\(\large[\frac{\frac{x+3}{8}}{\frac{(x^2 - 9)}{3}}\large]^{1}\)

The first quantity is raised to -1, then switch it in the reverse form.

\(\frac{3(x+3)}{8(x+3)(x-3)}\)

(x+3) cancel out

\(\frac{3}{8(x-3)}\)

Since x > 4, the denominator in Quantity A must be greater than 8; since the numerators in Quantity A and Quantity B are the same and the denominator in Quantity A is larger, Quantity B must be greater.

Hope now is clear

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