## College Algebra (10th Edition)

$x=2$
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 $x=2$.