Question:If 3x - 2y = x + 2y and 16x/hy = 8, what is the value of h?Enter your answer...

1. TRANSLATE the problem information

• Given information:
  • First equation: \(\mathrm{3x - 2y = x + 2y}\)
  • Second equation: \(\mathrm{\frac{16x}{hy} = 8}\)
  • Need to find: value of h (as integer)

2. INFER the solution strategy

• The first equation can give us a relationship between x and y
• This relationship can then be used in the second equation to solve for h
• Start by solving the linear equation for the x-y relationship

3. SIMPLIFY the first equation

• Start with: \(\mathrm{3x - 2y = x + 2y}\)
• Move x terms to left side: \(\mathrm{3x - x = 2y + 2y}\)
• Combine like terms: \(\mathrm{2x = 4y}\)
• Divide both sides by 2: \(\mathrm{x = 2y}\)

4. INFER how to use this relationship

• We now know \(\mathrm{x = 2y}\)
• This means we can substitute 2y for x in the second equation
• This will eliminate one variable and let us solve for h

5. SIMPLIFY by substitution and solving

• Start with: \(\mathrm{\frac{16x}{hy} = 8}\)
• Substitute \(\mathrm{x = 2y}\): \(\mathrm{\frac{16(2y)}{hy} = 8}\)
• Multiply in numerator: \(\mathrm{\frac{32y}{hy} = 8}\)
• Divide both numerator and denominator by y: \(\mathrm{\frac{32}{h} = 8}\)
• Solve for h: \(\mathrm{h = \frac{32}{8} = 4}\)

Answer: 4

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students treat the two equations as completely separate problems rather than recognizing that the first equation provides crucial information for solving the second equation.

They might try to solve \(\mathrm{\frac{16x}{hy} = 8}\) directly without using the x-y relationship, leading to an equation with too many unknowns. This causes confusion since they can't solve for specific values, leading them to guess or abandon the problem.

Second Most Common Error:

Poor SIMPLIFY execution: Students make algebraic errors when dividing the fraction \(\mathrm{\frac{32y}{hy}}\) by y, incorrectly thinking this gives \(\mathrm{\frac{32}{y}}\) instead of \(\mathrm{\frac{32}{h}}\).

This error comes from not properly understanding that when both numerator and denominator contain the same factor (y), it cancels out completely. This might lead them to get \(\mathrm{h = \frac{32y}{8}}\) or some other incorrect expression instead of \(\mathrm{h = 4}\).

The Bottom Line:

This problem tests whether students can see connections between equations in a system and execute multi-step algebraic simplification accurately. Success requires both strategic thinking (using one equation to inform the other) and careful algebraic manipulation.