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Unformatted text preview: (a)
=2
[Solution]
x + 2 x + 5x + 6
2
2x
(b)
= 4[Solution]
x +1
x +1
Solution
(a) 2
x
=2
x + 2 x + 5x + 6 The first step is to factor the denominators to get the LCD. © 2007 Paul Dawkins 68 http://tutorial.math.lamar.edu/terms.aspx College Algebra 2
x
=
x + 2 ( x + 2 )( x + 3)
So, the LCD is ( x + 2 ) ( x + 3) and we will need to avoid x = 2 and x = 3 so we don’t get
division by zero.
Here is the work for this problem. ö
2öæ
x
÷ ( x + 2 )( x + 3)
÷=ç
è x + 2 ø ç ( x + 2 )( x + 3) ÷
è
ø æ
( x + 2 )( x + 3) ç 2 ( x + 3) =  x
2x + 6 = x
3 x = 6
x = 2 So, we get a “solution” that is in the list of numbers that we need to avoid so we don’t get
division by zero and so we can’t use it as a solution. However, this is also the only possible
solution. That is okay. This just means that this equation has no solution.
[Return to Problems] (b) 2
2x
= 4x +1
x +1 The LCD for this equation is x + 1 and we will need to avoid x = 1 so we don’t get division by
zero. Here is the work for this equation. 2x ö
æ2...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.
 Spring '12
 MrVinh

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