## College Algebra 7th Edition

$x=4$
$\sqrt{2x-1}=\sqrt{3x-5}$ $(\sqrt{2x-1})^{2}=(\sqrt{3x-5})^{2}$ $2x-1=3x-5$ $-1+5=3x-2x$ $x=4$ We need to confirm that $x=4$ works in the original equation: Left side=$\sqrt{2(4)-1}=\sqrt{8-1}=\sqrt{7}$ Right side=$\sqrt{3(4)-5}=\sqrt{12-5}=\sqrt{7}$ Both sides agree, so the solutions works.