algrbra answers

Question

t2/3=4

 

5√x+1=3

 

2/3=2-  5x-3/x-1

 

√x+2-x=0

Answer 1

First equation shall read t^2/3=4 ? 

Then cube both sides first, you will get (t^2/3)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.