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)} \frac{(a - 5)}{(a - 5)} = \frac{7}{(a - 5)} \frac{(a + 5)}{(a + 5)}\\)
\\(2a^2 - a(a - 5) = 7(a + 5)\\)
\\(2a^2 - a^2 + 5a = 7a + 35\\)
\\(a^2 + 5a = 7a + 35\\)
\\(-7a -7a\\)
\\(a^2 - 2a = 35\\)
\\(-35 -35\\)
\\(a^2 - 2a - 35 = 0\\)
\\((a - 7)(a + 5) = 0\\)
\\(a - 7 = 0\\) \\(a + 5 = 0\\)
\\(a = 7\\) \\(a = -5\\)
multiple - choice question
listen carefully to my voice note before reading on.
if we substitute \\(a = 7\\) into any of the factors from the lcd that are highlighted in red and green, will the result come out to 0 for any of those denominators?
00:38 00:49
yes; \\(a = 7\\) doesnt cause any of the denominators to be 0.
no; \\(a = 7\\) doesnt cause any of the denominators to be 0.

Explanation:

Step1: Identify denominators' factors

The denominators come from factoring: \(a^2 - 25=(a + 5)(a - 5)\), so the denominators (from the factors in LCD) are \((a + 5)\) and \((a - 5)\).

Step2: Substitute \(a = 7\) into factors

• For \((a + 5)\): Substitute \(a = 7\), we get \(7+5 = 12

eq0\).

• For \((a - 5)\): Substitute \(a = 7\), we get \(7 - 5=2

eq0\).
So \(a = 7\) does not make any of the denominators (from the LCD factors) equal to 0.

Answer:

Yes; \(a = 7\) doesn't cause any of the denominators to be 0. (Assuming the first option is "Yes; \(a = 7\) doesn't cause any of the denominators to be 0" and the second has a typo, the correct one is the first option with that text.)