Anyone?? 2 Math problems Help PLEASE!!!?

Question

simplify these equations.


1. 2y/y^2- 25 - y/ y-5

2. 2n/ n^3 - 5n^2 + 2/ n^2 +5n

Answer 1

1) 2y/(y^2 -25) - y/(y-5)
Notice that y^2 - 25 factors into (y-5)(y+5).
Multiply the second term by (y+5) on the top and bottom to get a common denominator.
2y/(y^2 -25) - y/(y-5)
= 2y / [(y-5)(y+5)] - y(y+5) / [(y-5)(y+5)]
= [2y - y(y+5)] / [(y-5)(y+5)]
= [2y - y^2 - 5y] / [(y-5)(y+5)]
= (-y^2 - 3y) / [(y-5)(y+5)]

2) The second one is similar.
Factor each denominator.
n^3 - 5n^2 = n^2 (n-5)
n^2 + 5n = n(n+5)
The common denominator will be n^2 (n+5)(n-5)
The first term is missing (n+5) and the second term is missin n(n-5)
2n/(n^3 - 5n^2) + 2/(n^2 +5n)
= 2n(n+5) / [n^2 (n+5)(n-5)] + 2n(n-5) / [n^2 (n+5)(n-5)]
= [2n^2 + 10n + 2n^2 - 10n] / [n^2 (n+5)(n-5)]
= 4n^2 / [n^2 (n+5)(n-5)]
Cancel n^2
= 4 / [(n+5)(n-5)]

Answer 2

1)
2y / (y^2-25) - y / (y-5) => Factor the first term, make the 2nd match.

2y / (y-5)(y+5) - y(y+5) / (y-5)(y+5) = > Denominators match. Simplify and combine.

(2y - y^2 - 5y) / (y-5)(y+5) => Combine numerator y terms. I like to pull out the negative

(-1)(y^2 + 3y) / (y-5)(y+5) => Maybe prettier to factor the top

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

2)
2n / (n^3 - 5n^2) + 2 / (n^2 + 5n) => Pull the n's out of the first term

2 / (n^2 - 5n) + 2 / (n^2 + 5n) => Get a common denominator

2(n^2+5n) / (n^4 - 25n^2) + 2(n^2-5n) / (n^4-25n^2) => Combine

(2n^2 + 10n + 2n^2 - 10n) / (n^4-25n^2) => Drop the 10n, and pull out the n^2

(n^2)(4) / (n^2)(n^2-25) => Cancel.

4 / (n^2-25) => QED

Answer 3

1. 2y/y^2 - 25 - y/y-5 = 2/y - 25 - y/y-5 = 2(y-5) -25y(y-5) - y^2/y(y-5) = 2y-10-25y^2 + 125y -y^2 / y^2 - 5y
= (-26y^2 + 127y -10) / y^2-5y

2. 2n/n^3 -5n^2 + 2/n^2 +5n = 2/n^2 - 5n^2 + 2/n^2 +5n
= 4/n^2 - 5n^2 + 5n = (4 - 5n^4 + 5n^3) / n^2

Answer 4

Get a common denominator, which is y^3-5y^2 here

= 2(y-5)/y^2(y-5) - 25(y-5)/(y-5) - y(y-5)/(y-5)

= 2y-10/(y^3-5y^2) - (25y-125)y^2/(y^3-5y^2) - (y^2-5y)y^2/(y^3-5y^2)

= (2y - 10 - 25y^3 - 125y^2 - y^4 - 5y^3)/(y^3-5y^2)

= (y^4 -30y^3 -125y^2 + 2y - 10)/(y^3-5y^2)

Use the common denominator method for the next one as well.

Answer 5

1.2y/y^2- 25 - y/ y-5If no parentheses clear denom Y*(Y-5)/ Y*(Y-5) : 2Y-10 -25Y*(Y-5) –Y^2 = 2Y-10-25Y^2 + 125Y-Y^2 = 26Y^2 +127Y-10

Answer 6

1. 2y/(y+5)(y-5) - y/(y-5)
=2y/(y+5)(y-5) - y(y+5)((y+5)(y-5))
=(2y - y(y+5))/((y+5)(y-5))
=(2y - y^2 - 5y)/((y+5)(y-5))
=(-y^2 - 3y)/((y+5)(y-5))

I would solve number two but alas, I must be going for work....

Answer 7

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