Question

2x^2+y-3/4y^2-9 *3y+3/5y-5y^2

Answer 1

probably best if factored out first yes? let's start with the first numerator...

2y^2 + y - 3...only one factor pair of 2 and only one factor pair of 3 so it'll be quite easy to factor. we can see that 3 is negative so surely one of the binomials will have negative..soo... (2y+3)(y-1)

now the denominator...4y^2 -9...pretty easier..difference of squares so... (2y +3)(2y-3)


now the second numerator...3y + 3..only one way to factor this...3(y+1)

second denom...5y-5y^2 only one way too...5y(1 -y) let's make it (y-1) so it's cancelable...how? we multiply it by negative 1..so...it becomes -5y(y-1)

now we're ready to put them together

(2y+3)(y-1)/(2y+3)(2y-3) * (3(y+1)/-5y(y-1)

2y + 3 gets canceled...so will y-1..so...

1/2y-3 * 3(y+1)/-5y

combine...

3(y+1)/-5y(2y-3)

does it need to be distributed?

Answer 2

Thank you I factored the difference of squares but I didn't do it right so I couldn't simplify it any further

Answer 3

it's no biggie ^_^ happy to help...havent seen an algebra question in a while...t'was a good feeling to finally have something i can solve hahahah