An artist paints and sells square tiles. The selling price P, in dollars, of a painted tile is a linear...

1. TRANSLATE the problem information

• Given information:
  • Table showing side length s (inches) vs price P (dollars)
  • Need to find linear equation \(\mathrm{P = ms + b}\)
  • Points: \(\mathrm{(3, 8), (6, 18), (9, 28)}\)

2. INFER the approach

• Since this is linear, the slope between any two points should be the same
• Find slope first, then use any point to find the y-intercept
• Check our equation with all given points

3. SIMPLIFY to find the slope

• Using points \(\mathrm{(3, 8)}\) and \(\mathrm{(6, 18)}\):
  \(\mathrm{Slope = \frac{18 - 8}{6 - 3} = \frac{10}{3}}\)
• Verify with points \(\mathrm{(6, 18)}\) and \(\mathrm{(9, 28)}\):
  \(\mathrm{Slope = \frac{28 - 18}{9 - 6} = \frac{10}{3}}\) ✓

4. SIMPLIFY to find the y-intercept

• Use point \(\mathrm{(3, 8)}\) in equation \(\mathrm{P = \frac{10}{3}s + b}\):
  \(\mathrm{8 = \frac{10}{3}(3) + b}\)
  \(\mathrm{8 = 10 + b}\)
  \(\mathrm{b = -2}\)

5. INFER the final equation and verify

• Our equation: \(\mathrm{P = \frac{10}{3}s - 2}\)
• Check all points:
  • \(\mathrm{s = 3: P = 10 - 2 = 8}\) ✓
  • \(\mathrm{s = 6: P = 20 - 2 = 18}\) ✓
  • \(\mathrm{s = 9: P = 30 - 2 = 28}\) ✓

Answer: C. \(\mathrm{P = \frac{10}{3}s - 2}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make calculation errors when finding the slope, getting 3 instead of \(\mathrm{\frac{10}{3}}\).

They might calculate \(\mathrm{\frac{18-8}{6-3}}\) incorrectly as \(\mathrm{\frac{10}{3} = 3}\) (forgetting the division) or mishandle the fraction arithmetic. With slope = 3, they then substitute to get \(\mathrm{P = 3s + b}\), and using point \(\mathrm{(3,8)}\): \(\mathrm{8 = 3(3) + b}\), so \(\mathrm{b = -1}\). But this doesn't match exactly with any choice, so they might select the closest looking option.

This may lead them to select Choice A (\(\mathrm{P = 3s + 10}\)).

Second Most Common Error:

Poor INFER reasoning: Students assume the first y-value in the table (\(\mathrm{P = 8}\) when \(\mathrm{s = 3}\)) represents the y-intercept without understanding that y-intercept occurs when \(\mathrm{s = 0}\).

They correctly calculate slope as \(\mathrm{\frac{10}{3}}\), but then think "the y-intercept is 8 because that's the first price listed." They don't realize they need to substitute a point into \(\mathrm{P = \frac{10}{3}s + b}\) to solve for b.

This may lead them to select Choice B (\(\mathrm{P = \frac{10}{3}s + 8}\)).

The Bottom Line:

This problem tests whether students can systematically apply the slope-intercept process rather than making assumptions or calculation shortcuts. The key insight is that finding slope is only the first step—you must still use point substitution to find the actual y-intercept.