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August 1st, 2013, 03:59 AM  #1  
Newbie Joined: Aug 2013 Posts: 21 Thanks: 0  Practical Algebra: Reciprocals
Hello. I'm working my way through the book Practical Algebra. I'm now learning about reciprocals and I'm well and truly stuck. Usually when I don't understand something I can look at the answer and work out how they came to that answer but not in this case. The problem: 1 2/3k = 5, k = To clarify, that is 1 & 2/3k, not sure how to type it out. The answer they give is 3/5 * 5/3k = 3/5 * 5, k = 3 To solve this problem we are supposed to use the following rule: Quote:
If any of you could spare the time to help me understand this I would be very grateful.  
August 1st, 2013, 05:11 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,472 Thanks: 2039 
The coefficient of k is 1 & 2/3, i.e., 1 + 2/3, which is 5/3 and has reciprocal 3/5. Multiplying the equation by 3/5 gives (3/5)(5/3)k = (3/5)5, which simplifies to k = 3. 
August 1st, 2013, 05:17 AM  #3  
Newbie Joined: Aug 2013 Posts: 21 Thanks: 0  Re: Quote:
Ok thanks. I'm not sure how 1 + 2/3 makes 5/3 so I think maybe I should put the algebra book on hold while I learn more about fractions. Thanks again.  
August 1st, 2013, 06:22 AM  #4  
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Re: Re: Quote:
Alternatively, imagine two chocolate bars cut into thirds. You are given the whole of the first bar and 2/3 of the second bar, so how many thirds do you have?  

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