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January 3rd, 2017, 01:04 PM   #1
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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.
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January 3rd, 2017, 01:10 PM   #2
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$\dfrac{a b}{a+b}=c$

$a b = c(a+b)$

$a b - c b = c a$

$b(a - c) = c a$

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 $b = \begin{cases} \dfrac{a c}{a - c} & a \neq c \\ \\ x \in \mathbb{R} &a=c \end{cases}$
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January 3rd, 2017, 01:12 PM   #3
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(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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January 4th, 2017, 02:06 AM   #4
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$\dfrac{a×b}{a+b}=c$

Put values

$\dfrac{10×b}{10+b}=100$

multiply both sides by $(10+b)$

$10b=100(10+b)$

$10b=1000+100b)$

$10b-1000b= 1000$

$-90b=1000$

$b=\dfrac{1000}{-90b}$

$b=-\dfrac{100}{9}$
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January 4th, 2017, 08:32 AM   #5
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Quote:
Originally Posted by Impel Tutors View Post
(10*b) = 100 *(10+b)
10b = 100(10+b)
b = 10(10+b)
b = 100 +10b
-9b = 100
b = -100/9
That'll cost 9.99 ?
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