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Question
Find the algebraic sum of the following:
(2a+3b)/3a - (a-2b)/4b + (a^2+b^2)/ab
The common denominator is 12ab
(2a+3b)/3a = (4b)(2a+3b)/12ab = (8ab + 12b^2)/12ab
(a-2b)/4b = (3a)(a-2b)/12ab = (3a^2 - 6ab)/12ab
(a^2+b^2)/ab = (12a^2+12b^2)/12ab
So (2a+3b)/3a - (a-2b)/4b + (a^2+b^2)/ab =
(8ab + 12b^2)/12ab - (3a^2 - 6ab)/12ab + (12a^2+12b^2)/12ab =
(9a^2 + 14ab + 24b^2)/12ab
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| Comment | thanks Socrates | ||
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Socrates
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I can answer any questions from the standard four semester Calulus sequence. Derivatives, partial derivatives, chain rule, single and multiple integrals, change of variable, sequences and series, vector integration (Green`s Theorem, Stokes, and Gauss) and applications. Pre-Calculus, Linear Algebra and Finite Math questions are also welcome.
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Ph.D. in Mathematics and many years teaching undergraduate courses at three state universities.
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B.S. , M.S. , Ph.D.
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