Consider the equation \(\frac{\mathrm{x - 3}}{\mathrm{2x + 5}} = \mathrm{4(x - 3)}\), where x is a real number. Note that...

1. TRANSLATE the problem information

• Given equation: \(\frac{\mathrm{x - 3}}{\mathrm{2x + 5}} = \mathrm{4(x - 3)}\)
• APPLY CONSTRAINTS from the start: Since we have a fraction, the denominator cannot be zero
• Domain restriction: \(\mathrm{2x + 5} \neq \mathrm{0}\), so \(\mathrm{x} \neq \mathrm{-}\frac{\mathrm{5}}{\mathrm{2}}\)

2. CONSIDER ALL CASES for the common factor

• Notice that \(\mathrm{(x - 3)}\) appears on both sides of the equation
• This means we need to consider two cases:
  • Case 1: \(\mathrm{x - 3 = 0}\) (the factor equals zero)
  • Case 2: \(\mathrm{x - 3} \neq \mathrm{0}\) (the factor doesn't equal zero)

3. Solve Case 1: \(\mathrm{x - 3 = 0}\)

• If \(\mathrm{x = 3}\), then both sides of the original equation equal 0:
  • Left side: \(\frac{\mathrm{3 - 3}}{\mathrm{2(3) + 5}} = \frac{\mathrm{0}}{\mathrm{11}} = \mathrm{0}\)
  • Right side: \(\mathrm{4(3 - 3) = 4(0) = 0}\)
• APPLY CONSTRAINTS: Check domain restriction: \(\mathrm{x = 3} \neq \mathrm{-}\frac{\mathrm{5}}{\mathrm{2}}\) ✓
• So \(\mathrm{x = 3}\) is a valid solution

4. Solve Case 2: \(\mathrm{x - 3} \neq \mathrm{0}\)

• Since \(\mathrm{x - 3} \neq \mathrm{0}\), we can divide both sides by \(\mathrm{(x - 3)}\):
• SIMPLIFY: \(\frac{\mathrm{x - 3}}{\mathrm{2x + 5}} \div \mathrm{(x - 3)} = \mathrm{4(x - 3)} \div \mathrm{(x - 3)}\)
• This gives us: \(\frac{\mathrm{1}}{\mathrm{2x + 5}} = \mathrm{4}\)

5. SIMPLIFY the resulting equation

• Cross multiply: \(\mathrm{1 = 4(2x + 5)}\)
• Distribute: \(\mathrm{1 = 8x + 20}\)
• Solve: \(\mathrm{8x = 1 - 20 = -19}\)
• Therefore: \(\mathrm{x = -}\frac{\mathrm{19}}{\mathrm{8}}\)

6. APPLY CONSTRAINTS to verify the second solution

• Check domain restriction: \(\mathrm{2(-}\frac{\mathrm{19}}{\mathrm{8}}\mathrm{) + 5 = -}\frac{\mathrm{19}}{\mathrm{4}} + \frac{\mathrm{20}}{\mathrm{4}} = \frac{\mathrm{1}}{\mathrm{4}} \neq \mathrm{0}\) ✓
• Check in original equation:
  • Left side: \(\frac{\mathrm{(-}\frac{\mathrm{19}}{\mathrm{8}} \mathrm{- 3)}}{\mathrm{(}\frac{\mathrm{1}}{\mathrm{4}}\mathrm{)}} = \frac{\mathrm{(-}\frac{\mathrm{43}}{\mathrm{8}}\mathrm{)}}{\mathrm{(}\frac{\mathrm{1}}{\mathrm{4}}\mathrm{)}} = \mathrm{-}\frac{\mathrm{43}}{\mathrm{2}}\)
  • Right side: \(\mathrm{4(-}\frac{\mathrm{19}}{\mathrm{8}} \mathrm{- 3) = 4(-}\frac{\mathrm{43}}{\mathrm{8}}\mathrm{) = -}\frac{\mathrm{43}}{\mathrm{2}}\) ✓

7. Find the sum of all solutions

• Solutions: \(\mathrm{x = 3}\) and \(\mathrm{x = -}\frac{\mathrm{19}}{\mathrm{8}}\)
• Sum: \(\mathrm{3 + (-}\frac{\mathrm{19}}{\mathrm{8}}\mathrm{) = }\frac{\mathrm{24}}{\mathrm{8}} \mathrm{-} \frac{\mathrm{19}}{\mathrm{8}} = \frac{\mathrm{5}}{\mathrm{8}}\)

Answer: \(\frac{\mathrm{5}}{\mathrm{8}}\)

Alternative acceptable forms: \(\mathrm{0.625}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak CONSIDER ALL CASES skill: Students often divide both sides by \(\mathrm{(x - 3)}\) immediately without considering that \(\mathrm{x - 3}\) might equal zero.

They go straight to: \(\frac{\mathrm{1}}{\mathrm{2x + 5}} = \mathrm{4}\), solve to get \(\mathrm{x = -}\frac{\mathrm{19}}{\mathrm{8}}\), and stop there. This causes them to miss the solution \(\mathrm{x = 3}\) entirely. They would then calculate a sum using only one solution, getting an incorrect answer.

Second Most Common Error:

Poor SIMPLIFY execution: Students make arithmetic errors when cross-multiplying or combining fractions.

For example, they might incorrectly solve \(\mathrm{8x = -19}\) as \(\mathrm{x = -19 \div 8 = -}\frac{\mathrm{19}}{\mathrm{8}}\), but then make errors when adding fractions: \(\mathrm{3 + (-}\frac{\mathrm{19}}{\mathrm{8}}\mathrm{)}\). They might incorrectly calculate this as \(\mathrm{3 -} \frac{\mathrm{19}}{\mathrm{8}} = \frac{\mathrm{24}}{\mathrm{8}} \mathrm{-} \frac{\mathrm{19}}{\mathrm{8}} = \frac{\mathrm{5}}{\mathrm{8}}\), but get confused about signs or fraction arithmetic along the way.

The Bottom Line:

The key insight is recognizing that when both sides of an equation contain the same factor, you must consider the case where that factor equals zero separately. This is a critical step that many students skip, leading to incomplete solutions.