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September 21st, 2017, 11:07 PM  #1 
Newbie Joined: Sep 2017 From: xxxx Posts: 2 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: Southern California, USA Posts: 1,488 Thanks: 749 
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: 2 Thanks: 0 
Thank you! I feel ashamed for not seeing that! 
September 22nd, 2017, 02:56 PM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 10,667 Thanks: 697  

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