Unformatted text preview: (a) 1
5
= 1x +1
2x  4 The LCD for this problem is ( x + 1) ( 2 x  4 ) and we will need to avoid x = 1 and x = 2 to
make sure we don’t get division by zero. Here is the work for this equation. ( x + 1)( 2 x  4 ) æ
ç 1ö
5ö
æ
÷ = ( x + 1)( 2 x  4 ) ç1 ÷
è x +1ø
è 2x  4 ø
2 x  4 = ( x + 1)( 2 x  4 )  5 ( x + 1)
2 x  4 = 2 x2  2 x  4  5x  5
0 = 2 x2  9 x  5
0 = ( 2 x + 1)( x  5 ) So, it looks like the two solutions to this equation are, x= 1
2 and x=5 Notice as well that neither of these are the values of x that we needed to avoid and so both are
solutions.
[Return to Problems] 3
4 x
=
x 1 x 1
In this case the LCD is x  1 and we will need to avoid x = 1 so we don’t get division by zero.
(b) x + 3 + Here is the work for this problem. ( x  1) æ x + 3 +
ç 3 ö æ 4 xö
÷=ç
÷ ( x  1)
x 1 ø è x 1 ø è
( x  1)( x + 3) + 3 = 4  x
x2 + 2 x  3 + 3 = 4  x
x 2 + 3x  4 = 0 ( x  1)( x + 4 ) = 0
So, the quadratic that we factored and solved has two solutions, x = 1 and x = 4 . However,
when we...
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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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