Exercise 2: A Second Calibration Check — Possible Solution
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GIVEN
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world 3 -> pixel 160
world 6 -> pixel 310

STEP 1: SETTING UP THE SYSTEM
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3m + c = 160
6m + c = 310

STEP 2: GAUSSIAN ELIMINATION
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R2 -> R2 - 2 R1 (since row 2's leading coefficient, 6, is exactly
twice row 1's leading coefficient, 3):

  New row 2, m-coefficient: 6 - 2(3) = 0
  New row 2, c-coefficient: 1 - 2(1) = -1
  New row 2, RHS:           310 - 2(160) = 310 - 320 = -10

Resulting row: 0m - c = -10  ->  c = 10

STEP 3: BACK-SUBSTITUTION
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3m + c = 160
3m + 10 = 160
3m = 150
m = 50

STEP 4: VERIFYING AGAINST BOTH POINTS
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world = 3: 50(3) + 10 = 150 + 10 = 160  -- matches
world = 6: 50(6) + 10 = 300 + 10 = 310  -- matches

STEP 5: COMPARING TO STEP 5's OWN CALIBRATION
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This exercise's constants are m = 50, c = 10. Step 5's own worked
example found m = 50, c = 0. The SLOPE is identical between the two
calibrations (m = 50 in both cases), but the INTERCEPT is genuinely
different (10 here vs. 0 in Step 5) - meaning both calibration runs
agree on how many pixels each world unit is worth, but disagree on
where world position 0 actually lands on screen. This is a genuinely
different mapping overall (a different straight line), even though
one of its two defining constants happens to match.

WHY THIS WORKS AS AN ANSWER
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The system is solved using this chapter's own Step 5 elimination
technique, both original points are used to verify the final answer,
and the comparison against Step 5's own result is made precise by
checking the slope and intercept separately rather than just
declaring the two mappings "different" without saying exactly how.