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

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The equation to solve is $\left\lfloor \dfrac{25x-2}{4} \right\rfloor=\dfrac{13x+4}{3}$. Participants express appreciation for contributions to the solution process. The discussion emphasizes the importance of correctly interpreting the floor function in the equation. There is a collaborative atmosphere as users acknowledge each other's efforts. The focus remains on solving the equation effectively.
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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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