If 3x + 2y = 10 and 2xy = 7, then find the value of 3x − 2y (if 3x − 2y > 0).

Correct — A. The stem is printed as an image: given 3x + 2y = 10 and 2xy = 7, find 3x − 2y, where 3x − 2y > 0.

Use the identity (p + q)² − (p − q)² = 4pq, with p = 3x and q = 2y:
(3x + 2y)² − (3x − 2y)² = 4 × 3x × 2y = 24xy

Now substitute:
(3x − 2y)² = 10² − 24xy
2xy = 7, so xy = 3.5 and 24xy = 84
(3x − 2y)² = 100 − 84 = 16

3x − 2y = ±4, and the rider 3x − 2y > 0 keeps the positive root:
3x − 2y = 4 → option (a)

Why the others are wrong

• (b)
  6 — 6 squares to 36, which would need 24xy = 100 − 36 = 64 and so 2xy = 16⁄3 ≈ 5.3. The stem gives 2xy = 7, a larger product, so the difference is smaller than 6.
• (c)
  10 — 10 is 3x + 2y copied straight from the stem. It would need 24xy = 0, that is 2xy = 0, which contradicts 2xy = 7.
• (d)
  8 — 8 squares to 64, which would need 24xy = 36 and so 2xy = 3. The stem gives 2xy = 7, so 8 is too large.

Two symmetric expressions, p + q and p − q, are linked by one identity: (p + q)² − (p − q)² = 4pq. Give a solver any two of the three quantities and the third follows without ever finding p and q.

Here p = 3x and q = 2y, so 4pq = 4 × 3x × 2y = 24xy — the factor 24, not 4 and not 12, is where most of the marks are lost.

The stem hands you xy through 2xy = 7 precisely so that you never have to solve the pair. The moment you find yourself forming a quadratic in x, you have taken the long road.

Key facts

• (p + q)² − (p − q)² = 4pq, so (3x + 2y)² − (3x − 2y)² = 24xy.
• 2xy = 7 makes xy = 3.5, and 24xy = 84.
• (3x − 2y)² = 100 − 84 = 16, so 3x − 2y is 4 or −4.
• The stated condition 3x − 2y > 0 selects 4.

Common traps

• Subtracting 2xy = 7 where the identity needs 24xy = 84, which leaves 93 and no clean root.
• Building 4pq as 4xy or 12xy and forgetting that p = 3x and q = 2y carry coefficients of their own.
• Stopping at ±4 and ignoring the rider — the pair x = 1, y = 3.5 fits both equations and gives −4.

No directly related past PYQ was found.