Solving equation II

[anemone]

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

 

[kaliprasad]

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

Spoiler

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 $\frac{17}{13}$

 

[anemone]

Well done, kaliprasad!

Spoiler

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?

 

[kaliprasad]

Well done, kaliprasad!

Spoiler

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?

Spoiler

It is not required. I did it to make the RHS simpler but it did not help

 

[anemone]

Spoiler

It is not required. I did it to make the RHS simpler but it did not help

I see. Thanks for the reply, kali!