(y2-9)/4y times 12/3y+9 does it equal
y-3
No, it is not y-3
[(y2-9)/4y] * 12/(3y+9)
=(1/4)(y+3)(y-3)(1/y)*(1/3)*[4/(y+3)]
=(y-3)/(3y)
thank you
but that wasn't an answer choice. (y-3)/y was an answer choice though
Sorry, I made an algebra error factoring the "12"
=(1/4)(y+3)(y-3)(1/y)*(1/3)*[3*4/(y+3)]
=(y-3)/y
To solve this problem, we can start by simplifying the given expression [(y^2 - 9)/4y] * [12/(3y + 9)] step by step.
First, let's factor the numerator of the first fraction, y^2 - 9, which is a difference of squares. It can be factored as (y + 3)(y - 3). Now we have [(y + 3)(y - 3)/4y] * [12/(3y + 9)].
Next, we can simplify the second fraction, 12/(3y + 9), by factoring out the common factor of 3 in both the numerator and the denominator. This gives us 4/(y + 3).
Now, we can cancel out the common factor of (y + 3) between the two fractions, leaving us with [(y - 3)/4y] * (4/y).
After canceling out the common factors, we are left with (y - 3)/y.
So, the simplified expression is (y - 3)/y.
Please note that this answer is different from y - 3.