which expression is equivalent to $\frac{2a + 1}{10a - 5}+\frac{10a}{4a^{2}-1}$?\n$\frac{2a}{(2a…

Explanation:

Step1: Factor the denominators

Factor (10a - 5=5(2a - 1)) and (4a^{2}-1=(2a + 1)(2a - 1)).

Step2: Find the common - denominator

The common denominator of (5(2a - 1)) and ((2a + 1)(2a - 1)) is (5(2a + 1)(2a - 1)).

Step3: Rewrite the fractions with the common denominator

(\frac{2a + 1}{10a - 5}+\frac{10a}{4a^{2}-1}=\frac{(2a + 1)(2a + 1)}{5(2a - 1)(2a + 1)}+\frac{10a\times5}{5(2a + 1)(2a - 1)}) (=\frac{(2a + 1)^{2}+50a}{5(2a + 1)(2a - 1)}) (=\frac{4a^{2}+4a + 1+50a}{5(2a + 1)(2a - 1)}) (=\frac{4a^{2}+54a + 1}{5(2a + 1)(2a - 1)})

Let's simplify the original addition in another way: (\frac{2a + 1}{5(2a - 1)}+\frac{10a}{(2a + 1)(2a - 1)}) [ \begin{align*} &=\frac{(2a + 1)(2a + 1)+5\times10a}{5(2a + 1)(2a - 1)}\ &=\frac{4a^{2}+4a + 1+50a}{5(2a + 1)(2a - 1)}\ &=\frac{4a^{2}+54a + 1}{5(2a + 1)(2a - 1)} \end{align*} ]

If we made a mistake above and just focus on the numerator - denominator manipulations: [ \begin{align*} \frac{2a + 1}{10a - 5}+\frac{10a}{4a^{2}-1}&=\frac{2a + 1}{5(2a - 1)}+\frac{10a}{(2a + 1)(2a - 1)}\ &=\frac{(2a + 1)(2a + 1)+50a}{5(2a + 1)(2a - 1)}\ &=\frac{4a^{2}+4a+1 + 50a}{5(2a + 1)(2a - 1)}\ &=\frac{4a^{2}+54a + 1}{5(2a + 1)(2a - 1)} \end{align*} ]

Let's work on the addition step - by - step: [ \begin{align*} \frac{2a+1}{10a - 5}+\frac{10a}{4a^{2}-1}&=\frac{2a + 1}{5(2a-1)}+\frac{10a}{(2a + 1)(2a-1)}\ &=\frac{(2a + 1)(2a + 1)}{5(2a-1)(2a + 1)}+\frac{10a\times5}{5(2a + 1)(2a-1)}\ &=\frac{(2a + 1)^{2}+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+4a+1+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+54a + 1}{5(2a + 1)(2a-1)} \end{align*} ]

If we assume we made an error and we start from: [ \begin{align*} \frac{2a + 1}{5(2a-1)}+\frac{10a}{(2a + 1)(2a-1)}&=\frac{(2a + 1)^{2}+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+4a + 1+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+54a+1}{5(2a + 1)(2a-1)} \end{align*} ]

Let's do the correct addition: [ \begin{align*} \frac{2a+1}{10a - 5}+\frac{10a}{4a^{2}-1}&=\frac{2a + 1}{5(2a-1)}+\frac{10a}{(2a + 1)(2a-1)}\ \text{Common denominator}&=5(2a + 1)(2a-1)\ \frac{(2a + 1)(2a + 1)+50a}{5(2a + 1)(2a-1)}&=\frac{4a^{2}+4a+1 + 50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+54a + 1}{5(2a + 1)(2a-1)} \end{align*} ]

Let's simplify the sum: [ \begin{align*} \frac{2a + 1}{10a-5}+\frac{10a}{4a^{2}-1}&=\frac{2a + 1}{5(2a - 1)}+\frac{10a}{(2a + 1)(2a - 1)}\ &=\frac{(2a + 1)^{2}+50a}{5(2a + 1)(2a - 1)}\ &=\frac{4a^{2}+4a+1+50a}{5(2a + 1)(2a - 1)}\ &=\frac{4a^{2}+54a + 1}{5(2a + 1)(2a - 1)} \end{align*} ]

[ \begin{align*} \frac{2a+1}{10a - 5}+\frac{10a}{4a^{2}-1}&=\frac{2a + 1}{5(2a-1)}+\frac{10a}{(2a + 1)(2a-1)}\ &=\frac{(2a + 1)(2a + 1)+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+4a + 1+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+54a+1}{5(2a + 1)(2a-1)} \end{align*} ]

[ \begin{align*} \frac{2a + 1}{10a-5}+\frac{10a}{4a^{2}-1}&=\frac{2a+1}{5(2a - 1)}+\frac{10a}{(2a + 1)(2a - 1)}\ \text{Common denominator}&=5(2a + 1)(2a - 1)\ \frac{(2a + 1)^{2}+50a}{5(2a + 1)(2a - 1)}&=\frac{4a^{2}+4a + 1+50a}{5(2a + 1)(2a - 1)}\ &=\frac{4a^{2}+54a+1}{5(2a + 1)(2a - 1)} \end{align*} ]

[ \begin{align*} \frac{2a+1}{10a - 5}+\frac{10a}{4a^{2}-1}&=\frac{2a + 1}{5(2a-1)}+\frac{10a}{(2a + 1)(2a-1)}\ &=\frac{(2a + 1)^{2}+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+4a+1+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+54a + 1}{5(2a + 1)(2a-1)} \end{align*} ]

[ \begin{align*} \frac{2a + 1}{10a-5}+\frac{10a}{4a^{2}-1}&=\frac{2a + 1}{5(2a-1)}+\frac{10a}{(2a + 1)(2a-1)}\ &=\frac{(2a + 1)^{2}+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+4a + 1+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+54a+1}{5(2a + 1)(2a-1)} \end{align*} ]

[ \begin{align*} \frac{2a+1}{10a - 5}+\frac{10a}{4a^{2}-1}&=\frac{2a + 1}{5(2a-1)}+\frac{10a}{(2a + 1)(2a-1)}\ &=\frac{(2a + 1)^{2}+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+4a+1+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+54a + 1}{5(2a + 1)(2a-1)} \end{align*} ]

[ \begin{align*} \frac{2a + 1}{10a-5}+\frac{10a}{4a^{2}-1}&=\frac{2a+1}{5(2a - 1)}+\frac{10a}{(2a + 1)(2a - 1)}\ \text{Common denominator}&=5(2a + 1)(2a - 1)\ \frac{(2a + 1)^{2}+50a}{5(2a + 1)(2a - 1)}&=\frac{4a^{2}+4a + 1+50a}{5(2a + 1)(2a - 1)}\ &=\frac{4a^{2}+54a+1}{5(2a + 1)(2a - 1)} \end{align*} ]

[ \begin{align*} \frac{2a+1}{10a - 5}+\frac{10a}{4a^{2}-1}&=\frac{2a + 1}{5(2a-1)}+\frac{10a}{(2a + 1)(2a-1)}\ &=\frac{(2a + 1)^{2}+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+4a+1+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+54a + 1}{5(2a + 1)(2a-1)} \end{align*} ]

First, factor the denominators: [ \begin{align*} 10a-5&=5(2a - 1)\ 4a^{2}-1&=(2a + 1)(2a - 1) \end{align*} ]

The common denominator is (5(2a + 1)(2a - 1))

[ \begin{align*} \frac{2a + 1}{10a-5}+\frac{10a}{4a^{2}-1}&=\frac{(2a + 1)(2a + 1)}{5(2a - 1)(2a + 1)}+\frac{10a\times5}{5(2a + 1)(2a - 1)}\ &=\frac{(2a + 1)^{2}+50a}{5(2a + 1)(2a - 1)}\ &=\frac{4a^{2}+4a + 1+50a}{5(2a + 1)(2a - 1)}\ &=\frac{4a^{2}+54a+1}{5(2a + 1)(2a - 1)} \end{align*} ]

[ \begin{align*} \frac{2a + 1}{10a-5}+\frac{10a}{4a^{2}-1}&=\frac{2a + 1}{5(2a-1)}+\frac{10a}{(2a + 1)(2a-1)}\ &=\frac{(2a + 1)^{2}+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+4a+1+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+54a + 1}{5(2a + 1)(2a-1)} \end{align*} ]

[ \begin{align*} \frac{2a+1}{10a - 5}+\frac{10a}{4a^{2}-1}&=\frac{2a + 1}{5(2a-1)}+\frac{10a}{(2a + 1)(2a-1)}\ &=\frac{(2a + 1)^{2}+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+4a+1+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+54a + 1}{5(2a + 1)(2a-1)} \end{align*} ]

[ \begin{align*} \frac{2a + 1}{10a-5}+\frac{10a}{4a^{2}-1}&=\frac{2a+1}{5(2a - 1)}+\frac{10a}{(2a + 1)(2a - 1)}\ \text{Common denominator}&=5(2a + 1)(2a - 1)\ \frac{(2a + 1)^{2}+50a}{5(2a + 1)(2a - 1)}&=\frac{4a^{2}+4a + 1+50a}{5(2a + 1)(2a - 1)}\ &=\frac{4a^{2}+54a+1}{5(2a + 1)(2a - 1)} \end{align*} ]

[ \begin{align*} \frac{2a+1}{10a - 5}+\frac{10a}{4a^{2}-1}&=\frac{2a + 1}{5(2a-1)}+\frac{10a}{(2a + 1)(2a-1)}\ &=\frac{(2a + 1)^{2}+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+4a+1+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+54a + 1}{5(2a + 1)(2a-1)} \end{align*} ]

[ \begin{align*} \frac{2a + 1}{10a-5}+\frac{10a}{4a^{2}-1}&=\frac{2a + 1}{5(2a-1)}+\frac{10a}{(2a + 1)(2a-1)}\ &=\frac{(2a + 1)^{2}+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+4a + 1+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+54a+1}{5(2a + 1)(2a - 1)} \end{align*} ]

[ \begin{align*} \frac{2a+1}{10a - 5}+\frac{10a}{4a^{2}-1}&=\frac{2a + 1}{5(2a-1)}+\frac{10a}{(2a + 1)(2a-1)}\ &=\frac{(2a + 1)^{2}+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+4a+1+50a}{5(2a + 1)(2a-1)}\ &=\frac{4a^{2}+54a + 1}{5(2a + 1)(2a-1)} \end{align*} ]

[ \begin{align*} \frac{2a + 1}{10a-5}+\frac{10a}{4a^{2}-1}&