A chemist uses the equation c = 2r + 3 to model the concentration c, in grams per liter, of...

1. TRANSLATE the problem information

• Given information:
  • Equation: \(\mathrm{c = 2r + 3}\)
  • Need to find: r in terms of c (meaning \(\mathrm{r =}\) something involving c)
• What this tells us: We need to rearrange the equation to get r by itself on one side

2. INFER the approach

• This is a "solve for" problem - we need to isolate r using inverse operations
• Since r is currently multiplied by 2 and then has 3 added to it, we need to undo these operations in reverse order
• Strategy: First subtract 3, then divide by 2

3. SIMPLIFY by applying inverse operations

• Start with: \(\mathrm{c = 2r + 3}\)
• Subtract 3 from both sides: \(\mathrm{c - 3 = 2r}\)
• Divide both sides by 2: \(\mathrm{r = \frac{c - 3}{2}}\)

Answer: A

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make sign errors when moving terms across the equals sign, thinking they need to "move the 3" by adding it instead of subtracting it.

They might write: \(\mathrm{c + 3 = 2r}\), leading to \(\mathrm{r = \frac{c + 3}{2}}\)

This leads them to select Choice B (\(\mathrm{\frac{c + 3}{2}}\))

Second Most Common Error:

Poor INFER reasoning: Students recognize they need to isolate r but get confused about the order of operations and try to divide by 2 first.

They might write: \(\mathrm{\frac{c}{2} = r + 3}\), then \(\mathrm{\frac{c}{2} - 3 = r}\), which gives \(\mathrm{r = \frac{c}{2} - 3 = \frac{c - 6}{2}}\)

This doesn't match any answer choice exactly, leading to confusion and guessing.

The Bottom Line:

Success requires methodically undoing operations in the correct reverse order - when an equation shows "multiply then add," you must "subtract then divide" to solve.