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how to solve y=a*x+b,y=c*x+d (a,b,c,d are known constant)in labview?

how to solve y=a*x+b,y=c*x+d (a,b,c,d are known constant)in labview? any help is appreciated.
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Message 1 of 9
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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.

Message Edited by Ravens Fan on 04-07-2010 11:03 PM
Message 2 of 9
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homework?
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Message 3 of 9
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Hi,

 To solve the linear equations,you can refer the mathematics/linear equations pallette and use  solve linear equations.vi.

Thanks and regards,

srikrishnaNF

Regards,
Srikrishna


Message 4 of 9
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Charles ,

 

This is an example:

algebra.jpg

Message 5 of 9
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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

????????????????????

 

- tbob

Inventor of the WORM Global
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Message 6 of 9
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NaN

 

Lynn 

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Message 7 of 9
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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

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Message 8 of 9
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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.

 

- tbob

Inventor of the WORM Global
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