Equation Containing Floor Function on both sides

Hello, does anyone know how to solve an equation containing an unknown inside a floor function in both sides?

Consider this example:

$\lfloor 2x + 3 \rfloor = \lfloor 5x + 1 \rfloor$ Thanks in advance!

#Algebra

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

I'm considering $5x+1>2x+3$ , therefore

From the equation given: $5x+1-2x-3<1$ $3x-2<1$ $3x<3$ $\boxed{x<1}$ If $5x+1<2x+3$ , then

From the equation given: $2x+3-5x-1<1$ $2-3x<1$ $\boxed{3x<1}$ I have worked out this much only

Zakir Husain - 1 year ago

Hint: Show that $-1 < x < 1$ .

Hint 2: What happens to $\lfloor 2x\rfloor$ and $\lfloor 5x \rfloor$ when $x$ increases by $\frac1{10}$ .

Hint 3: Check for the intervals $[-1, -1 + 1/10), [-1 + 1/10, -1 + 2/10), \ldots , [8/10, 9/10), [9/10, 1)$ .

Pi Han Goh - 1 year ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$