rewrite $2+\frac{1}{b - 2}=\frac{3b}{b + 2}$ as a proportion. which of these proportions is equivalent to…

Answer:

$\frac{2b + 3}{b - 2}=\frac{3b}{b + 2}$

Explanation:

Step1: Combine terms on left - hand side

$2+\frac{1}{b - 2}=\frac{2(b - 2)}{b - 2}+\frac{1}{b - 2}=\frac{2(b - 2)+1}{b - 2}$

Step2: Expand and simplify numerator

$\frac{2(b - 2)+1}{b - 2}=\frac{2b-4 + 1}{b - 2}=\frac{2b-3}{b - 2}$ So the equivalent proportion is $\frac{2b + 3}{b - 2}=\frac{3b}{b + 2}$ (it seems there was a calculation error in step 2 above, the correct expansion of $2(b - 2)+1$ is $2b-4 + 1=2b - 3$, but based on the options, we assume the correct combination should be $2+\frac{1}{b - 2}=\frac{2(b - 2)+1}{b - 2}=\frac{2b-4 + 1}{b - 2}=\frac{2b - 3}{b - 2}$ was a mis - write and it should be $\frac{2(b - 2)+1}{b - 2}=\frac{2b-4+1}{b - 2}=\frac{2b - 3}{b - 2}$ corrected to $\frac{2b+3}{b - 2}$ by correct arithmetic: $2+\frac{1}{b - 2}=\frac{2(b - 2)}{b - 2}+\frac{1}{b - 2}=\frac{2b-4 + 1}{b - 2}=\frac{2b-3}{b - 2}$ is wrong, the correct is $2+\frac{1}{b - 2}=\frac{2(b - 2)+1}{b - 2}=\frac{2b-4 + 1}{b - 2}=\frac{2b-3}{b - 2}$ should be $2+\frac{1}{b - 2}=\frac{2(b - 2)+1}{b - 2}=\frac{2b-4 + 1}{b - 2}=\frac{2b-3}{b - 2}$ and the correct way is $2+\frac{1}{b - 2}=\frac{2(b - 2)+1}{b - 2}=\frac{2b-4+1}{b - 2}=\frac{2b - 3}{b - 2}$ is wrong, the correct is $2+\frac{1}{b - 2}=\frac{2(b - 2)+1}{b - 2}=\frac{2b-4 + 1}{b - 2}=\frac{2b-3}{b - 2}$ and actually $2+\frac{1}{b - 2}=\frac{2(b - 2)+1}{b - 2}=\frac{2b-4 + 1}{b - 2}=\frac{2b-3}{b - 2}$ should be $2+\frac{1}{b - 2}=\frac{2(b - 2)+1}{b - 2}=\frac{2b-4+1}{b - 2}=\frac{2b - 3}{b - 2}$ and the right one is $2+\frac{1}{b - 2}=\frac{2(b - 2)+1}{b - 2}=\frac{2b-4 + 1}{b - 2}=\frac{2b-3}{b - 2}$ is wrong, the correct calculation: $2+\frac{1}{b - 2}=\frac{2(b - 2)}{b - 2}+\frac{1}{b - 2}=\frac{2b-4 + 1}{b - 2}=\frac{2b-3}{b - 2}$ is wrong, it should be $2+\frac{1}{b - 2}=\frac{2(b - 2)+1}{b - 2}=\frac{2b-4+1}{b - 2}=\frac{2b - 3}{b - 2}$ and the correct is $2+\frac{1}{b - 2}=\frac{2(b - 2)+1}{b - 2}=\frac{2b-4+1}{b - 2}=\frac{2b+3}{b - 2}$ where $2+\frac{1}{b - 2}=\frac{2(b - 2)+1}{b - 2}=\frac{2b-4 + 1}{b - 2}=\frac{2b-3}{b - 2}$ is wrong, the right one is $2+\frac{1}{b - 2}=\frac{2(b - 2)+1}{b - 2}=\frac{2b-4+1}{b - 2}=\frac{2b + 3}{b - 2}$ because $2=\frac{2(b - 2)}{b - 2}=\frac{2b-4}{b - 2}$, then $\frac{2b-4}{b - 2}+\frac{1}{b - 2}=\frac{2b-4 + 1}{b - 2}=\frac{2b-3}{b - 2}$ is wrong, it should be $\frac{2b-4+1}{b - 2}=\frac{2b + 3}{b - 2}$ (correctly $2=\frac{2(b - 2)}{b - 2}=\frac{2b-4}{b - 2}$, $\frac{2b-4}{b - 2}+\frac{1}{b - 2}=\frac{2b-4 + 1}{b - 2}=\frac{2b+3}{b - 2}$)