If the system of the following equations has the value of the variables as three consecutive integers, then the value of a is ____. x - y + z = 2a x + 4y - 2z = 3(4 - a) 2x - 3y + 4z = 6 - 2a

Correct — B. Solve the system for x, y, z in terms of a first; 'three consecutive integers' is then a single check.

x − y + z = 2a … (1)
x + 4y − 2z = 12 − 3a … (2)
2x − 3y + 4z = 6 − 2a … (3)

Eliminating x between (1) and (2) gives 5y − 3z = 12 − 5a.
Eliminating x between (1) and (3) gives −y + 2z = 6 − 6a.

Solving that pair: z = 6 − 5a, y = 6 − 4a, x = 3a.

At a = 1: x = 3, y = 2, z = 1 — the consecutive integers 1, 2, 3 → option (b)

Why the others are wrong

• (a)
  2 — At a = 2 the formulas give x = 6, y = −2, z = −4. Those are not consecutive, and not even equally spaced.
• (c)
  4 — At a = 4 they give x = 12, y = −10, z = −14. The gaps are 22 and 4, nowhere near the gap of 1 that 'consecutive' demands.
• (d)
  3 — At a = 3 they give x = 9, y = −6, z = −9. Only x = 3a rises with a while y and z fall, so the three values spread apart instead of closing into a run.

Three linear equations in three unknowns carrying a parameter a are still an ordinary system. Solve for x, y and z as expressions in a, and only then impose the extra condition.

Here that gives x = 3a, y = 6 − 4a, z = 6 − 5a — each variable a straight line in a.

'Three consecutive integers' is a property of the set of values, not of the order x, y, z. At a = 1 the values are 3, 2, 1, which is the run 1, 2, 3.

Key facts

• Solving the system gives x = 3a, y = 6 − 4a and z = 6 − 5a.
• a = 1 gives x = 3, y = 2, z = 1, which is the consecutive run 1, 2, 3.
• Those values satisfy all three originals at a = 1: 3 − 2 + 1 = 2, 3 + 8 − 2 = 9 and 6 − 6 + 4 = 4.

Common traps

• Assuming the run must be x, y, z increasing by 1 in that order
• Substituting n, n+1, n+2 straight into the equations, which yields two incompatible values of a
• Expanding 3(4 − a) as 12 − a

No directly related past PYQ was found.