rewrite the expression [4x>-4/ 9y>-2] > -3/2, with positive exponent

Question

> equals to the power of, I cannot get it on my keyboard. Thanks!! I need to know every step, because I cannot figure out how to get all the exponents to be positive...

Answer 1

Your question reads: (4x^-4/9y^-2)^-3/2, I think. Negative exponents are just reciprocals or inverses, so x^-4 means 1/(x^4). 1/y^-2 means y^2. The coefficients are not affected by the inversions, so 4x^-4 is 4/x^4 and 1/9y^-2=y^2/9. The contents of the brackets can now be written 4y^2/(9x^4) or 4/9*(y^2/x^4). But there's a negative power outside the brackets so we have 1/((...)^3/2). 3/2 is the same as (square root) cubed. The square root of 4/9 is 2/3, and the square root of a power is half the power, so the square root of  y^2/x^4 is y/x^2. Nearly there. Finally we have to cube the result: (2y/3x^2)^3=8y^3/27x^6 and then invert it because of the negative power: 27x^6/8y^3.

Answer 2

Given [(4x>-4)/(9y>-2)]>(-3/2)
(x<1)/(y<2/9)<3/2

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