Solve the system: x + y + z = 6, x - y = 2, y + z = 3. What is the value of x?

Question

Accepted Answer

x = 3. From the third equation, y + z = 3, we can substitute into the first equation. Since x + y + z = 6 and y + z = 3, we get x + 3 = 6, so x = 3. We can find the other variables to verify: from x - y = 2 and x = 3, we get y = 1. From y + z = 3 and y = 1, we get z = 2.

Verification: x + y + z = 3 + 1 + 2 = 6. x - y = 3 - 1 = 2. y + z = 1 + 2 = 3. All three equations are satisfied. The solution is (x, y, z) = (3, 1, 2). This problem rewarded recognizing that y + z in the first equation could be directly replaced using the third equation.

Systems of three equations in three variables appear on the CLEP exam. While they can be solved systematically using elimination or substitution (reducing to a two-variable system, then a one-variable equation), always look for shortcuts. Here, noticing that the first and third equations share the expression y + z saves significant computation. On the exam, strategic observation often reduces a seemingly complex problem to a simple substitution.

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