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April 24, 2014

April 24, 2014

Posted by **JoniAnne** on Wednesday, October 20, 2010 at 11:50am.

Solve 3 > sqrt(3x)

3x >=o

x>=o/3

x>=0

square both sides and you have 3x < 3^2

3x,9

x<9/3

x<3

answer is 0<=x<3 correct or not?

3(cube) sqrt(2-5) = 3

cube both sides = 2-5x = 27

-5x = 25

x = -5, correct?

Thank you

- Algebra-Please check -
**Henry**, Friday, October 22, 2010 at 7:14pm1. Correct.

2. 3^3*sqrt(2 - 5x) = 3,

27*sqrt(2 - 5x) = 3,

Divide both sides by 27:

sqrt(2 - 5x) = 3/27 = 1/9,

Square both sides and get:

2 - 5x = 1/81,

-5x = 1/81 - 2,

-5x = 1/81 - 162/81,

-5x = -161/81,

Multiply both sides by -1/5:

x = (-161 / 81) * (-1/5) = 161 / 405.

CHECK:

3^3*sqrt(2 - (5*161/405)) =

27*sqrt(2 - 1.9877) =

27*sqrt(0.01235) =

27 * 0.11111 = 3.

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