|
Very nice, put that away for safe keeping! Thanks
Everyone has a photographic memory; some just don't have film. Steven Wright
|
|
|
|
|
Wolfram alpha also does this for you:
http://www.wolframalpha.com/input/?i=solve+x%2F(x%2Bb)+%3D+c%2Fd
Matlab also has similar tools but that costs a lot of money.
|
|
|
|
|
Thanks, tried out the wolfram alpha.  Too bad they also charge to see the steps.
"Go forth into the source" - Neal Morse
|
|
|
|
|
It must be all of the algebra I taught over the years.... burned in my skull.
Now, can you derive the equation of a line given two points?
Muahahaha 
Charlie Gilley
<italic>Stuck in a dysfunctional matrix from which I must escape...
"Where liberty dwells, there is my country." B. Franklin, 1783
“They who can give up essential liberty to obtain a little temporary safety deserve neither liberty nor safety.” BF, 1759
|
|
|
|
|
charlieg wrote: Now, can you derive the equation of a line given two points?
Without looking slope = rise/run so m = (y2-y1)/(x2-x1)  (probably not right at all)
Now you're making my brain hurt again! btw, after revisiting my original problem, and thinking a little harder, I was able to work it out. 
"Go forth into the source" - Neal Morse
|
|
|
|
|
It was rhetorical Slope is correct, but the y intercept?
I worked with a coder who was a wizard in C++. But he had no idea how to code a linear function to calculate a simulated value...
anyway, good math links
Charlie Gilley
<italic>Stuck in a dysfunctional matrix from which I must escape...
"Where liberty dwells, there is my country." B. Franklin, 1783
“They who can give up essential liberty to obtain a little temporary safety deserve neither liberty nor safety.” BF, 1759
|
|
|
|
|
 Thanks for the link. Bookmarked
M.D.V. 
If something has a solution... Why do we have to worry about?. If it has no solution... For what reason do we have to worry about?
Help me to understand what I'm saying, and I'll explain it better to you
Rating helpful answers is nice, but saying thanks can be even nicer.
|
|
|
|
|
It's trivial https://codeproject.global.ssl.fastly.net/script/Forums/Images/smiley_wink.gif, even after almost 60 years away from Grammar School in England. The steps are, using C formulae:
a / (a + b) = c / d
Multiply both sides by (a + b)
a = (a + b) * c / d
Collect the terms in a
a - a * c / d = b * c / d
Collect the coefficients of a
a * (1 - c / d) = b * c / d
Divide both sides by (1 - c / d)
a = b * c / d / (1 - c / d)
Simplify
a = b * c / (d - c)
Hope that's right, solved before you would have found the program
In those good old days you got the algebra beaten into you
Show-off Jerry
|
|
|
|
|
... and c needs to be not equal to d ...
|
|
|
|
|
I am so grateful for a post from someone who doesn't feel being innumerate is a way to be part of some team.
Like the slope-of-a-line question. dy/dx - WTF's the difficulty with high school math?
Something akin to pride as a consequence of membership in a pool of those who "can't" seems to be spreading from the TV and Radio to CP. Maybe the problem with poor code isn't only from the mass-production hordes now being unleashed?
Alas.
| Ravings en masse^ |
|---|
| "The difference between genius and stupidity is that genius has its limits." - Albert Einstein | | "If you are searching for perfection in others, then you seek disappointment. If you are seek perfection in yourself, then you will find failure." - Balboos HaGadol Mar 2010 |
|
|
|
|
|
Probably took me the same amount of time by a slightly different route:
a/(a+b)=c/d
Multiply both sides by (a+b) and by d
ad = c(a+b)
Expand
ad = ac + bc
Subtract ac from both sides
ad - ac = bc
Collect the coefficients of a
a(d-c) = bc
Divide both sides by (d-c)
a = bc/(d-c)
Oh, if only all the equations were linear.
|
|
|
|
|
Probably took me the same amount of time by a slightly different route:
You're right, I thought of this after having written the route.
Jerry
|
|
|
|
|
I got a slightly different (if not simpler) answer.
a/(a+b) = c/d
a = c/d*(a+b)
a*d/c = a+b
a*d/c -a = b
a*(d/c -1) = b
a=b/(d/c-1)
Which if you multiply the numerator & denominator by c you'll get the answer the machine reports.
|
|
|
|
|
a/(a+b) = c/d
a = c(a+b)/d
a = ca/d + cb/d
a - ca/d = cb/d
ad - ac = cb
a(d-c) = cb
a = cb/(d-c) A slightly different version from Jerry's. I had 32 credit-hours of math in college - calculus, differential equations, and matrix algebra. All of it's gone, with the space in my head now used for old movie lines. The algebra's stayed around, as it seems to come in handy now and then.
Software Zen: delete this;
|
|
|
|
|
As a math minor I suddenly understand the disdain for cliffs notes...
LOL
|
|
|
|
|
kmoorevs wrote: a/(a+b) = c/d where I was solving for the variable a.
It's a lot easier to see the solution when one writes it like:
a c
------ = ---
a + b d
I'm a firm believer that visualizing (or expressing) the problem correctly is 90% of the solution.
Given the above, cross multiply to get:
ad = c (a+b)
then:
ad = ca + cb
ad - ca = cb
a(d-c) = cb
cb
a = ---------
d-c
Latest Article - A Concise Overview of Threads
Learning to code with python is like learning to swim with those little arm floaties. It gives you undeserved confidence and will eventually drown you. - DangerBunny
Artificial intelligence is the only remedy for natural stupidity. - CDP1802
|
|
|
|
|
|
oh wow, thanks. helped prove that the following does simplify
(A / (B * C) ) * B
simplifies to
A / C
|
|
|
|
|
Message Removed
modified 12-Aug-18 7:04am.
|
|
|
|
|
Message Removed
modified 12-Aug-18 7:04am.
|
|
|
|
|
Message Removed
modified 12-Aug-18 7:03am.
|
|
|
|
|
Message Removed
modified 12-Aug-18 7:03am.
|
|
|
|
|
Message Removed
modified 12-Aug-18 7:03am.
|
|
|
|
|
Message Removed
modified 12-Aug-18 7:03am.
|
|
|
|
|
Message Removed
modified 12-Aug-18 7:03am.
|
|
|
|
General 
News 
Suggestion 
Question 
Bug 
Answer 
Joke 
Praise 
Rant 
Admin 
Submit an article or tip
