5x + 14y = 45 10x + 7y = 27 The solution to the given system of equations is \((\mathrm{x},...

1. TRANSLATE the problem information

• Given system of equations:
  • \(5\mathrm{x} + 14\mathrm{y} = 45\)
  • \(10\mathrm{x} + 7\mathrm{y} = 27\)
• Need to find: the value of \(\mathrm{xy}\) (the product of x and y)

2. INFER the solution strategy

• Since we need both x and y values, we need to solve the system
• Notice the x-coefficients are 5 and 10 - elimination method would work well
• Strategy: Multiply first equation by 2 to get matching x-coefficients, then eliminate

3. SIMPLIFY by preparing for elimination

• Multiply the first equation by 2:
  \(2(5\mathrm{x} + 14\mathrm{y}) = 2(45)\)
  \(10\mathrm{x} + 28\mathrm{y} = 90\)
• Now we have:
  \(10\mathrm{x} + 28\mathrm{y} = 90\)
  \(10\mathrm{x} + 7\mathrm{y} = 27\)

4. SIMPLIFY by eliminating x

• Subtract the second equation from the first:
  \((10\mathrm{x} + 28\mathrm{y}) - (10\mathrm{x} + 7\mathrm{y}) = 90 - 27\)
  \(10\mathrm{x} + 28\mathrm{y} - 10\mathrm{x} - 7\mathrm{y} = 63\)
  \(21\mathrm{y} = 63\)
• Therefore: \(\mathrm{y} = 3\)

5. SIMPLIFY by finding x through substitution

• Substitute \(\mathrm{y} = 3\) into either original equation. Using \(10\mathrm{x} + 7\mathrm{y} = 27\):
  \(10\mathrm{x} + 7(3) = 27\)
  \(10\mathrm{x} + 21 = 27\)
  \(10\mathrm{x} = 6\)
  \(\mathrm{x} = \frac{6}{10} = \frac{3}{5}\)

6. SIMPLIFY to find the final answer

• Calculate the product: \(\mathrm{xy} = \left(\frac{3}{5}\right)(3) = \frac{9}{5} = 1.8\)

Answer: \(\frac{9}{5}\), 1.8, or \(1\frac{4}{5}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students attempt elimination but don't recognize they need to create matching coefficients first. They try to directly subtract \(5\mathrm{x} + 14\mathrm{y} = 45\) from \(10\mathrm{x} + 7\mathrm{y} = 27\), which leads to \(-5\mathrm{x} - 7\mathrm{y} = -18\). This creates a more complex system that doesn't eliminate either variable cleanly, causing confusion and often leading to guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly set up elimination but make arithmetic errors when multiplying the first equation by 2 or when subtracting equations. For example, they might get \(10\mathrm{x} + 28\mathrm{y} = 90\) correctly but then subtract incorrectly to get something like \(28\mathrm{y} - 7\mathrm{y} = 20\mathrm{y}\) instead of \(21\mathrm{y}\), leading to \(\mathrm{y} = \frac{63}{20}\) instead of \(\mathrm{y} = 3\). This cascades into wrong x-values and ultimately an incorrect product.

The Bottom Line:

This problem tests whether students can strategically plan their elimination approach AND execute multi-step algebraic manipulations accurately. The key insight is recognizing that creating matching coefficients (by multiplying strategically) makes elimination much cleaner than trying to work with the original coefficients.