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Solving equation II

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anemone

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Feb 14, 2012
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Solve the equation $\left\lfloor \dfrac{25x-2}{4} \right\rfloor=\dfrac{13x+4}{3}$.
 

kaliprasad

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Mar 31, 2013
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Solve the equation $\left\lfloor \dfrac{25x-2}{4} \right\rfloor=\dfrac{13x+4}{3}$.
Subtract 1 from both sides
$\lfloor\frac{25x-6}{4}\rfloor = \frac{13x+1}{3}$
So $\frac{13x + 1}{3}$ = integer say n
So x = $\frac{3n-1}{13}$
So we get
$\lfloor\frac{25\frac{3n-1}{13}-6}{4}\rfloor = n$
Or
$\lfloor\frac{75n-103}{52}\rfloor = n$
Or 23n-103= > 0 and < 52
Or 103 <= 23n < 155
Or $\frac{103}{23} <=n < \frac{155}{23}$
So n = 5 or 6
Hence x = $\frac{14}{13}$ or $\frac{17}{13}$
 
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anemone

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Feb 14, 2012
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Well done, kaliprasad! :)

But, if you don't mind me asking, I see that we could make the substitution right from the start, why would you do so only after subtracting both sides of the equation by 1?:confused:
 

kaliprasad

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Mar 31, 2013
1,309
Well done, kaliprasad! :)

But, if you don't mind me asking, I see that we could make the substitution right from the start, why would you do so only after subtracting both sides of the equation by 1?:confused:
It is not required. I did it to make the RHS simpler but it did not help :eek:
 
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anemone

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