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September 22nd, 2017, 12:07 AM   #1
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Simple Algebra need help

Hello all,

I simply do not understand the following equation:
$\displaystyle 80−c=80b+40b−40 - c/ 5 $
$\displaystyle 120B + -1/5c = -c + 120 $

Then you have to add 1/5c to both sides of the equation

$\displaystyle 120B+−1/5c+1/5c=−c+120+1/5c$

Resulting in
$\displaystyle 120B = -4/5c + 120$

answer:
$\displaystyle B = -1/150c + 1$

Can someone explain to me:
1: how is the -4/5c calculated? If I rearrange it I seem to not come up with -4/5c
2: How is the -1/150c calculated? (i know you have to divide by 120..) But how does -4/5c / 120 result in 150c? --> 5/4 = 1,25 * 120?

I really appreciate any effort
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September 22nd, 2017, 12:13 AM   #2
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1:$ (-c) + \dfrac 1 5 c = \left(-\dfrac 5 5\right)c + \dfrac 1 5 c = -\dfrac 4 5 c$

2: $120B = -\dfrac 4 5 c + 120$

divide both sides by 120

$B = -\dfrac{4}{120\cdot 5}c + 1$

$B = -\dfrac{1}{30\cdot 5}c + 1 $

$B = -\dfrac{1}{150}c + 1$
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September 22nd, 2017, 08:46 AM   #3
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Thank you!
I feel ashamed for not seeing that!

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September 22nd, 2017, 03:56 PM   #4
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Quote:
Originally Posted by New123 View Post
I feel ashamed for not seeing that!
You shouldn't: you'll never do it gain:
best way to learn
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