SOLUTION: I am not sure if I am doing this right. Solve for K. a/b=b/a+1/k (bak) a/b=(bak)b/a+1/k(bak) a^2k=b^2k+ba a^2k-b^2k=ba k(a^2-b^2)=ab k=ab/a^2-b^2

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Question 121989: I am not sure if I am doing this right. Solve for K.
a/b=b/a+1/k
(bak) a/b=(bak)b/a+1/k(bak)
a^2k=b^2k+ba
a^2k-b^2k=ba
k(a^2-b^2)=ab
k=ab/a^2-b^2
Found 2 solutions by oscargut, bucky:
Answer by oscargut(2103) (Show Source):
You can put this solution on YOUR website!
Yes you are right,
obs: a^2-b^2 must be not equal to 0
Other way
a/b=b/a+1/k then (a/b)-(b/a)=1/k then (a^2-b^2)/ab = 1/k
then k= (ab)/(a^2-b^2)

Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website!
Your method of solving this is correct, and your answer is correct.
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However, you should write your answer as:
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k = ab/(a^2 - b^2)
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The parentheses in the bottom are important, because without them the rules of algebra say
that the way you wrote the answer (that is, without the parentheses) the answer should be read as:
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and this is not what you intended, I'm sure. With the parentheses surrounding all the terms in
the denominator, the answer is correctly interpreted to be:
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Just a comment to get you to think about the way a reader could interpret your answer incorrectly.
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Good job on getting your way through the problem ... You obviously understand what you are
doing.
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