\(5(\mathrm{x} + 7) = 15(\mathrm{x} - 17)(\mathrm{x} + 7)\) What is the sum of the solutions to the given equation?...

1. TRANSLATE the problem information

• Given equation: \(5(x + 7) = 15(x - 17)(x + 7)\)
• Need to find: Sum of all solutions

2. INFER the approach

• Notice both sides contain the factor \((x + 7)\)
• Strategy: Move everything to one side and factor out the common term
• This will create a product equal to zero, allowing use of zero product property

3. SIMPLIFY by rearranging

• Subtract \(5(x + 7)\) from both sides:

\(0 = 15(x - 17)(x + 7) - 5(x + 7)\)

4. SIMPLIFY by factoring out common factor

• Factor out \((x + 7)\):

\(0 = (x + 7)[15(x - 17) - 5]\)

• SIMPLIFY the bracketed expression:

\(0 = (x + 7)[15x - 255 - 5]\)
\(0 = (x + 7)(15x - 260)\)

5. INFER using zero product property

• Since the product equals zero, either factor can be zero:

\(x + 7 = 0\) OR \(15x - 260 = 0\)

6. SIMPLIFY each equation

• From \(x + 7 = 0\): \(x = -7\)
• From \(15x - 260 = 0\): \(x = \frac{260}{15} = \frac{52}{3}\)

7. SIMPLIFY to find the sum

• Sum \(= -7 + \frac{52}{3} = \frac{-21}{3} + \frac{52}{3} = \frac{31}{3}\)
• Convert to decimal (use calculator): \(\frac{31}{3} = 10.33...\)

Answer: \(\frac{31}{3}\) or 10.33

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY skill: Students often struggle with factoring out the common factor \((x + 7)\) correctly. They might try to divide both sides by \((x + 7)\) directly, which loses the solution \(x = -7\) since division by zero is undefined when \(x = -7\). Or they make algebraic mistakes when computing \(15(x - 17) - 5\), getting terms like \(15x - 250\) instead of \(15x - 260\).

This leads to incorrect solutions and therefore an incorrect sum, causing confusion and potentially random guessing.

Second Most Common Error:

Inadequate SIMPLIFY execution: Students correctly factor out \((x + 7)\) and apply zero product property, but make arithmetic errors when adding the final fractions. They might incorrectly compute \(-7 + \frac{52}{3}\) by adding \(-7 + 52 = 45\), then dividing by 3 to get 15, instead of converting \(-7\) to \(\frac{-21}{3}\) first.

This may lead them to select an incorrect numerical answer if multiple choice options are available.

The Bottom Line:

This problem tests whether students can recognize and properly factor common algebraic expressions while maintaining precision in fraction arithmetic. The key insight is seeing the common factor \((x + 7)\) and knowing that factoring it out (rather than dividing by it) preserves all solutions.