please help?

Question

I have trouble with these if you could please either show me how to do them or show me a site to help

1)subtract. Express your answer in simplest form (2x/x-5)-(10/x-5)

2) simplify (m^4)^3(m^2)^9/(m^3)^2

3)solve x^2-7x+6=0

4) factor by grouping
35x^3+30x^2-7x-6

5) subtract. Write your answer in simplest form.
(y-2/y+3)-(y/2y+6)

6) solve
(5/x+4)+3=2/x+4

7) factor completely
3(x-2)^2-3(x-2)-6

8)solve
(x/3)-(1/4)=3x-8/12

Answer 1

1) The denominators are both x-5, so you can just subtract the numerators:

(2x/x-5)-(10/x-5) =
(2x - 10)/(x - 5) =
2 (x - 5) / (x - 5) =
2

It turned out that the numerator was twice the denominator, so the answer is just "2".

Note that you're not writing the problems technically correctly: 2x/x-5 is really calculated as (2x/x) - 5, which would be just -3, because division takes precedence over subtraction... but I'm pretty sure you meant 2x/(x-5). Remember the precedence of operators and use parentheses where necessary.

2) You need to know that an exponentiation of an exponentiation equals multiplying the exponents. And that multiplying adds exponents, and dividing subtracts them...

(m^4)^3 (m^2)^9 / (m^3)^2 =
m^(4*3) m^(2*9) / m^(3*2) =
m^12 m^18 / m^6 =
m^(12 + 18 - 6) =
m^24

3) You can factor this one by looking at it. Since the coefficient of x is negative and the constant is positive, you want two numbers whose product is +6 and sum is -7 (-1 and -6)

x^2 - 7x + 6 = 0
(x - 1)(x - 6) = 0

The solutions are x=1 and x=6

4) Look for a multiple of the two right-hand terms (7x+6) in the two left-hand terms:

35x^3+30x^2-7x-6 =
5x^2 (7x + 6) - (7x + 6) =
(5x^2 - 1)(7x + 6)

5) Need to put that into least common denominator, which is 2y + 6 (since that's twice y+3):

(y-2/y+3)-(y/2y+6) =
2(y-2)/2(y+3) - (y/(2y+6)) =
(2y-4)/(2y+6) - y/(2y+6) =
(y - 4)/(2y + 6) =
(y - 4)/2(y+3)

Again, assumes that you mean (y-2)/(y+3) when you write (y-2/y+3).

6) Multiply by x+4 to get it out of the denominator:

(5/x+4)+3=2/x+4
5 + 3(x+4) = 2
3x + 17 = 2
3x = -15
x = -5

Again, this assumes the right-hand term is "2/(x+4)".

7)
3(x-2)^2-3(x-2)-6
3(x^2 - 4x +4) - 3x + 6 - 6
3x^2 - 12x + 12 - 3x
3(x^2 -5x + 4)
3 (x-4)(x-1)

8) Multiply by 12 to get rid of all the denominators:

(x/3)-(1/4)=3x-8/12
4x - 3 = 3x - 8
x = -5

Note that this assumes "3x-8/12" is meant as "(3x-8)/12" and not "3x-(8/12)" which is the more technically correct interpretation as written.

One other suggestion: If you want people to help and give detailed answers, you should post each of your eight problems as a separate question. Very few people are going to do the work to write eight different detailed explanations for one question for you.

Answer 2

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Answer 3

Of course you're having trouble with them. You were given a book and you aren't reading it. You're playing around on the computer. And sleeping through class, if you made it there at all. Did you ever consider that people who might be willing to help those in need are deliberately feeding YOU false answers simply because they like to laugh at people who don't want to do the work?

Answer 4

1. 2
2. m^24
3. x= 6 and 1
5. y-4/2y-6
6 x= -5
7.x= -9
8. x=3/32