The formula F = 9/5C + 32 converts temperature from degrees Celsius (C) to degrees Fahrenheit (F). For what value...

1. TRANSLATE the problem information

• Given information:
  • Formula: \(\mathrm{F = \frac{9}{5}C + 32}\)
  • We want \(\mathrm{F = 50}\)
  • Need to find: the value of C

• What this tells us: We need to substitute \(\mathrm{F = 50}\) into the formula and solve for C

2. TRANSLATE the setup into an equation

• Substitute \(\mathrm{F = 50}\) into the formula:
  \(\mathrm{50 = \frac{9}{5}C + 32}\)

• This gives us a linear equation in C that we can solve

3. SIMPLIFY by isolating the variable term

• Subtract 32 from both sides:
  \(\mathrm{50 - 32 = \frac{9}{5}C}\)
  \(\mathrm{18 = \frac{9}{5}C}\)

• Now we have the coefficient \(\mathrm{\frac{9}{5}}\) times C equals 18

4. SIMPLIFY to find C

• Multiply both sides by 5/9 (the reciprocal of 9/5):
  \(\mathrm{C = 18 \times \frac{5}{9}}\)
  \(\mathrm{C = \frac{18 \times 5}{9} = \frac{90}{9} = 10}\)

Answer: \(\mathrm{C = 10}\) (or \(\mathrm{10°C}\))

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Making arithmetic errors when working with fractions

Students often struggle with multiplying \(\mathrm{18 \times \frac{5}{9}}\), either:

• Forgetting to multiply by the reciprocal and instead dividing by 5/9
• Making calculation errors: getting \(\mathrm{18 \times 5 = 80}\) instead of 90, leading to \(\mathrm{C = \frac{80}{9} \approx 8.9}\)
• Not simplifying \(\mathrm{\frac{90}{9}}\) to 10

This leads to confusion and incorrect numerical answers.

Second Most Common Error:

Poor TRANSLATE reasoning: Setting up the wrong equation

Some students substitute \(\mathrm{C = 50}\) instead of \(\mathrm{F = 50}\), creating:
\(\mathrm{F = \frac{9}{5}(50) + 32 = 90 + 32 = 122}\)

They then think the answer is 122, completely missing that they were supposed to find C, not F.

The Bottom Line:

This problem tests whether students can work backwards from a formula they're given. The key challenge is maintaining accuracy through multiple algebraic steps while working with fractions.