Question

redo 1st nine weeks
6 of 40
simplify the expression.
\\(\frac{1 + \frac{1}{2 - b}}{1 - \frac{1}{2 + b}}\\)
\\(\frac{1 + \frac{1}{2 - b}}{1 - \frac{1}{2 + b}}=\square\\)

Explanation:

Step1: Combine numerator terms

First, find a common - denominator for the numerator $1+\frac{1}{2 - b}$. The common denominator is $2 - b$, so $1+\frac{1}{2 - b}=\frac{2 - b}{2 - b}+\frac{1}{2 - b}=\frac{2 - b + 1}{2 - b}=\frac{3 - b}{2 - b}$.

Step2: Combine denominator terms

Next, find a common - denominator for the denominator $1-\frac{1}{2 + b}$. The common denominator is $2 + b$, so $1-\frac{1}{2 + b}=\frac{2 + b}{2 + b}-\frac{1}{2 + b}=\frac{2 + b-1}{2 + b}=\frac{1 + b}{2 + b}$.

Step3: Rewrite the original expression

The original expression $\frac{1+\frac{1}{2 - b}}{1-\frac{1}{2 + b}}$ becomes $\frac{\frac{3 - b}{2 - b}}{\frac{1 + b}{2 + b}}$.

Step4: Divide by a fraction

Dividing by a fraction is the same as multiplying by its reciprocal. So $\frac{\frac{3 - b}{2 - b}}{\frac{1 + b}{2 + b}}=\frac{3 - b}{2 - b}\times\frac{2 + b}{1 + b}=\frac{(3 - b)(2 + b)}{(2 - b)(1 + b)}$.

Step5: Expand the numerator and denominator

Expand the numerator: $(3 - b)(2 + b)=6+3b-2b - b^{2}=6 + b - b^{2}$.
Expand the denominator: $(2 - b)(1 + b)=2+2b - b - b^{2}=2 + b - b^{2}$.
So the simplified expression is $\frac{6 + b - b^{2}}{2 + b - b^{2}}$.

Answer:

$\frac{6 + b - b^{2}}{2 + b - b^{2}}$