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Intermediate Algebra

OPENSTAX

$Math Studyset Vaia Explanations$ Math

OER 2017 Edition · 1346 pages

ISBN: 9780998625720

Chapter 7: Q 437 (page 737)

In the following exercise solve $\frac{1}{b - 2} + \frac{1}{b + 2} = \frac{3}{b^{2} - 4}$

Short Answer

Expert verified

The solution is $b = \frac{3}{2}$

Step by step solution

01

Step 1. Given Information

$\frac{1}{b - 2} + \frac{1}{b + 2} = \frac{3}{b^{2} - 4}$

02

Step 2. Solve.

The least common denominator of all denominators in the equation is $(b - 2) (b + 2)$

Clear the fraction by multiplying both sides of the equation by the LCD.

$(b - 2) (b + 2) (\frac{1}{b - 2} + \frac{1}{b + 2}) = (b - 2) (b + 2) (\frac{3}{b^{2} - 4}) ((b - 2) (b + 2) \frac{1}{b - 2} + (b - 2) (b + 2) \frac{1}{b + 2}) = (b - 2) (b + 2) (\frac{3}{b^{2} - 4}) (b + 2) + (b - 2) = 3 2 b = 3 b = \frac{3}{2}$

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See all solutions

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