Question 178954
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It looks like you mean:


*[tex \Large 1 = \frac {3}{x + 2} + \frac {1}{x - 2}]


The Lowest Common Denominator means the smallest number or expression that can be evenly divisible by <i><b>all</b></i> of the denominators.  If the denominators in question have factors in common, then the LCD is something smaller than the simple product of all the denominators. In this case, none of the three denominators, i.e. 1, <i>x + 2</i>, and <i>x - 2</i> have any factors in common.  (I included 1 as a denominator because we are going to have to look at the 1 on the left as the fraction *[tex \Large {1 \over 1}].)


Therefore, the LCD is the product of all the denominators:


*[tex \Large (1)(x + 2)(x - 2) = x^2 - 4] (Use FOIL to verify)


Now you need to apply the LCD so that all of the fractions in the problem have the same denominator.  The process is to multiply each fractional expression by 1 in a form that will supply the missing factor(s) to each denominator:


*[tex \Large \left(\frac {1}{1} \right) \left (\frac {x^2 - 4}{x^2 - 4}\right) = \left[\left(\frac {3}{x + 2}\right) \left(\frac{x - 2}{x - 2}\right)\right] + \left[\left(\frac {1}{x - 2}\right) \left(\frac{x + 2}{x + 2}\right)\right]]


Simplify:


*[tex \Large \frac {x^2 - 4}{x^2 - 4} = \frac {3(x - 2) + (x + 2)}{x^2 - 4}]


Multiply both sides by *[tex \Large x^2 - 4]


*[tex \Large x^2 - 4 = 3(x - 2) + (x + 2)]


Distribute and collect like terms, getting everything on the right and leaving 0 on the left:


*[tex \Large x^2 - 4 = 3x - 6 + x + 2]


*[tex \Large x^2 - 4x = 0]


Factor:


*[tex \Large x(x - 4) =0]


Use the Zero Product Rule (*[tex \Large ab = 0] if and only if *[tex \Large a = 0] or *[tex \Large b = 0])


*[tex \Large x = 0 \text { or } \math x - 4 = 0 \text { } \rightarrow \text { } \math x = 4]


Check the answers against excluded values.  Either 2 or -2 would result in a zero denominator in the original equation, therefore both 2 and -2 must be excluded from any potential solution set.  In this case, neither of the excluded values are potential solutions.


Check the answers to ensure that no extraneous roots were introduced.



*[tex \Large 1 = \frac {3}{0 + 2} + \frac {1}{0 - 2} \text { } \rightarrow \text { } 1 = \frac {3}{2} + \frac {1}{-2} = \frac {2}{2} = 1], Answer checks


*[tex \Large 1 = \frac {3}{4 + 2} + \frac {1}{4 - 2} \text { } \rightarrow \text { } 1 = \frac {3}{6} + \frac {1}{2} = \frac {1}{2} + \frac {1}{2} = 1], Answer checks



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