Question

cphs : advanced algebra: concepts and connections - block (27.0831030)
rational equations
what is the solution to the equation \\( \frac{1}{h - 5} + \frac{2}{h + 5} = \frac{16}{h^2 - 25} \\)?
\\( h = 5 \\)
\\( h = \frac{11}{3} \\)
\\( h = 7 \\)
\\( h = \frac{21}{2} \\)

Explanation:

Step1: Factor the denominator

Notice that \(h^2 - 25=(h - 5)(h + 5)\) by the difference of squares formula \(a^2-b^2=(a - b)(a + b)\). The equation is \(\frac{1}{h - 5}+\frac{2}{h + 5}=\frac{16}{(h - 5)(h + 5)}\).

Step2: Multiply through by the LCD

The least common denominator (LCD) of the fractions is \((h - 5)(h + 5)\). Multiply each term by \((h - 5)(h + 5)\) to eliminate the denominators:

$$
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$$

Step3: Simplify and solve for \(h\)

First, expand the left - hand side: \(h + 5+2h-10 = 16\).
Combine like terms: \(3h-5 = 16\).
Add 5 to both sides: \(3h=16 + 5=21\).
Divide both sides by 3: \(h=\frac{21}{3}=7\).
We need to check for extraneous solutions. The original equation has restrictions \(h
eq5\) and \(h
eq - 5\) (since these values would make the denominators zero). When \(h = 7\), the denominators \(h-5=2\), \(h + 5 = 12\), and \(h^2-25=49 - 25 = 24\) are all non - zero. So \(h = 7\) is a valid solution.

Answer:

\(h = 7\)