Question

Can someone please help me with this math problem? Please.

Simplify the expression:(2x-3y)^2/(3y-2x)^2 Ava simplified the expression and said that the value is -1. (2x-3y)^2/(3y-2x)^2 (2x-3y)^2/-(2x-3y)^2 and then she has both of them crossed out (2x-3y)^2/-1(2x=3y)^2 =-1 Is Ava's solution correct? If not, find and explain her mistake. Give the correct solution.

Answer 1

incorrect

Explanation:

Answer 2

Ava is incorrect

Explanation:

Given

`((2x - 3y)^2)/((3y - 2x)^2)`

`Ava: ((2x - 3y)^2)/((3y - 2x)^2) = -1`

Required

Determine if Ava is correct or not

Ava's solution is incorrect and the proof is as follows;

`((2x - 3y)^2)/((3y - 2x)^2)`

Apply the following law of indices;

`(a^n)/(b^n) = ((a)/(b))^n`

`((2x - 3y)^2)/((3y - 2x)^2) = [(2x - 3y)/(3y - 2x)]^2`

Rewrite the numerator

`((2x - 3y)^2)/((3y - 2x)^2) = [(-(3y - 2x))/(3y - 2x)]^2`

Cross out the common factor of the numerator and denominator

`((2x - 3y)^2)/((3y - 2x)^2) = [(-1)/(1)]^2`

`((2x - 3y)^2)/((3y - 2x)^2) = [-1]^2`

`((2x - 3y)^2)/((3y - 2x)^2) = 1`

Hence, the solution is 1; not -1

Her mistake is that, she did not apply the square on -1, after she factorize the denominator