Vincen wrote:Is a > c?
(1) b > d
(2) ab^2 - b > b^2c - d
how can I use both statements together to get an answer?
Statement 1: b>d
Rephrased:
b - d > 0.
Statement 2: ab² - b > cb² - d
ab² - cb² > b - d
b²(a-c) > b-d.
Since b²(a-c) > b-d and b-d > 0, we get:
b²(a-c) > b-d > 0
b²(a-c) > 0.
The inequality in blue is valid only if b is nonzero.
Since the square of any nonzero value must be POSITIVE, b² > 0.
Since b² > 0, we safely both sides of the blue inequality by b²:
b²(a-c)/b² > 0/b²
a-c > 0
a > c.
Thus, the answer to the question stem is YES.
SUFFICIENT.
The correct answer is
C.
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