Question

$[x - 1] = \sqrt{3} \times [2x - 3]$
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Answer 1

$x-1=\sqrt{3} (2x-3)\\x-1=2\sqrt{3x} -3\sqrt{3} \\x-1-2\sqrt{3x} =-3\sqrt{3} \\x-2\sqrt{3x} =-3\sqrt{3} +1\\(1-2\sqrt{3} )x=1-3\sqrt{3}\\\frac{(1-2\sqrt{3} )x}{1-2\sqrt{3} } =\frac{1-3\sqrt{3} }{1-2\sqrt{3} } \\x=\frac{1-3\sqrt{3} }{1-2\sqrt{3} } =\frac{\sqrt{3}+17 }{11}$

Answer 2

$x - 1 = \sqrt{3} - \times (2x - 3)$

$x - 1 = 2 \sqrt{3} x - 3 \sqrt{3}$

$x - 2 \sqrt{3} x = - 3 \sqrt{3} + 1$

$(1 - 2 \sqrt{3} )x = - 3 \sqrt{3} + 1$

$x = \frac{ - 3 \sqrt{3} + 1}{1 - 2 \sqrt{3} }$

$x = \frac{ - 3 \sqrt{3} - 18 + 1 + 2 \sqrt{3} }{ - 11}$

$x = \frac{ - \sqrt{3} - 17}{ - 11}$

$x = - \frac{ - \sqrt{3} - 17 }{11}$

$x = \frac{ \sqrt{3} + 17 }{11}$