Math Help? PLEASE? 5 Stars

1. Solve. 3/x + 5/x = -4

*x = - 1/2
*x = 1/2
*x = -2
*x = 2

2. Solve. 1/3x+9 - 2/x+3 = 2

*x = 1
*x = -23/6
*x = -5/2
*x = -5

3. Solve. 8/x + 15/x^2 = -1

*x = 5, –3
*x = –5, –3
*x = –5, 3
*x = 5, 3

4. The temperature T of a given mass of gas varies inversely with its volume V. The temperature of 90 cm^3 of a certain gas is 30°C. What will be the temperature of the gas when it is compressed to a volume of 18 cm^3?

*6°C
*150°C
*120°C
*75°C

5. What are the excluded values of the function? y = 7/2x + 48

*0.04
*24
*–24
*–0.04

Please and Thank you. I've been stuck on these all day and I have more work to do after this. :(

Some of the problems (#2 and #5) are not presented clearly, so I have to make assumptions on some of them, based on studying the answer choices.

1) 3/x + 5/x = -4
3 + 5 = -4x, from multiplying both sides by the least common denominator x
x = (3 + 5)/(-4) = -2, from dividing both sides by 4.

Yes, x = -2 is valid (not extraneous) since no denominators become zero.

2) I will need to assume that you mean 1/(3x+9) - 2/(x+3) = 2; don't forget parentheses.
1/(3x+9) - 2/(x+3) = 2
1/(3(x+3)) - 2/(x+3) = 2, from factoring any denominators that are factorable
1 - 3(2) = 2(3)(x+3), from multiplying both sides by the least common denominator 3(x+3)
-5 = 6(x+3)
-5/6 = x+3, from dividing by sides by 6
x = -5/6 - 3 = -5/6 - 18/6 = -23/6, from subtracting 3 from both sides.

Yes, x = -23/6 is valid (not extraneous) since no denominators become zero.

3) 8/x + 15/x^2 = -1
8x + 15 = -x^2 from multiplying both sides by the least common denominator x^2
x^2 + 8x + 15 = 0, from adding x^2 to both sides
(x + 3)(x + 5) = 0, from factoring the left side
x + 3 = 0 or x + 5 = 0
x = -3 or x = -5.

Yes, x = -3, x = -5, are both valid (not extraneous) since no denominators become zero.