Answer:
Ava's solution was not correct.
The correct solution is 1.
Step-by-step explanation:
The expression Ava was to simplify(as seen in the attachment) is:
[tex]\dfrac{(2x-3y)^2}{(3y-2x)^2}[/tex]
[tex]3y-2x=-2x+3y=-(2x-3y)[/tex]
On substitution of the result above into the denominator, we have:
[tex]\dfrac{(2x-3y)^2}{(-(2x-3y))^2}\\=\dfrac{(2x-3y)^2}{(-1)^2(2x-3y)^2}\\\\\text{Canceling out the like term} (2x-3y)^2, \text{we have:}\\=\dfrac{1}{(-1)^2}\\\\=1[/tex]
Therefore:
Ava did not take the square of the negative sign. This led her to have a negative result.
Answer:
Ava correctly factored the denominator:
(3y - 2x) = [-(2x - 3y)] 2
But when she took the negative outside the braces, she did not keep the (-1) squared.
The correct solution is 1.
Step-by-step explanation:
used this and it was correct
