Which expression is equivalent to \( \frac{2 a+1}{10 a-5} \div \frac{10 a}{4 a^{2}-1} \) ? \( \frac{2 a}{(2 a-1)^{2}} \) \( \frac{50 a}{(2 a+1)^{2}} \) \( \frac{(2 a-1)^{2}}{2 a} \) \( \frac{(2 a+1)^{2}}{50 a} \)

To simplify the expression \( \frac{2 a+1}{10 a-5} \div \frac{10 a}{4 a^{2}-1} \), we start by converting the division into multiplication by the reciprocal of the second fraction: \[ = \frac{2 a + 1}{10 a - 5} \times \frac{4 a^2 - 1}{10 a} \] Next, we can simplify each component. The denominator \( 10 a - 5 \) can be factored as \( 5(2a - 1) \), and \( 4 a^2 - 1 \) is a difference of squares that can be factored as \( (2a - 1)(2a + 1) \). Putting these together, we have: \[ = \frac{(2 a + 1) \cdot (2a - 1)(2a + 1)}{5(2a - 1) \cdot 10a} \] Now, we can cancel \( 2a - 1 \) from the numerator and the denominator: \[ = \frac{(2a + 1)(2a + 1)}{5 \cdot 10a} \] This simplifies further to: \[ = \frac{(2a + 1)^2}{50a} \] So, the expression simplifies to: \(\frac{(2 a+1)^{2}}{50 a}\), which is one of the options! Are you curious to see how this concept shows up in real-life applications, like engineering challenges or economic models? This kind of algebra is often used in calculating ratios, optimizing functions, and even in designs where precision is key—think stability in bridges or the efficiency of machinery! And hey, if you're ever unsure as you simplify expressions, don’t jump the gun! Double check each factoring step and cancellation. It's a common mistake to overlook factors that can simplify the process quicker. Taking your time can prevent errors that lead you down the wrong path!