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Solve the system of equations: 2xy + y² - 4x - 3y + 2 = 0 and xy + 3y² - 2x - 14y + 16 = 0.

1. (x, y) = (3, 2) and (3/2,...) only
2. x = 3, y = 2
3. x = 2, y = 3
4. no real solution

Correct answer: x = 3, y = 2

Solution

Solve the first equation for x in terms of y, then substitute. Checking y = 2 (which makes the coefficient of x special) and the reduced linear case yields the consistent real solution x = 3, y = 2.

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