Unformatted text preview: rks identical to the previous one so
let’s do the work. ( 2 )( y  3) 2 æ
ö æ 10  y ö
5
2
÷ 2 y  3)
ç
÷=ç
2 ( )(
ç 2 ( y  3) ÷ ç ( y  3 ) ÷
è
øè
ø
5 ( y  3) = 2 (10  y )
5 y  15 = 20  2 y
7 y = 35
y=5 Now the solution is not y = 3 so we won’t get division by zero with the solution which is a good
thing. Finally, let’s do a quick verification. 5
10  5
?
=2
2 ( 5)  6 5  6 ( 5 ) + 9
55
=
44 OK So we did the work correctly.
[Return to Problems] (d) 2z
3
=
+2
z + 3 z  10 In this case it looks like the LCD is ( z + 3) ( z  10 ) and it also looks like we will need to avoid z = 3 and z = 10 to make sure that we don’t get division by zero.
Let’s get started on the work for this problem. ( z + 3)( z  10 ) æ
ç 2z ö æ 3
ö
+ 2 ÷ ( z + 3)( z  10 )
÷=ç
è z + 3 ø è z  10
ø 2 z ( z  10 ) = 3 ( z + 3) + 2 ( z + 3)( z  10 )
2 z 2  20 z = 3z + 9 + 2 ( z 2  7 z  30 ) At this point let’s pause and acknowledge that we’ve got a z2 in the work here. Do not get excited
about that. Sometimes these will show up temporarily in these problems. You should only worry
about it...
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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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