The equation \(\mathrm{h = \frac{9(v-273.15)}{5} + 32}\) gives the corresponding temperature h, in degrees Fahrenheit, of any substance that has...

1. TRANSLATE the problem information

• Given information:
  • Temperature conversion formula: \(\mathrm{h = \frac{9(v-273.15)}{5} + 32}\)
  • Fahrenheit temperature: \(\mathrm{h = 467.33°F}\)
  • Need to find: v (temperature in kelvins)

2. INFER the solution strategy

• Since we know h and need to find v, we must solve for v by isolating it
• We'll work backwards through the order of operations, undoing each operation in reverse sequence
• The formula shows: v → subtract 273.15 → multiply by 9 → divide by 5 → add 32 → get h
• To reverse this: h → subtract 32 → multiply by 5 → divide by 9 → add 273.15 → get v

3. SIMPLIFY by substituting and solving step-by-step

• Substitute h = 467.33:

\(\mathrm{467.33 = \frac{9(v-273.15)}{5} + 32}\)

• Subtract 32 from both sides:

\(\mathrm{467.33 - 32 = \frac{9(v-273.15)}{5}}\)

\(\mathrm{435.33 = \frac{9(v-273.15)}{5}}\)

• Multiply both sides by 5 (use calculator):

\(\mathrm{435.33 \times 5 = 9(v-273.15)}\)

\(\mathrm{2176.65 = 9(v-273.15)}\)

• Divide both sides by 9 (use calculator):

\(\mathrm{2176.65 \div 9 = v-273.15}\)

\(\mathrm{241.85 = v-273.15}\)

• Add 273.15 to both sides (use calculator):

\(\mathrm{241.85 + 273.15 = v}\)

\(\mathrm{515 = v}\)

Answer: 515 (or 515.0, or 515 kelvins)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make arithmetic errors when working with decimals, especially in the multiplication and division steps.

For example, they might calculate \(\mathrm{435.33 \times 5}\) incorrectly as 2176.5 instead of 2176.65, or make errors in \(\mathrm{2176.65 \div 9}\). These small arithmetic mistakes compound through the remaining steps, leading to an incorrect final answer. This causes confusion and often leads to guessing among reasonable temperature values.

Second Most Common Error:

Incomplete SIMPLIFY process: Students solve partway through the problem but don't complete all steps to fully isolate v.

They might successfully get to \(\mathrm{241.85 = v - 273.15}\) but then forget the final step of adding 273.15 to both sides. This leads them to incorrectly think the answer is 241.85 instead of 515, which seems reasonable as a temperature value but is wrong.

The Bottom Line:

This problem requires sustained attention through multiple decimal calculations. Students who rush or lose focus during the algebraic manipulation often make small errors that completely derail their solution, even when they understand the correct approach.