Question

lesson 25 review - solving rational equations
② (\frac{2a^2}{a^2 - 25} - \frac{a}{(a + 5)} = \frac{7}{(a - 5)})
(lcd = (a + 5)(a - 5))
① factor denominators
② find lcd
③ multiply to create lcd
④ solve using numerator
⑤ check for extraneous solutions
(\frac{2a^2}{(a + 5)(a - 5)} - \frac{a}{(a + 5)} cdot \frac{(a - 5)}{(a - 5)} = \frac{7}{(a - 5)} cdot \frac{(a + 5)}{(a + 5)})
(2a^2 - a(a - 5) = 7(a + 5))
multiple - choice question
this is a powerful yet smooth transition.
she realized that once all fractions contain the lcd in them, we can literally just
□ leave the lcd behind, and basically just bring down only the terms in the numerators.
□ leave the lcd behind, and basically just bring down only the terms in the denominators.
rewatch

Explanation:

To solve a rational equation, after finding the LCD and multiplying each term by the LCD to eliminate denominators, we work with the numerators. The first option says we leave the LCD behind and bring down numerator terms, which matches the process (since multiplying by LCD cancels denominators, leaving numerator expressions). The second option is incorrect as we don't bring down denominator terms.

Answer:

The first option (the one with "leave the LCD behind, and basically just bring down only the terms in the numerators")