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First equation shall read t2/3=4 ? 
Then cube both sides first, you will get (t2/3)3=43=64 or t2=64; This equation has two solutions, t=±√64=±8;
2) 5√x+1=3 ⇔ 5√x=3-1=2 ⇔ √x=2/5 ⇔ x=(2/5)2=4/25;
3) 2/3=2-(5x-3)/(x-1); ⇔ (5x-3)/(x-1)=4/3 Cross-multiply to get 3(5x-3)=4(x-1) ⇔ 15x-9=4x-4 ⇔11x=5 ⇔ x=5/11;
4) √x+2-x=0 √x=z--change the variable.
z+2-z2=0 or z2-z-2=0. Solve by factoring, z2-z-2=0 ⇔ (z-2)(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.