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 January 3rd, 2017, 01:04 PM #1 Newbie   Joined: Jan 2017 From: uk Posts: 1 Thanks: 0 inverting an equation Hi, I have this equation that I need to invert: (a * b) / (a + b) = c. I want to use a value in c, say 100, and have a value in a, say 10, to find out the answer to b, instead. Last edited by skipjack; January 3rd, 2017 at 02:19 PM.
 January 3rd, 2017, 01:10 PM #2 Senior Member     Joined: Sep 2015 From: Southern California, USA Posts: 1,493 Thanks: 752 $\dfrac{a b}{a+b}=c$ $a b = c(a+b)$ $a b - c b = c a$ $b(a - c) = c a$ if $a \neq c$ $b = \dfrac{a c}{a - c}$ if $a = c$ then $b(0) = c^2$ $0 = c^2$ $c = a = 0$ and $b$ can be any value at all. so $b = \begin{cases} \dfrac{a c}{a - c} & a \neq c \\ \\ x \in \mathbb{R} &a=c \end{cases}$
 January 3rd, 2017, 01:12 PM #3 Newbie   Joined: Nov 2016 From: Canada Posts: 13 Thanks: 3 Math Focus: All (10*b) = 100 *(10+b) 10b = 100(10+b) b = 10(10+b) b = 100 +10b -9b = 100 b = -100/9 Thanks from Davil
 January 4th, 2017, 02:06 AM #4 Member   Joined: Sep 2016 From: India Posts: 88 Thanks: 30 $\dfrac{a×b}{a+b}=c$ Put values $\dfrac{10×b}{10+b}=100$ multiply both sides by $(10+b)$ $10b=100(10+b)$ $10b=1000+100b)$ $10b-1000b= 1000$ $-90b=1000$ $b=\dfrac{1000}{-90b}$ $b=-\dfrac{100}{9}$ Thanks from Davil
January 4th, 2017, 08:32 AM   #5
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 Originally Posted by Impel Tutors (10*b) = 100 *(10+b) 10b = 100(10+b) b = 10(10+b) b = 100 +10b -9b = 100 b = -100/9
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