Question

# Simplify the complex fraction.$\frac { \frac { 3 } { b ^ { 2 } } - \frac { 4 } { b } + 1 } { 1 - \frac { 1 } { b } - \frac { 6 } { b ^ { 2 } } }$

Solution

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Step 1
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\begin{align*} & \dfrac{ \frac{3}{b^{2}} - \frac{4}{b} + \frac{1}{1} }{ \frac{1}{1} - \frac{1}{b} - \frac{6}{b^{2}} } && \text{write 1 as fraction}\\ \\ & \dfrac{ \frac{3}{b^{2}} - \frac{4}{b} \cdot \frac{b}{b} + \frac{1}{1} \cdot \frac{b^{2}}{b^{2}} }{ \frac{1}{1} \cdot \frac{b^{2}}{b^{2}} - \frac{1}{b} \cdot - \frac{b}{b} - \frac{6}{b^{2}} } && \text{build rational expression}\\ \\ & \dfrac{ \frac{3}{b^{2}} - \frac{4b}{b^{2}} + \frac{b^{2}}{b^{2}} }{ \frac{b^{2}}{b^{2}} - \frac{b}{b^{2}} - \frac{6}{b^{2}} } && \text{multiply numerators and denominators}\\ \\ & \dfrac{ \frac{3-4b+b^{2}}{b^{2}} }{ \frac{b^{2} - b - 6}{b^{2}} } && \text{perform operations in the numerator}\\ \\ & \dfrac{3-4b+b^{2}}{b^{2}} \div \dfrac{b^{2}-b-6}{b^{2}} && \text{write as division}\\ \\ & \dfrac{3-4b+b^{2}}{b^{2}} \cdot \dfrac{b^{2}}{b^{2}-b-6} && \text{multiply the first by the reciprocal of second}\\ \\ & \dfrac{ \left(3-4b+b^{2} \right)b^{2} }{ b^{2} \left(b^{2} - b - 6 \right) } && \text{multiply the numerators and denominators}\\ \\ & \dfrac{ \left(3-b \right) \left(1-b \right) b^{2} }{ b^{2} \left(b+2 \right) \left(b-3 \right) } && \text{factor out completely}\\ \\ & \dfrac{ -1 \cancel{\left(3-b \right)} \left(1-b \right) \cancel{b^{2}} }{ \cancel{b^{2}} \left(b+2 \right) \cancel{\left(b-3 \right)} } && \text{remove common factors}\\ \\ & \boldsymbol{\dfrac{b-1}{b+2}} && \text{simplify}\\ \\ \end{align*}

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