Question: The total cost C (in dollars) for producing x items is given by C = 6/7x + 8/7. How...

1. TRANSLATE the problem setup

• Given information:
  • Cost equation: \(\mathrm{C = \frac{6}{7}x + \frac{8}{7}}\)
  • Target cost: \(\mathrm{C = \$8}\)
  • Need to find: x (number of items)

• What this tells us: We need to substitute \(\mathrm{C = 8}\) into the equation and solve for x

2. SIMPLIFY by substituting and isolating the variable

• Substitute \(\mathrm{C = 8}\): \(\mathrm{8 = \frac{6}{7}x + \frac{8}{7}}\)

• Subtract \(\mathrm{\frac{8}{7}}\) from both sides to isolate the x term:
  \(\mathrm{8 - \frac{8}{7} = \frac{6}{7}x}\)

3. SIMPLIFY the left side using common denominators

• Convert 8 to sevenths: \(\mathrm{8 = \frac{56}{7}}\)
• Now subtract: \(\mathrm{\frac{56}{7} - \frac{8}{7} = \frac{48}{7}}\)
• So we have: \(\mathrm{\frac{48}{7} = \frac{6}{7}x}\)

4. SIMPLIFY to solve for x

• Multiply both sides by \(\mathrm{\frac{7}{6}}\) (the reciprocal of \(\mathrm{\frac{6}{7}}\)):
  \(\mathrm{x = \frac{48}{7} \times \frac{7}{6}}\)
  \(\mathrm{x = \frac{48}{6}}\)
  \(\mathrm{x = 8}\)

• Verify:
  \(\mathrm{C = \frac{6}{7}(8) + \frac{8}{7}}\)
  \(\mathrm{C = \frac{48}{7} + \frac{8}{7}}\)
  \(\mathrm{C = \frac{56}{7}}\)
  \(\mathrm{C = 8}\) ✓

Answer: 8

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students often make fraction arithmetic errors, particularly when subtracting mixed numbers and fractions or when multiplying by reciprocals.

For example, they might incorrectly compute \(\mathrm{8 - \frac{8}{7}}\) as 0 (thinking \(\mathrm{\frac{8}{7}}\) cancels with the 8), or make errors converting 8 to the fraction \(\mathrm{\frac{56}{7}}\). Some students might also struggle with multiplying \(\mathrm{\frac{48}{7} \times \frac{7}{6}}\), possibly getting confused about which fraction goes in the numerator versus denominator.

These calculation errors lead to incorrect values for x, causing confusion and potentially random answer selection.

Second Most Common Error:

Poor TRANSLATE reasoning: Some students might set up the equation incorrectly by not properly understanding what 'total cost equals $8' means in the context of the given formula.

They might try to solve \(\mathrm{C = \frac{6}{7}x + \frac{8}{7} = 8}\) as if it were asking for when the expression equals zero, or they might get confused about which variable to solve for.

This leads to working with the wrong equation setup from the start, making their entire solution process incorrect.

The Bottom Line:

This problem tests whether students can handle the combination of algebraic thinking (setting up and solving linear equations) with careful fraction arithmetic - a skill combination that requires both conceptual understanding and computational precision.