Home Competitions AHSME 1960 Problem 14
If a a a and b b b are real numbers, the equation 3 x − 5 + a = b x + 1 3x-5+a=bx+1 3 x − 5 + a = b x + 1 has a unique solution x x x [The symbol a ≠ 0 a \neq 0 a = 0 means that a a a is different from zero]:
(A) for all a and b(B) if a ≠ 2b(C) if a ≠ 6 (D) if b ≠ 0 (E) if b ≠ 3 \text{(A) for all a and b} \text{(B) if a }\neq\text{2b} \text{(C) if a }\neq 6 \\ \text{(D) if b }\neq 0 \text{(E) if b }\neq 3 (A) for all a and b (B) if a = 2b (C) if a = 6 (D) if b = 0 (E) if b = 3
Dataset text NuminaMath-1.5 . Quoted with source.
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