Simplify (39(x − 3)/3 + 39)/13

2024 Handbook

2024 Test A Math · Question 10 of 57

Question

Simplify:

(39(x−3)3)+3913\frac{\left(\frac{39(x - 3)}{3}\right) + 39}{13}

  • Axx
  • Bx−2x - 2
  • C13x−3613x - 36
  • Dx+3613x + \frac{36}{13}
Show the answer

Correct answer: A

(39(x−3)3)+3913\frac{\left(\frac{39(x - 3)}{3}\right) + 39}{13}

Divide the first 39 by 3.

13(x−3)+3913\frac{13(x - 3) + 39}{13}

Distribute 13 through (x−3)(x - 3).

13x−39+3913\frac{13x - 39 + 39}{13}

In the numerator, the terms −39 and 39 cancel out.

13x13\frac{13x}{13}

Finally, divide the numerator and denominator by 13 to get x.

Question analysis

An algebra question that asks students to simplify a stacked fraction: 39 times (x − 3) over 3, plus 39, all divided by 13. It tests dividing, distributing and combining like terms in the right order. The nested fraction bars make it easy to divide only some of the terms or to distribute to only one.

NYS standards
  • 7.EE.1 linear expressions with rational coefficients
  • 6.EE.3 generating equivalent expressions
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students expand and simplify linear expressions such as 13(x − 3) + 39. This wraps that work inside two fraction bars, which adds reading load and chances to misstep but no new mathematics.
Estimated difficulty
Straightforward: two steps or one careful reading.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 16+, 6 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The expression looks heavier than it is: every number is a multiple of 13, and the constants cancel to leave a single term. What makes it risky is scope, knowing which terms each fraction bar and each multiplier covers. Each wrong choice comes from stretching or shrinking that scope by one term, so working outward from the innermost parentheses, one line per step, is the safe way.

  1. Error-prone arithmetic moderate Working it out

    Two fraction bars are nested: 39 ÷ 3 applies only to the first term, the 13 must reach both x and −3, and only then does the whole numerator divide by 13. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. Many steps minor Working it out

    Four moves come in a fixed order, divide 39 by 3, distribute 13, cancel −39 with +39, divide by 13, and each is a chance to apply a step to the wrong terms. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Work from the inside out: divide 39 by 3, distribute through the parentheses, combine the constants in the numerator, and only then divide everything by 13.

Traps

  • A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
  • Order of operations: Working left to right, or distributing a minus sign wrongly, gives a choice.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Distributive property
a(b−c)=a×b−a×ca(b - c) = a \times b - a \times c
a is the number outside the parentheses. b and c are the terms inside. Multiply a by each term inside, and keep the minus sign between them. It works because a groups of (b − c) are a groups of b, take away a groups of c.
  1. Sort out the expression

    You need to simplify the big fraction, which means write it in its shortest form. It has a small fraction inside its top. The top is 39(x−3)3+39\frac{39(x - 3)}{3} + 39, and the bottom is 13. Work from the inside out, one step at a time.

  2. Divide the first 39 by 3

    The small fraction bar is only under 39(x−3)39(x - 3). So divide just that 39 by 3.

    39(x−3)3=13(x−3)\frac{39(x - 3)}{3} = 13(x - 3)

    Now the top is 13(x−3)+3913(x - 3) + 39. Watch out: the small bar does not reach the added 39. Dividing that 39 by 3 as well leads to x−2x - 2, choice B.

  3. Multiply out the parentheses

    a(b−c)=a×b−a×ca(b - c) = a \times b - a \times c

    13(x−3)=13×x−13×3=13x−3913(x - 3) = 13 \times x - 13 \times 3 = 13x - 39

    Watch out: multiply the 3 by 13 too, not just the x. Leaving −3 as it is leads to choice D.

  4. Combine the plain numbers on top

    13x−39+39=13x13x - 39 + 39 = 13x

    −39 and 39 add up to 0, so they cancel. The top is now just 13x, which means 13 times x.

  5. Divide by the bottom

    Now divide the whole top by 13. Dividing by 13 undoes times 13.

    13x13=x\frac{13x}{13} = x

    Watch out: the long bar is under the whole top. Dividing only the last 39 by 13 gives 13x−3613x - 36, choice C. So the answer is A, xx.

Check your answer

Try x = 5. Then x−3=2x - 3 = 2 and 39×2=7839 \times 2 = 78. Next, 78÷3=2678 \div 3 = 26 and 26+39=6526 + 39 = 65. Last, 65÷13=565 \div 13 = 5. You get back 5, the same as x. So x is right.

The original DOE question

Question 67 as printed on page 78 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 78, question 67. Open page 78 of the handbook (PDF)