If frac{ax + b}{(3x + 4)^2} = frac{1}{3x + 4} | vedclass

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(B, C) Given the equation: $\frac{ax + b}{(3x + 4)^2} = \frac{1}{3x + 4} - \frac{3}{(3x + 4)^2}$ To combine the terms on the right side,find a common

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$b = 4$

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(B, C) Given the equation: $\frac{ax + b}{(3x + 4)^2} = \frac{1}{3x + 4} - \frac{3}{(3x + 4)^2}$ To combine the terms on the right side,find a common denominator,which is $(3x + 4)^2$: $\frac{ax + b}{(3x + 4)^2} = \frac{1(3x + 4) - 3}{(3x + 4)^2}$ $\frac{ax + b}{(3x + 4)^2} = \frac{3x + 4 - 3}{(3x + 4)^2}$ $\frac{ax + b}{(3x + 4)^2} = \frac{3x + 1}{(3x + 4)^2}$ By comparing the numerators on both sides,we get: $ax + b = 3x + 1$ Comparing the coefficients of $x$ and the constant terms: $a = 3$ and $b = 1$ Therefore,both $b = 1$ and $a = 3$ are correct statements.

If $\frac{ax + b}{(3x + 4)^2} = \frac{1}{3x + 4} - \frac{3}{(3x + 4)^2}$,then:

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