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Bigred2008
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Updated on: 08 Dec 2010, 19:18
Originally posted by Bigred2008 on 08 Dec 2010, 19:00.
Last edited by Bigred2008 on 08 Dec 2010, 19:18, edited 1 time in total.
Last edited by Bigred2008 on 08 Dec 2010, 19:18, edited 1 time in total.
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I was working in Book 3 of MGMAT and one of the problems confused me, I was able to solve for it, but the explanation the book gave was confusing but looked like something I may be able to add to my repertoire if I learn it. Can you provide any assistance?
Solve for A using these two equations
\(\frac{B+A}{2A}\)=2
\(a+b=8\)
I was able to solve for this but MGMAT gave an interesting answer for it, and was wondering if someone could help me understand it. They gave their answer as Divide the first equation by the second
\(\frac{B+A}{2A/A+B}\) =\(\frac{2}{8}\) Sorry for the mess, couldn't get this to post neatly
Which reduces to
\(\frac{B+A}{2A(a+b)}\)=\(\frac{2}{8}\)
Which when reduced ends up 2A=4 and A=2
Should you divide equations like that and any suggestions to avoid that in the future?
Solve for A using these two equations
\(\frac{B+A}{2A}\)=2
\(a+b=8\)
I was able to solve for this but MGMAT gave an interesting answer for it, and was wondering if someone could help me understand it. They gave their answer as Divide the first equation by the second
\(\frac{B+A}{2A/A+B}\) =\(\frac{2}{8}\) Sorry for the mess, couldn't get this to post neatly
Which reduces to
\(\frac{B+A}{2A(a+b)}\)=\(\frac{2}{8}\)
Which when reduced ends up 2A=4 and A=2
Should you divide equations like that and any suggestions to avoid that in the future?
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whiplash2411
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08 Dec 2010, 19:09
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Bigred2008
It really depends on each individual equation. Another way of thinking about the solution the book has is to think about substituting the value of a + b, which you know, into the first equation, thereby leaving only one unknown variable. Hence you can easily solve for a.
Bigred2008
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08 Dec 2010, 19:20
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Thanks, you're a pretty big source of knowledge for my math problems in these forums. But your response makes sense, I was just wondering.
Here, the (a+b) chunk is common to both, and if you saw that you could combine the equations that way to eliminate variables (the book solution is the multiplication version of the stack & add/subtract method for systems of equation), awesome. If not-- no worries--substitution will still work, and it's a waste of time to pull your hair out if you don't ID a "cool" way to do a problem quickly.
