3. x + 5 = 105x = 10  x = 5 - 5 - 5 5 5(  x ) 2 = ( 5 ) 2 x = 5 x = 2 x = 25 (5) + 5 = 105(2) = 10  25 = 5 10 = 10 10 = 10 5 = 5

5. Solve: a 1/2 = 8 a 1 = a 8 [(-2x+6) 1/5 ] 5 = [(-8+10x) 1/5 ] 5 -2x + 6 = -8 + 10x -12x = -14 x = 7/6

6. Raise both sides to the 9 th power. Raise both sides to the 2/3 power. -4x – 6 = -2 + 3x -7x = 4 x = -4/7 Check : [-4(-4/7)-6] 1/9 =[(-2+3(-4/7)] 1/9 [(3x-8) 3/2 ] 2/3 = 8 2/3 3x-8 = 4 3x = 12 x = 4 Check : [3(4)-8] 3/2 =8 (4) 3/2 =8 8 = 8

7. After you add the radical to the Right Side, Square Both Sides 3x+2 = 5x -10 -2x = -12 x = 6 Check:  (3*6 + 2) =  (5*6-10)  (20) =  (20) Square Both Sides x 2 = -3x + 40 Solve the quadratic equation! x 2 + 3x – 40 = 0 (x+8)(x-5) = 0 x = -8 and 5 Check: (Use the Original Equation) -8 =  (-3*-8+40) -8 =  44 5=  (-3*5+40) 5=  25 The only solution is x = 5. x=-8 is an extraneous solution! Example 9 Example 10 Example 9Example 10