Intermediate Algebra help

H

high point

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Feb 26, 2010
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Hi, I am studying for a test on Monday and I came across this algebra problem. I have the question: 3m = 9k[sup:2csipxex]2[/sup:2csipxex]. The answer for that is: 3k[sup:2csipxex]2[/sup:2csipxex] – m = 1/3.

I do not understand how to get the 1/3.

Thanks in advance.
 
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\(\displaystyle 3m = 9k^2\)

\(\displaystyle 3m - 9k^2 = 0\)

\(\displaystyle (3)(m-3k^2) = 0\)

\(\displaystyle (3)(3k^2 - m) = 0\)

\(\displaystyle 3k^2 - m = 0\)

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Crap, I am sorry, I mistyped my question. The question was: 3m = 9k[sup:1tw6vj5n]2[/sup:1tw6vj5n] - 1. The answer to that is: 3k[sup:1tw6vj5n]2[/sup:1tw6vj5n] – m = 1/3

How do get to the (1/3)?

Thank you.
 
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Do the same steps I've showed you take 3 as a common factor and you'll get the answer . . . :wink:

\(\displaystyle (3)(3k^2 - m) = 1\)

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Thank you so much Aladdin. This was confusing for me for a while. What threw me off was solving for a real number and not a variable, even though to solve it was like solving for a variable.
 
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