Please help me with this question?

Question

Simplify , leaving denominator in factorised form:

(1 divided by (y -2)) + (3 divided by (y squared + y - 6))

It looks alot harder written like that but it's supposed to be like 1 over y-2 and then add 3 divided by y squared + y -6 but i didn't know how to write that!!!!!! If you could please tell me how you got the answer it would be great!!!! thank you!!! Also apparently it's easier if you factorise the (y squared + y -6) bit first.

Sorry i forgot to say that the final answer has to be a fraction

Answer 1

First, factorise the y²+y-6 using the quadratic equation resolution:
y = (-b ± sqr(b² - 4ac)) / 2a
y = (-1 ± sqr(1 - 4.1.-6)) / 2.1
y1 = -3, y2 = 2, so the factors of the second denominator are
(y + 3) and (y - 2).
Wow! Look! the first denominator is (y - 2)!
So we just have to multiply the first fraction by (y+3)!
[1 * (y+3) / (y-2) * (y+3)] + [3 / (y-2) * (y+3)]
We have a common denominator, so we just have to add the numerators... That you can do...

Answer 2

You are really making this more difficult than it has to be. If you think of this as a simple problem of adding fractions with a one letter denominator, then back substituting, it becomes a simple matter.

Let 1/(y-2) = 1/a

The fact that the instructions told you to leave the final denominator in factored form implies, in this problem anyway, that the denominator of the second fraction is some multiple of the denominator of the first fraction.

Therefore, we can write the second fraction like this:

3/(y^2+y-6) = 3/ab

Adding the two fractions together then yields something like this:

1/(y-2) + 3/(y^2+y-6) = 1/a + 3/ab = b/ab + 3/ab = (b+3)/ab.

Notice we don't have to multiply the numerator "3" in the second fraction by "a" because the denominator of its fraction already contains that factor.

Now it remains only to find "b." To do this, all you must do is divide (y^2+y-6) by (y-2). If you do this longhand, you will get (y+3).

Back substituting into the above expression, (b+3)/ab, gets us this result:

[(y+3)+3]/(y-2)(y+3) = (y+6)/(y-2)(y+3).

Answer 3

1/(y-2) + 3/(y^2+y-6)

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

{(y+3) + 3}/(y-2)(y+3)

(y+6)/(y^2+y-6)

Answer 4

1/Y - 2 + 3/ (y² + y - 6)

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

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

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

y + 6 / (y - 2) (y + 3)

The answer is y + 6 / (y - 2)(y + 3)

- - - - - - - -s-

Answer 5

Are they really using the word 'factorise' now? Good grief.

When you factor Y^2 +Y-6, you get (Y+3) and (Y-2)

Now you can rearrange
(1/(y-2)) * (1 + (3/(y+3))

You can take it further, but it really doesnt simplify anything.

Answer 6

=1/(y-2) + 3/(y^2+y-6)

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

by using a least common denominator

=((y+3)+3)/((y+3)(y-2))

final answer is (y+6)/((y+3)(y-2))

Answer 7

1/(y-2) + 3/(y^2 +y - 6)
= 1/(y-2) + 3/((y-2)(y+3)}
=y+3 +3/(y-2)(y+3)
= y+6/(y-2)(y+3) Ans..

Answer 8

1/(y-2) + 3/(y+3)(y-2)
go to
1 + 3(y-2)/(y+3)(y-2)
then (y-2)'s cancel to give
1 + 3/(y+3)
then times by by y+3
(y+3) + 3
i think

Answer 9

1/(y-2)+3/(y^2+y-6)
(factorize denominator)
=!/y-2)+3/(y+3)(y-2)
{(y+3)/(y+3)(y-2)=1/(y-2)}
{3/(y^2+y-6)=3/(y+3)(y-2)}
=(y+3)+3/(y+3)(y-2)
=(y+6)/(y+3)(y-2)

i hope that this helps

Answer 10

More homework. you should have done already. not left for others to do for you see me after class