Question

What is the product in simplest form? State any restrictions on the variable. y^2/y-3 * y^2-y-6/y^2+1y

Answer 1

Well, you want to usually factor these problems as much as possible and see what cancels. The right fraction can be factored quite a bit. So starting with the numerator of the right fraction, I can factor y^2 - y - 6 into (y-3)(y+2). The bottom part of that fraction you can factor out a y and make it into y(y+1). Now if you multiply the two fractionstogether you can see what cancels.
[(y^2)(y-3)(y+2)]/[(y-3)(y)(y+1)] so the whole y-3 cancelsand one of the y's on top cancels with the one on bottom. So that leaves you with y(y+2)/(y+1), which you can leave as is. As for the restrictions, your restrictions are numbers that make eitherof the two beginning fractions = 0. For the first fraction, I set y - 3 = 0 and see that when y is 3, I am undefined. If I take y^2 + y, I can factor it and set each of its factors equal to 0. so y^2 + y becomes y(y+1). Setting those 2 factors to 0, I hav y = 0 and y+1 = 0. Meaning solving for y in those two gives me 0 and -1. So in total, I cannot have y equal to 0, -1, or 3.

Answer 2

thank you so much.