Answer:
Ava is incorrect
Step-by-step explanation:
Given
[tex]\frac{(2x - 3y)^2}{(3y - 2x)^2}[/tex]
[tex]Ava: \frac{(2x - 3y)^2}{(3y - 2x)^2} = -1[/tex]
Required
Determine if Ava is correct or not
Ava's solution is incorrect and the proof is as follows;
[tex]\frac{(2x - 3y)^2}{(3y - 2x)^2}[/tex]
Apply the following law of indices;
[tex]\frac{a^n}{b^n} = (\frac{a}{b})^n[/tex]
[tex]\frac{(2x - 3y)^2}{(3y - 2x)^2} = [\frac{2x - 3y}{3y - 2x}]^2[/tex]
Rewrite the numerator
[tex]\frac{(2x - 3y)^2}{(3y - 2x)^2} = [\frac{-(3y - 2x)}{3y - 2x}]^2[/tex]
Cross out the common factor of the numerator and denominator
[tex]\frac{(2x - 3y)^2}{(3y - 2x)^2} = [\frac{-1}{1}]^2[/tex]
[tex]\frac{(2x - 3y)^2}{(3y - 2x)^2} = [-1]^2[/tex]
[tex]\frac{(2x - 3y)^2}{(3y - 2x)^2} = 1[/tex]
Hence, the solution is 1; not -1
Her mistake is that, she did not apply the square on -1, after she factorize the denominator
