Answer
$x=2$
Work Step by Step
We solve:
$\sqrt{2x-3}+x=3$
$\sqrt{2x-3}=3-x$
$2x-3=(3-x)^2$
$2x-3=9-6x+x^{2}$
$-3-9=x^2-6x-2x$
$x^{2}-8x+12=0$
$(x-2)(x-6)=0$
$(x-2)=0$ or $(x-6)=0$
$x=2$ or $x=6$
However, the solution $x=6$ does not work in the original equation, so we throw this solution out:
$\sqrt{2*6-3}+6=\sqrt{12-3}+6=\sqrt{9}+6=3+6=9\neq 3$
Thus the only solution is $x=2$.
