![\sqrt{3x + 1} - \sqrt{2 - x} = \sqrt{2x - 1} \\ \\ squaring \: both \: sides \\ \\ ( \sqrt{3x + 1} - \sqrt{2 - x})^{2} = ( \sqrt{2x - 1} ) ^{2} \\ \\ 3x + 1 + 2 - x - 2( \sqrt{3x + 1} ) ( \sqrt{2 - x} ) = 2x - 1 \\ \\ \cancel{ 2x} + 3 - 2( \sqrt{3x + 1} ) ( \sqrt{2 - x} ) = \cancel{ 2x} - 1 \\ \\ 3 - 2( \sqrt{3x + 1} ) ( \sqrt{2 - x} ) = - 1 \\ \\ - 2( \sqrt{3x + 1} ) ( \sqrt{2 - x} ) = - 1 - 3 \\ \\ - 2( \sqrt{3x + 1} ) ( \sqrt{2 - x} ) = - 4 \\ \\ ( \sqrt{3x + 1} ) ( \sqrt{2 - x} ) = 2 \\ \\ squaring \: both \: sides \: again \\ { [( \sqrt{3x + 1} ) ( \sqrt{2 - x} )] }^{2} = {2}^{2} \\ \\ (3x + 1)(2 - x) = 4 \\ \\ 6x - 3 {x}^{2} + 2 - x = 4 \\ \\ - 3 {x}^{2} + 5x - 2 = 0 \\ \\ 3 {x}^{2} - 5x + 2 = 0 \\ \\ 3 {x}^{2} - 3x - 2x + 2 = 0 \\ \\ 3x(x - 1) - 2(x - 1) = 0 \\ \\ (x - 1)(3x - 2) = 0 \\ \\ x - 1 = 0, \: \: 3x - 2 = 0 \\ \\ x = 1, \: \: x = \frac{2}{3}](/tpl/images/1227/4035/0dff0.png)
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