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If a + b + c = 7/12, 3a - 4b + 5c = 3/4, and 7a - 11b - 13c = -7/12, what is the value of a + c?

Difficulty: Hard

Correct Answer: 5/12

Explanation:


Introduction / Context:
This question tests solving a system of three linear equations in three variables and then finding a required combination (a + c) without necessarily computing everything in the most complicated way. The key is to eliminate b using the first equation and reduce the system to two equations in a and c.


Given Data / Assumptions:

  • a + b + c = 7/12
  • 3a - 4b + 5c = 3/4
  • 7a - 11b - 13c = -7/12
  • Find a + c


Concept / Approach:
Express b from the first equation, substitute into the other two equations, solve the resulting two-equation system in a and c, then add a + c.


Step-by-Step Solution:

Step 1: From a + b + c = 7/12, write b = 7/12 - a - c. Step 2: Substitute into 3a - 4b + 5c = 3/4. Step 3: 3a - 4(7/12 - a - c) + 5c = 3/4. Step 4: 3a - 28/12 + 4a + 4c + 5c = 9/12 => 7a + 9c = 37/12. Step 5: Substitute b into 7a - 11b - 13c = -7/12. Step 6: 7a - 11(7/12 - a - c) - 13c = -7/12. Step 7: 7a - 77/12 + 11a + 11c - 13c = -7/12 => 18a - 2c = 35/6. Step 8: Divide by 2: 9a - c = 35/12 => c = 9a - 35/12. Step 9: Put c into 7a + 9c = 37/12: 7a + 9(9a - 35/12) = 37/12. Step 10: 7a + 81a - 315/12 = 37/12 => 88a = 352/12 => a = 1/3. Step 11: c = 9*(1/3) - 35/12 = 3 - 35/12 = 1/12. Step 12: a + c = 1/3 + 1/12 = 5/12.


Verification / Alternative check:
You can back-substitute to confirm all three equations are satisfied; the arithmetic matches exactly, so the derived a + c is reliable.


Why Other Options Are Wrong:

1/2, 1/4, 3/4, 7/12: these typically occur from fraction handling mistakes (especially converting 3/4 to 9/12 and distributing negatives with -11b).


Common Pitfalls:
Sign errors while substituting b = 7/12 - a - c, or mixing denominators (12 and 6) without converting carefully.


Final Answer:
5/12

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