For what value of w is 4w = 2w − 8

2018 Handbook

2018 Test A Math · Question 3 of 57

Also in 2026 Practice Test A Math, question 3; 2025 Practice Test A Math, question 3.

Question

For what value of ww is 4w=2w−84w=2w-8?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: -4

4w=2w−84w=2w-8

2w=−82w=-8

w=−4w=-4

Question analysis

A one-line algebra grid-in: find the value of w that makes 4w equal to 2w minus 8. Students solve a linear equation with the variable on both sides by collecting the w terms and dividing. Errors come from handling the subtraction sign or stopping before w stands alone.

NYS standards
  • 8.EE.7b linear equations with rational coefficients
  • 7.NS.2b rational number division
Grade level
Grade 8 content, by NYS standard
Compared with grade 7
At grade 7 Equations with the variable on both sides are formally a grade 8 topic, but this one takes two short steps with whole numbers. It asks about as much as a typical grade 7 two-step equation.
Estimated difficulty
Routine: one familiar step, little to misread.
Solution
2 steps
Reading level
Question and choices: about grade 7, 7 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

One of the easiest questions on the test: collect the w terms on one side and divide. Errors come from carelessness rather than method, such as dropping the negative sign or gridding −8 after the first step because it already looks like a finished answer. Substituting the result back into both sides of the equation takes seconds and catches both.

  1. Error-prone arithmetic minor Working it out

    The only constant is −8, so the answer is negative; dividing −8 by the coefficient 2 is the step where the sign is most often lost. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

  2. An answer in an unexpected form minor Choosing the answer

    The grid-in needs the negative sign filled in its own column, and there are no choices to show that the answer lies below zero. (Enright, Morley & Sheehan, 2002)

Strategy: Subtract 2w from both sides so the variable appears only once, divide by the remaining coefficient, and check by substituting into both sides.

Traps

  • Negatives and direction: Arithmetic with negative numbers, the order of a subtraction, or direction on a number line. (Moving a term across the equals sign without changing its sign is a faulty procedure.)
  • Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Keeping an equation balanced
if a=b, then a−c=b−c\text{if } a = b \text{, then } a - c = b - c
a and b are the two sides of an equation, and c is any number or term. An equation is like a balanced scale. If you subtract the same thing from both sides, it stays balanced. The same is true if you add the same thing, or divide both sides by the same number that is not zero.
  1. Sort out the equation

    An equation says two things are equal. You need the value of w that makes 4w4w equal to 2w−82w - 8. Here 4w means 4 times w, and 2w means 2 times w. The answer may be a negative number, so keep track of the signs.

  2. Get w on one side

    w is on both sides. Subtract 2w from both sides. Then w is only on the left, and the equation stays balanced.

    4w−2w=2w−8−2w4w - 2w = 2w - 8 - 2w

    2w=−82w = -8

    Now 2 times w is −8.

  3. Divide both sides by 2

    To undo times 2, divide both sides by 2. A negative number divided by a positive number is negative.

    w=−8÷2=−4w = -8 \div 2 = -4

    Watch out: keep the minus sign. Dropping it gives 4. Also, don’t stop at 2w=−82w = -8 and grid −8. So you would grid -4, with the minus sign in its own column.

Check your answer

Put −4 back in for w. Left side: 4×(−4)=−164 \times (-4) = -16. Right side: 2×(−4)−8=−8−8=−162 \times (-4) - 8 = -8 - 8 = -16. Both sides are −16, so −4 works.

The original DOE question

Question 60 as printed on page 81 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 81, question 60. Open page 81 of the handbook (PDF)