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https://imgur.com/a/JhMY5K2

I'm having trouble using LaTeX, uploading a photo from my computer, and making an account. Nothing works. The answer is 0 btw but I'm not sure how.

 

 

Pic added by Melody

 

 Feb 27, 2022
edited by Melody  Feb 27, 2022
 #1
avatar+117104 
0

I have played with it but I do not know how to solve it either.    frown

 Feb 27, 2022
 #2
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+1

I sort of understand the solution now. Here is the answer. 

https://imgur.com/a/HdMHSYG

 

Thanks for your help

Guest Feb 27, 2022
 #3
avatar+117104 
+1

Here is the answer you have provided.

 

 

It lost me after the first sentence ...   angry

 Feb 27, 2022
 #4
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+1

I don't understand that answer, but algebraically it works out easily enough.

We need

 \(\displaystyle \frac{a(ax+b)/(cx+d)+b}{c(ax+b)/(cx+d)+d}\equiv x.\)

Clearing the fraction within the fraction on the lhs,

\(\displaystyle \frac{a(ax+b)+b(cx+d)}{c(ax+b)+d(cx+d)}\equiv x. \)

For this to work we need three things to happen.

(1) the constant term on the top line should be zero, ie.     ab + bd = b(a + d) = 0,

(2) the x coefficient on the bottom line should be zero, ie.  ca + dc = c(a + d) = 0,

(3) The resulting fraction should cancel down to leave just x.

Applying (1) and (2) reduces the lhs to

\(\displaystyle \frac{x(a^{2}+bc)}{cb+d^{2}}\) 

and this cancels down if  

\(\displaystyle a^{2}=d^{2}.\)

 

\(\displaystyle a+d=0,( \text{ so }a=-d),\) meets all of the requirements.

 Feb 28, 2022
 #5
avatar+117104 
0

Thanks,

I should of thought of that but I didn't.   angry

Thanks for showing me   cool  laugh

Melody  Feb 28, 2022
 #6
avatar
+1

Thank you! That broke down easily. surprise

Guest Mar 1, 2022

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