Hi Kathy, yes, the procedure is very similar, except that you have to indicate where the double or triple bond is. When you begin by finding and numbering the carbons in the main chain, always start from the end that has the double or triple bond closest to it, not which just has a group closest to it. So that simplifies things.
CH3CH2CH2CH=CH2 This is 1-pentene.
The "1" means the double bond starts on carbon 1 of the main chain. We also changed the name ending from "ane" to "ene" since this is an alkene.
CH3
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CH3CHCH2CHCH2CH2CH=CH2 This is 5,7-dimethyl-1-octene.
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CH3
Getting the right number of hydrogens on the carbons takes a little practice when writing condensed structural formulas like these. When in doubt, draw your structure showing all of the bonds, C-to-C and C-to-H (remember, always have four bonds on carbon!).
When the double bond is not on the end, you have cis and trans isomeric possibilities:
CH3 CH3 CH3 H CH3 H
\ / \ / \ /
C=C C=C C=C
/ \ / \ / \
H H H CH3 CH3 H
cis-2-butene trans-2-butene 2-methyl-1-propene
(has no cis or trans isomers)
cis = "on the same side" of the double bond, trans = "on opposite sides" of the double bond as drawn above.
Alkynes follow the same rules, but there are no cis-trans isomers. (BTW the words cis and trans are alway italicized. 
) CH3CH2C
CH CH3C
CCH3
1-butyne 2-butyne
Since these are alkynes, the parent name ends in "yne".
CH2CH3
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CH3CH2CHCH2CHC
CCH2CH3 This is 7-ethyl-5-methyl-3-nonyne.
|
CH3
That's the general idea! 
Steve
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