rational expressions math help please???

Question

Could anyone help me with these math problems...Thank you soooooo much if anyone could help me..

1.3a + 5b / 9a^2 - 25b^2
2. 7m – 14 / 5m times 4m^3 / m-2

3. y + 3 / y^2 – 9 times y^2 -5y + 6 / y – 3

4. y^2 – y – 20 / y^2 + 7y + 12 divide y^2 -10y +25/ y^2 + 6y + 9

5. 2x + x^2 / 4x – 5 divide 4x^2 + 2x^3 / 16x – 20

Answer 1

1.3a + 5b / 9a² - 25b²
3a + 5b / (3a - 5b)(3a + 5b) =
Cancelling (3a + 5b):
1 over 3a - 5b
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2. 7m – 14 / 5m times 4m³ / m-2
[7(m-2)/5m] * [4m³/m-2] =
Cancelling (m-2):
[7 over 5] * [4m³] = 28m² over 5
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3. y + 3 / y² – 9 times y² -5y + 6 / y – 3
[(y+3)/(y+3)(y-3)] * [(y-2)(y-3)]/(y-3) =
Cancelling (y-3)and (y+3):
...y-2
= ----
...y-3
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4. y² – y – 20 / y² + 7y + 12 divide y² -10y +25/ y² + 6y + 9
[(y-5)(y+4)] .....[(y-5)(y-5)]
--------------- : ------------------- =
[(y+4)(y+3)]...[(y+3)(y+3)]
Cancelling :
Solution:
.. y+ 3
= ------
...y - 5
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5. 2x + x² / 4x – 5 divide 4x² + 2x³ / 16x – 20
2x - x² .... 4x² - 2x³
--------- : -------------
4x - 5 ..... 16x - 20

x(2 - x) .... 2x²(2 - x)
--------- : ---------------- =
4x - 5 ....... 4(4x - 5)
Inverting the second fraction and operation sign:

Cancelling too:
x(2 - x) .... 4(4x - 5)
--------- x ---------------- = 2 over x
4x - 5 ....... 2x²(2 - x)
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Answer 2

1. 3a + 5b / (9a^2 - 25b^2) = 3a + 5b / (3a + 5b) * (3a - 5b) = 1 / (3a - 5b)

2. 7m – 14 / 5m times 4m^3 / m-2 = (7m – 14) *(4m^3 ) / [5m*(m-2)] = 7 *(m-2)* 4m^3 / [(m-2) * 5m] = 28m^2 / 5

3. y + 3 / y^2 – 9 times y^2 -5y + 6 / y – 3 = (y + 3) * (y^2 -5y + 6) / [( y^2 – 9)*( y – 3)] = (y+3)*(y-3)*(y-2) / [(y+3)*(y-3)*(y-3)] = (y-2) / (y-3)

4. y^2 – y – 20 / y^2 + 7y + 12 divide y^2 -10y +25/ y^2 + 6y + 9 = (y^2 – y – 20)*(y^2 + 6y + 9 ) / [(y^2 + 7y + 12 )*(y^2 -10y +25)] = (y-5)*(y+4)*(y+3)*(y+3) / [(y+4)*(y+3)*(y-5)*(y-5)] = (y+3) / (y-5)

5. 2x + x^2 / 4x – 5 divide 4x^2 + 2x^3 / 16x – 20 = (2x + x^2)*(16x – 20) / [(4x – 5 )*(4x^2 + 2x^3)] = 4*(4x-5)*(2x+x^2) / [(4x-5)*(2x+x^2)*2x] = 2/x

Answer 3

I'm going to make the big assumption that everying on either side of the division sign is grouped together. Otherwise there's no way to tell from the way you've written things whether, for example, 7m – 14 / 5m is 7m - (14/5m) or is (7m-14)/5m.

1. (3a + 5b) / (9a^2 - 25b^2)
Remember that (a+b)(a-b)=(a^2 - b^2). So you can factor the bottom into:
(3a + 5b) / (3a + 5b)(3a - 5b)
You can cancel a term from the top and bottom and get
1 / (3a - 5b)

2, 3. Remember that (a/b) * (c/d) is (ac / bd). After combining, factor out each quadratic and try to cancel terms.

4, 5. Remember that (a/b) / (c/d) is (a/b) * (d/c).

Answer 4

You mean to put parentheses in here, don't you!!
e.g. it should be
4. . (y^2 – y – 20) / (y^2 + 7y + 12) divided by (y^2 -10y +25)/ (y^2 + 6y + 9)

It's all about factorising. These factor into

[(y-5)(y+4)]/[(y+3)(y+4)] and

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

The first reduces (when you cancel the common factor (y+4)) to
(y+5)/(y+3)
and to divide by the second, you multiply by its reciprocal, so the whole thing is equal to
[(y+5)/(y+3)] multiplied by

[(y+3)(y+3)]/[(y+5)(y+5)]

Now that first (y+5) -- on top -- cancels one of the (y+5)(y+5) on the bottom. And the first (y+3) on the bottom cancels one of the (y+3)(y+3) on top, leaving the final result

(y+3)/(y+5)

They're all basically like this. Try the rest yourself. Hope your factorising is good, otherwise you should practise that first.

Answer 5

Factor anything you can. Then see if you can cross-divide or reduce.

(3a + 5b)/ (3a+5b)(3a-5b)
1/(3a-5b)

On numbers 4 and 5, you'll have to invert the second fraction.
(y-5)(y+4)/(y+3)(y+4) * (y+3)(y+3)/(y-5)(y-5)
(y-5)/(y+3) * (y+3)(y+3)/(y-5)(y-5)
(y-5)/1 * (y+3)/(y-5)(y-5)
(y+3)/(y-5)

Answer 6

here are the answers:

1. 1/(3a-5b)
2. 28m/5
3. (y-2)/(y-3)
4. (y+3)/(y-5)
5. 2/x