Alg3.5.3

Solve $\sqrt{x+1}+3=\sqrt{3x+4}\,$

$\left(\sqrt{x+1}+3\right)^2=\left(\sqrt{3x+4}\right)^2\,$

$x+1+6\sqrt{x+1}+9=3x+4\,$

$6\sqrt{x+1}=2x-6\,$

$3\sqrt{x+1}=x-3\,$

$\left(3\sqrt{x+1}\right)^2=(x-3)^2\,$

$9(x+1)=x^2-6x+9\,$

$9x+9=x^2-6x+9\,$

$x^2-15x=0\,$

$x(x-15)=0\,$

$x=15,0\,$

Check all your answers and you'll see that only $x=15\,$ works.