How many solutions does the equation \(10(15\mathrm{x} - 9) = -15(6 - 10\mathrm{x})\) have?

1. SIMPLIFY the equation by applying the distributive property

Given equation: \(10(15\mathrm{x} - 9) = -15(6 - 10\mathrm{x})\)

• Left side: \(10(15\mathrm{x} - 9) = 10(15\mathrm{x}) + 10(-9) = 150\mathrm{x} - 90\)

• Right side: \(-15(6 - 10\mathrm{x}) = -15(6) + (-15)(-10\mathrm{x}) = -90 + 150\mathrm{x}\)

• After distribution: \(150\mathrm{x} - 90 = -90 + 150\mathrm{x}\)

2. SIMPLIFY further by rearranging terms

• Rewrite right side: \(150\mathrm{x} - 90 = 150\mathrm{x} - 90\)

• Notice: Both sides are now identical!

3. INFER what this identity means

• When both sides of an equation are exactly the same, we have an identity
• An identity is true for ANY value of x you substitute
• This means the original equation has infinitely many solutions

Answer: C. Infinitely many

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students correctly simplify to get \(150\mathrm{x} - 90 = 150\mathrm{x} - 90\), but then think "the x's cancel out, so there's no solution."

They mistakenly reason that since the variable "disappears," there must be no value of x that works. However, the opposite is true - when both sides become identical, EVERY value of x works. This confusion about what an identity means may lead them to select Choice D (Zero solutions).

Second Most Common Error:

Inadequate SIMPLIFY execution: Students make sign errors when distributing, particularly with \(-15(6 - 10\mathrm{x})\), forgetting that \(-15 \times (-10\mathrm{x}) = +150\mathrm{x}\).

This leads to incorrect expressions like \(150\mathrm{x} - 90 = -90 - 150\mathrm{x}\), which would actually have exactly one solution. This may lead them to select Choice A (Exactly one).

The Bottom Line:

This problem tests whether students understand the difference between "no solution" (contradiction) and "infinitely many solutions" (identity). The key insight is recognizing that when algebraic manipulation makes both sides identical, every possible x-value satisfies the original equation.