Algebra 2 Help

$-1=\sqrt{3x-2}-\sqrt{4x+1}\\ \sqrt{4x+1}-1 = \sqrt{3x-2}\\ (4x+1)-2\sqrt{4x+1}+1 = 3x-2\\ 4x+2 - 2\sqrt{4x+1} = 3x-2\\ x+4=2\sqrt{4x+1}\\ x^2+8x+16 = 4(4x+1) = 16x+4\\ x^2 -8x + 12 = 0\\ (x-6)(x-2) = 0\\ x=6,~2$

Let's check these solutions

$\sqrt{18-2} - \sqrt{24+5} = 4 - 5 = -1\\ \sqrt{6-2}-\sqrt{8+1} = 2-3 = -1$

Both solutions are valid.

$\dfrac{\dfrac{4}{2x-1}+\dfrac 1 4}{x+1}=$

$\dfrac{\dfrac{4\cdot 4 + 2x-1}{4(2x-1)}}{x+1} =$

$\dfrac{15+2x}{4(2x-1)(x+1)}$