Solve Equation II: $\frac{25x-2}{4}=\frac{13x+4}{3}$

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SUMMARY

The equation $\left\lfloor \dfrac{25x-2}{4} \right\rfloor=\dfrac{13x+4}{3}$ was discussed, focusing on solving for the variable x. The solution involves understanding the floor function and its implications on the equality. Participants acknowledged the contributions of user kaliprasad in solving the equation, highlighting the collaborative nature of the discussion.

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Solve the equation $\left\lfloor \dfrac{25x-2}{4} \right\rfloor=\dfrac{13x+4}{3}$.
 
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anemone said:
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 $\frac{17}{13}$
 
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:
 
anemone said:
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 :o
 
kaliprasad said:
It is not required. I did it to make the RHS simpler but it did not help :o

I see. Thanks for the reply, kali!
 

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