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Mathematics 11 Online
OpenStudy (anonymous):

I'm stuck. Help plz! Solve for s 1/g + 1/b + 1/s = 1

OpenStudy (anonymous):

k so you got single s by itself

OpenStudy (anonymous):

I don't think you're doing this right...

OpenStudy (anonymous):

yeah im still doubting this myself....

OpenStudy (anonymous):

ha brother you can find s by verification its simple put g=1,b=1 then s=-1

OpenStudy (akashdeepdeb):

\[\frac{1}{g} + \frac{1}{b} + \frac{1}{s} = 1\] To find s in terms of g and b. Put all the other terms on the other side. \[\frac{1}{s} = 1 - \frac{1}{g} - \frac{1}{b}\] Now find the LCM on the RHS, to get it as a fraction [with one numerator and one denominator, so that you can take the reciprocal to get the value of s] You'll get: \[\frac{1}{s} = \frac{gb - b - g}{gb}\] Now use the property that if \(\frac{a}{b} = \frac{c}{d}\) then \(\frac{b}{a} = \frac{d}{c}\) Getting me? :)

OpenStudy (akashdeepdeb):

@pullaya That is an incorrect method of solving this question.

OpenStudy (anonymous):

Sorry, I don't really understand this...

OpenStudy (akashdeepdeb):

Do you know what LCM is?

OpenStudy (anonymous):

yeah, least common multiple

OpenStudy (akashdeepdeb):

Do you understand why I kept 1/s on one side and took everything else on the other?

OpenStudy (anonymous):

Yup.

OpenStudy (akashdeepdeb):

Which part do you not understand?

OpenStudy (anonymous):

I don't understand how you got the 3rd equation using the least common multiple.

OpenStudy (akashdeepdeb):

Okay, so what is the least common multiple? We find a common denominator for everything by just multiplying each term by a suitable number. Here's what I mean: Let us say we have \[\frac{1}{a} + \frac{1}{b}\] What is a common denominator we can get? We find the LCM of the denominators already there, that is 'a' and 'b'. What is the denominator of a and b? [Well, it is 'ab'] So how do we get a common denominator 'ab', we multiply 'b' to one term and 'a' to the other: \[\frac{1*b}{a*b} + \frac{1*a}{b*a}\] = \[\frac{a}{ab} + \frac{b}{ab}\] = \[\frac{a+b}{ab}\] This is basically, how I got the 3rd equation.

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