Adding Rational Equations #2

[Deo3560]

OK, I know I have been asking quite a few of these questions, but I'm at a total loss, so could someone please help. =)

Question:
5/(x^3-y^3)+3/(x^2+xy+y^2)

Answer:
(3x-3y+5)/((x-y)(x^2+xy+y^2))

I am also having problems with one more question if you feel like answering it for me, so here it is:
~~~~~~(THIS IS SUBTRACTION)~~~~~~

Question:
(4a)/(a^2-ab-2b^2)-(6b)/(a^2+4ab+3b^2)

Answer:
(2(2a^2+3ab+6b^2))/((a+b)(a-2b)(a+3b))

Don't worry about the second one if you don't want, I may be able to figure it out if you can help me understand the first equation.

So if you could help me out here, I cannot understand this question.

Thanks,
~~~Deo3560~~~

 

[Deleted member 4993]

OK, I know I have been asking quite a few of these questions, but I'm at a total loss, so could someone please help. =)

Question:
5/(x^3-y^3)+3/(x^2+xy+y^2)

Factorize x[sup:zh0j2jle]3[/sup:zh0j2jle] - y[sup:zh0j2jle]3[/sup:zh0j2jle]

Look at the given-answer carefully - that'll give you a clue

Answer:
(3x-3y+5)/((x-y)(x^2+xy+y^2))

I am also having problems with one more question if you feel like answering it for me, so here it is:
~~~~~~(THIS IS SUBTRACTION)~~~~~~

Question:
(4a)/(a^2-ab-2b^2)-(6b)/(a^2+4ab+3b^2)

Factorize the denominators

Look at the given-answer carefully - that'll give you a clue


Answer:
(2(2a^2+3ab+6b^2))/((a+b)(a-2b)(a+3b))

Don't worry about the second one if you don't want, I may be able to figure it out if you can help me understand the first equation.

So if you could help me out here, I cannot understand this question.

Thanks,
~~~Deo3560~~~

 

[mmm4444bot]

Can you add 5/x + 3/y ?

You need a common denominator (i.e., the bottom of both fractions must be the same, before adding the numerators), yes ?

5y/(xy) + 3x/(xy)

(5y + 3x)/(xy)

If you don't understand my example, then you need to back up; otherwise, it works the same way in your posted exercises. You need to get a common denominator, before adding or subtracting numerators.

Follow Subhotosh's suggestion for the first exercise, and factor that difference of cubes. Then look to see how you can get the denominators to be the same.

 

[Deo3560]

No, you cannot add 5/x and 3/y, because they are not like denominators.

But Thanks to Subhotosh Khan, I understand the problem.

Here are the steps,

5/(x^3-y^3)+3/(x^2+xy+y^2)

Multiply the numerator and denominator of 3/(x^2+xy+y^2) by (x-y).
Which comes out to be:

5/(x^3-y^3)+(3(x-y))/(x-y)(x^2+xy+y^2)

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

Then Finally you add them together:

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


Thank for everyones help, I learned a lot!

Best Regards,
Deo3560