Question

(y^2-y-12)/(y^2-6y+8)/(y^2+2y-15)/(y^2-7y+10)

Answer 1

For this one we'll have to factor out all these equations. \[\large {y^2-y-12 = (x-4)(x+3)\\y^2-6y+8 = (x-4)(x-2)}\]\[\large {y^2-7y+10 = (x-5)(x-2)\\y^2+2y-15= (x+5)(x-3)}\]

Answer 2

\[\large \frac{(y^2 -y-12)}{(y^2-6y+8)}\cdot \frac{(y^2 -7y+10)}{(y^2+2y-15)}\]

Answer 3

I have no idea how my posts flipped order...

Answer 4

\[\large \frac{\cancel{(x-4)}(x+3)}{\cancel{(x-4)(x-2)}} \cdot \frac{(x-5)\cancel{(x-2)}}{(x+5)(x-3)}\]

Answer 5

\[\large \frac{(x+3)(x-5)}{(x+5)(x-3)}\] how expand these and reduce/simplify.