First equation shall read t^{2/3}=4 ?
Then cube both sides first, you will get (t^{2/3})^{3}=4^{3}=64 or t^{2}=64; This equation has two solutions, t=±√64=±8;
2) 5√x+1=3 ⇔ 5√x=31=2 ⇔ √x=2/5 ⇔ x=(2/5)^{2}=4/25;
3) 2/3=2(5x3)/(x1); ⇔ (5x3)/(x1)=4/3 Crossmultiply to get 3(5x3)=4(x1) ⇔ 15x9=4x4 ⇔11x=5 ⇔ x=5/11;
4) √x+2x=0 √x=zchange the variable.
z+2z^{2}=0 or z^{2}z2=0. Solve by factoring, z^{2}z2=0 ⇔ (z2)(z+1)=0 ⇔ z=2 or z=1. Since z=√x, z=1 shall be thrown out. So we get √x=2 or x=4.
9/10/2013

Kirill Z.