The given equation relates the distinct positive numbers v, w, and x. Which equation correctly expresses w in terms of...

1. TRANSLATE the problem information

• Given equation: \(\mathrm{v = -w/(150x)}\)
• Need to find: \(\mathrm{w}\) expressed in terms of \(\mathrm{v}\) and \(\mathrm{x}\)
• This means we need \(\mathrm{w}\) isolated on one side, with only \(\mathrm{v}\) and \(\mathrm{x}\) on the other side

2. INFER the solution strategy

• Currently \(\mathrm{w}\) is in the denominator of a fraction with a negative sign
• To isolate \(\mathrm{w}\), we need to "undo" the division by \(\mathrm{150x}\) and the negative sign
• Strategy: Multiply both sides by \(\mathrm{-150x}\) to cancel out the fraction

3. SIMPLIFY by applying the multiplication

• Multiply both sides by \(\mathrm{-150x}\):

\(\mathrm{v \times (-150x) = (-w/(150x)) \times (-150x)}\)

• Left side: \(\mathrm{-150vx}\)
• Right side: The \(\mathrm{-150x}\) cancels with the denominator \(\mathrm{150x}\), leaving us with \(\mathrm{-w \times (-1) = w}\)
• Result: \(\mathrm{-150vx = w}\)

4. SIMPLIFY to final form

• Rearrange: \(\mathrm{w = -150vx}\)

Answer: A. \(\mathrm{w = -150vx}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize what to multiply by to isolate \(\mathrm{w}\) effectively.

Many students see the fraction and think they should just multiply by \(\mathrm{150x}\) (forgetting the negative), or they get confused about which operation undoes division. This leads to incorrect algebraic steps and wrong expressions.

This may lead them to select Choice B (\(\mathrm{w = -150v/x}\)) or causes confusion leading to guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students make sign errors when multiplying by negative values.

When multiplying both sides by \(\mathrm{-150x}\), students often lose track of the negative signs, either forgetting to include the negative in their multiplication or making errors when combining negative signs. The expression \(\mathrm{(-w/(150x)) \times (-150x)}\) requires careful attention to sign changes.

This may lead them to select Choice D (\(\mathrm{w = v + 150x}\)) due to sign confusion.

The Bottom Line:

This problem tests algebraic manipulation skills with fractions and negative signs. Success requires both strategic thinking about how to isolate variables and careful execution with signs during multiplication.