Given (x+y-8)/2 = (x+2y-14)/3 = (3x+y-12)/11. Then x, y are

When three algebraic expressions are set equal to a common value, equate them pairwise to form a system of linear equations in the unknowns, then solve that system simultaneously.

1. Equate the first two expressions: (x+y-8)/2 = (x+2y-14)/3. Cross-multiplying: 3(x+y-8) = 2(x+2y-14), i.e., 3x+3y-24 = 2x+4y-28, which simplifies to x - y = -4 ... (i)

2. Equate the last two expressions: (x+2y-14)/3 = (3x+y-12)/11. Cross-multiplying: 11(x+2y-14) = 3(3x+y-12), i.e., 11x+22y-154 = 9x+3y-36, which simplifies to 2x + 19y = 118 ... (ii)

3. Multiply equation (i) by 2: 2x - 2y = -8. Subtracting this from equation (ii): (2x+19y) - (2x-2y) = 118 - (-8), giving 21y = 126, so y = 6.

4. Substitute y = 6 into equation (i): x - 6 = -4, so x = 2.

Substituting x = 2, y = 6 back into all three original expressions confirms the pair is consistent: (2+6-8)/2 = 0, (2+12-14)/3 = 0, and (6+6-12)/11 = 0 — all three equal 0, satisfying the given system simultaneously.

Hence x = 2 and y = 6.