the constants. Choice D (

16L + 8

) makes both errors: adding instead of subtracting the

L

terms, and incorrectly getting a positive 8 for the constant term. Remember that when subtracting expressions in parentheses, you must distribute the negative sign to every term inside. A common mistake is forgetting to change the sign of the constant term. Practice these problems by writing out each step clearly to avoid sign errors.

Question 20

Simplify the following expression completely: $7p + 2(p - 4)$

$9p - 8$ (correct answer)
$9p - 4$
$7p - 8$
$1p$

Explanation: 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.