Nice way to do an old problem

[Steven G]

Nice way to solve an old problem

Suppose you are asked to solve x/(x+6) =1/2. It would be nice to be able to just equate the numerator (or denominator) and in this case just say x=1 (or x+6=2,...). Most instructor will eventually get the equation 2x=x+6 and continue from there (hopefully saying that 2x=x+X, so x must be 6!)
I prefer to realize that x and (x+6) differ by 6 and that 1 and 2 differ by 1. I want the 1 and 2 to 'differ by 6'. So I multiply 1/2 by 6/6 and know have x/(x+6) = 6/12. So x=6.

Another example: (2x+3)/(2x+11) = 3/5.
1) note on the left side the bottom term is 8 more than the top.
2)We want the same difference on the rhs. So (3/5)(4/4) = 12/20.
3)Then say 2x+3=12 or 2x+11 = 20.
2x+3=12=> 2x=9 =>x=9/2.

Can we have a new section for different techniques for old problems?

 

[stapel]

Can we have a new section for different techniques for old problems?

If you want a place to post your articles, maybe try getting your own web space. Domain names are cheap, as is hosting. But this section (and this forum) aren't really the place for that. Thank you for your consideration.

 

[Steven G]

Sure, that's nice, but how often do you get something that easy to handle?
1 in 1000 cases?

What if you have (3x + 4)(2x + 2) = 5/3 ?
(3x + 4) - (2x + 2) = x + 2 ;
so we want numerator to be x+2 higher than denominator:
5/3[(x+2)/2][(x+2)/2] = [(5x+10)/2] / [(3x+6)/2]
so:
(3x + 4)(2x + 2) = [(5x+10)/2] / [(3x+6)/2]

3x + 4 = (5x + 10) / 2
solve: x = 2

However, cross multiplication would solve way quicker:
(3x + 4)(2x + 2) = 5 / 3
10x + 10 = 9x + 12
x = 2

In answer to your question how often do you get something that easy to handle, I think it is more often then you think. In textbooks (ok, I am talking about textbooks problems) when they have these type problems (ax+b)/(cx+d) =e/f they generally have small integer values for a and c. As a result you have a=c more frequently then you think. I am glad that you think it is a nice technique.