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 Post subject: Arithmetic pg. 74 Question #58Posted: Tue Apr 10, 2012 12:55 pm

Joined: Tue Apr 10, 2012 12:44 pm
Posts: 1
Guys,

I came to the answer of this questions from my own methods but I fear it might consume too much time in a test scenario and I was having a little trouble interpreting the solution given in the answer page. The way I solved it was using one variable and done as so:

[(5/7x + 3) / (x - 9)] = 3/2

The book suggested using two variables and arriving at the conclusions

5x + 3 = 3y and 7x - 9 = 2y

I must not have my head on straight but how do you derive that x=3 and y=6 after calculating those two equations?

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 Post subject: Re: Arithmetic pg. 74 Question #58Posted: Tue Apr 10, 2012 4:09 pm

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
Here is the best way to do it.

B/R = 5/7
7B = 5R
7B - 5R = 0

(B + 3)/(R - 9) = 3/2
2B + 6 = 3R - 27
2B - 3R = - 33

14B - 10R = 0
14B - 21R = - 231

11R = 231

R = 231/11 = 21

21 - 9 = 12 pens after the change

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