Solving ((A-B)/B)+((A-C)/C)=1 for A in terms of B & C
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What are your thoughts? What have you tried? For instance, you first found a common denominator for the left hand side, added the fractions... and then what? Please comply with the rules as shown in the Read Before Postinghttps://www.freemathhelp.com/forum/threads/41536-Read-Before-Posting!! thread that's stickied at the top of every sub-forum, and share with us all of the work you've done on this problem, even the parts you know for sure are wrong. Thank you.
((A-B)/B)+((A-C)/C)=1
(A/B)-(B/B)+(A/C)-(C/C)=1
(A/B)-1+(A/C)-1=1
(A/B)+(A/C)=3
A((1/B)+(1/C))=3
A=3/((1/B)+(1/C))
Am I allowed to take the entire ((1/B)+(1/C)) term and place it in the denominator in the last step? Thank you very much for your time
(A/B)-(B/B)+(A/C)-(C/C)=1
(A/B)-1+(A/C)-1=1
(A/B)+(A/C)=3
A((1/B)+(1/C))=3
A=3/((1/B)+(1/C))
Am I allowed to take the entire ((1/B)+(1/C)) term and place it in the denominator in the last step? Thank you very much for your time
Yes you are. That term is simply a number that you do not yet know so you can divide by it.
Because this is for personal use, I am going to go quickly. A great simplifier in algebra is to eliminate fractions as soon as possible by multiplying both sides of the equation by the product of the denominators of every fraction in the equation.
\(\displaystyle \dfrac{a - b}{b} + \dfrac{a - c}{c} = 1 \implies\)
\(\displaystyle bc * \left ( \dfrac{a - b}{b} + \dfrac{a - c}{c} \right ) = bc * 1 \implies \)
\(\displaystyle ac - bc + ab - bc = bc \implies\)
\(\displaystyle ac + ab= 3bc \implies\)
\(\displaystyle a(b + c) = 3bc \implies\)
\(\displaystyle a = \dfrac{3bc}{b + c}\).
You may wonder at the different results. They are equivalent.
\(\displaystyle \dfrac{3}{\dfrac{1}{b} + \dfrac{1}{c}} = 1 * \dfrac{3}{\dfrac{1}{b} + \dfrac{1}{c}} =\)
\(\displaystyle \dfrac{bc}{bc} * \dfrac{3}{\dfrac{1}{b} + \dfrac{1}{c}} = \dfrac{3bc}{\dfrac{bc}{b} + \dfrac{bc}{c}} = \)
\(\displaystyle \dfrac{3bc}{c + b} = \dfrac{3bc}{b + c}.\)
