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 September 21st, 2017, 11:07 PM #1 Newbie   Joined: Sep 2017 From: xxxx Posts: 8 Thanks: 0 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
 September 21st, 2017, 11:13 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,790 Thanks: 923 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 $B = -\dfrac{4}{120\cdot 5}c + 1$ $B = -\dfrac{1}{30\cdot 5}c + 1$ $B = -\dfrac{1}{150}c + 1$
 September 22nd, 2017, 07:46 AM #3 Newbie   Joined: Sep 2017 From: xxxx Posts: 8 Thanks: 0 Thank you! I feel ashamed for not seeing that!
September 22nd, 2017, 02:56 PM   #4
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 Originally Posted by New123 I feel ashamed for not seeing that!
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