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August 1st, 2013, 03:59 AM   #1
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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:
 To solve an equation for an unknown having a fractional coefficient, multiply both numbers by the reciprocal of the fractional coefficient. For example, if 3/4x = 6, multply both sides by 4/3 (the reciprocal of 3/4). This gives 4/3 * 3/4 = 4/3 * 6, or x = 8.
Ok I don't really understand this at all, I did manage to get the 3 questions before this correct although I'm not really sure what is going on, but I just don't understand were the 3/5 comes from as it's not in the question at all. I thought it would use 3/2 as that is the reciprocal of 2/3.

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: 20,931 Thanks: 2205 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
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Re:

Quote:
 Originally Posted by skipjack 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.

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
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Re: Re:

Quote:
 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.
Well, how many thirds are there in 1? 1 = 3/3 (surely), so 1 + 2/3 = 3/3 + 2/3 = 5/3.

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? Tags algebra, practical, reciprocals Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post gelatine1 Algebra 4 March 23rd, 2014 01:24 AM Jim Advanced Statistics 4 September 9th, 2011 07:46 AM julieta Algebra 1 March 18th, 2011 04:31 AM zolden Number Theory 0 December 14th, 2008 01:32 PM Infinity Number Theory 13 July 21st, 2007 08:35 PM

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