04-07-2010 09:36 PM
04-07-2010 10:03 PM - edited 04-07-2010 10:03 PM
What are you looking for, the intersection of 2 lines? Trying to find an x,y pair that satisfies both equations?
That is basic algebra.
ax+b=cx+d
(a-c)x=d-b
x=(d-b)/(a-c)
Now you have an x, put it back into either original equation to get y.
04-07-2010 10:52 PM
04-07-2010 11:15 PM
04-07-2010 11:28 PM
Charles ,
This is an example:
04-08-2010 11:37 AM
Does anyone think Labview could solve this problem:
let a = b
then a^2 = ab (multiply both sides by a)
then a^2 - b^2 = ab - b^2 (subtract b squared from both sides)
then (a+b)(a-b) = b(a-b) (factoring)
then a+b = b (divide both sides by a-b)
so if a=1, then b=1
and we get 1+1=1
or 2=1
????????????????????
04-08-2010 12:35 PM
NaN
Lynn
04-08-2010 11:02 PM
tbob wrote:Does anyone think Labview could solve this problem:
let a = b
then a^2 = ab (multiply both sides by a)
then a^2 - b^2 = ab - b^2 (subtract b squared from both sides)
then (a+b)(a-b) = b(a-b) (factoring)
then a+b = b (divide both sides by a-b)
so if a=1, then b=1
and we get 1+1=1
or 2=1
????????????????????
The problem is here
(a+b)(a-b) = b(a-b) (factoring)
Since a=b, dividing both sides by (a-b) will give NaN as mentioned by Lynn.
Mathan
04-09-2010 10:11 AM
Give both of those men a cigar. That's t he answer. You would be surprised at how many people, even math wizzes, can't find the answer.