Find the value of K for which equation x - Ky = 2, 3x + 2y = -5 has unique…

A system of two linear equations a1x + b1y = c1 and a2x + b2y = c2 has a unique solution if and only if the coefficient ratios differ: a1/a2 ≠ b1/b2 (equivalently, a1b2 − a2b1 ≠ 0). When the ratios are equal, the two lines are parallel or coincident, giving no solution or infinitely many solutions — never exactly one.

Applying this to the given pair of equations, x − Ky = 2 and 3x + 2y = −5:

1. Write both equations in the form ax + by = c: from x − Ky = 2, a1 = 1 and b1 = −K; from 3x + 2y = −5, a2 = 3 and b2 = 2.

2. Apply the unique-solution condition a1/a2 ≠ b1/b2: 1/3 ≠ −K/2.

3. Cross-multiply (valid for a “not-equal-to” relation regardless of sign, since it is unaffected by multiplying both sides by a nonzero number): 1×2 ≠ 3×(−K), that is, 2 ≠ −3K.

4. Divide both sides by −3 (a “not-equal-to” relation is unaffected by dividing by any nonzero number — unlike an inequality such as < or >, there is no direction to flip): K ≠ −2/3.

As an independent check, substitute K = −2/3 into the original pair: the first equation becomes x + (2/3)y = 2, and multiplying by 3 gives 3x + 2y = 6 — which contradicts the second equation 3x + 2y = −5. The two lines are exactly parallel (no common point) at this value, confirming K = −2/3 must be excluded.

Hence the system has a unique solution for every real value of K except K = −2/3.