HELP witth algebra!?

Question

this is algebra one. you just have to solve the rational expressions and thats all.

and i'm not cheating, i'm just doing this cause i've been trying to work these out for like ever and i can't get it. :[[

1. x/3 - 2/3 = 1/x

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

3. x/x+4-4/x-4=x^2+16/x^2-16

4.x^2/x^2-4=x/x+2-2x/2-x

THANKS

Answer 1

1. Since the x and 2 have the same denominator you can combine them to say (x-2)/3 and then cross multiply. You multiply 3x1 and (x-2)(x) and you'll get x^2 - 2x = 3. If you make the equation equal to 0 you'll have a quadratic and by subtracting the 3 you'll get zero. It'll say x^2 - 2x - 3 = 0

At this point you FOIL and it will say (x + 1)(x - 3) = 0

Make both equal to 0 to get:

x+1= 0 and x-3=0

Solve for x and you'll get

x = -1 and x = 3

2. The left side of the second equal is kind of confusing because the way it's written the 3's cancel out. Like you have 1/x + 3 but not 1/(x+3) which I think you may mean to have? If that's not what you mean then the three's will cancel out on the left side and you'll be left with 1/x + 1/x and that's just 2/x. From that point you can do the same thing I did with the first one and cross multiply. It'll say after you cross multiply:

x=2x^2 - 18. I'm also assuming x isn't being raised to 2-9 but parenthesis aren't there where there should be. It's okay though..

Like with the other one, you make one side equal to zero. It's easier to subract the X from the left side and then you'll get:

2x^2 - 18 - x = 0

It's easier to solve if you put the x in the middle so..

2x^2 - x - 18= 0

From this point I believe you'd have to use the quadratic equation so it might have been that you wrote the equation wrong on here. I'll try it the other way. If this is what you meant (with the (1/x)+ 3 and not 1/(x+3)) then the quadratic equation is:

-b +/- sqr root of: [(b^2) - (4ac)] all of it divided by 2.

Now, if the problem is supposed to say:

1/(x+3) + 1/(x-3) = (1/x^2) - 9

You do need to find a common denominator on the left side. You have to multiply top and bottom of 1/(x+3) with (x-3) and vice versa with 1/(x-3) multiplied by x+3. After doing so you'll have a common denominator. You'll end up with:

[(xt3)+(x-3)]/[(x^2)-9]. At that point, you can cross multiply and it can get messy but you follow the order of operations, you'll be fine.

I left you with some work to do on your own but after explaining the first one and giving you most of the second (whichever version it is) you should figure it out. The last 2 problems are probably done just the same. Good luck!

Answer 2

1) 3

(3 divided by 3) - (2 divided by 3) = 1 divided by 3  Or simply put in fractions

3 thirds - 2 thirds = 1 third

2) simplify and turn problem into this -

2/x+3-3 = 1/x^2 -9 then take it further to make it 2/x=1/x^2-9. then work the problem out from there

For problems 3 and 4, it doesn't matter what x equals when it comes to the x/x and x^2/x^2 fractions.  Both of those will end up being 1, as anything divided by itself will always be 1.

which will make the equations, respectively,

1+4-4/x-4 = x^2+16/x^2-16

and

1-4 = 1+2-2x/2-x

I gave you first answer, hope you can get the others!

BTW - I'm going on the assumption that the fractions are as they appear (i.e. #2's 1/x + 3 does not mean that x+3 is the total denominator... if it is, that really changes things!)

And let me know if I've goofed.  I'm running low on sleep.

Answer 3

x/3 - 2/3 = 1/x

(x - 2)/3 = 1/x

x(x - 2) = 3

x^2 - 2x - 3 = 0

(x - 3)(x + 1) = 0

x = 3 ...OR... x = -1

i assume your question is...

1/(x + 3) +  1/(x - 3) = 1/(x^2 - 9)

[1(x - 3)]/[(x + 3)(x - 3)] + [1(x + 3)]/[(x - 3)(x + 3)] = 1/(x^2 - 9)

(x - 3 + x + 3)/(x^2 - 9) = 1/(x^2 - 9)

2x/(x^2 - 9) = 1/(x^2 - 9)

2x = 1

x = 1/2

x/(x + 4) - 4/(x - 4) = (x^2 + 16)/(x^2 - 16)

x/(x + 4) - 4/(x - 4) = (x^2 + 16)/(x^2 - 16)

[x(x - 4)]/[(x+4)(x-4)] - [4(x + 4)]/[(x-4)(x+4)] = (x^2 + 16)/(x^2 - 16)

(x^2 - 4x - 4x - 16)/(x^2 - 16) = (x^2 + 16)/(x^2 - 16)

x^2 - 8x - 16 = x^2 + 16

x^2 - x^2 - 8x = 16 + 16

-8x = 32

x = 32/-8

x = -4

x^2/(x^2 - 4) = x/(x + 2) - 2x/(2-x)

x^2/(x^2 - 4) = [x(2 - x)]/[(x + 2)(2 - x)] - [2x(x + 2)]/[(2 - x)(x + 2)]

x^2/(x^2 - 4) = (2x - x^2)/(-x^2 + 4) - (2x^2 + 4x)/(-x^2 + 4)

x^2/(x^2 - 4) = (2x - x^2 - 2x^2 - 4x)/(-x^2 + 4)

x^2/(x^2 - 4) = (-3x^2 - 2x)/(-x^2 + 4)

-x^2/(-x^2 + 4) = (-3x^2 - 2x)/(-x^2 + 4)

-x^2 = -3x^2 - 2x

-x^2 + 3x^2 + 2x = 0

2x^2 + 2x = 0

2x(x + 1) = 0

2x = 0

x = 0

...OR...

x + 1 = 0

x = -1

Answer 4

Question 1

1) find common denominators

x(x)/3(x) - 2(x)/3(x) = 1(3)/x(3)

2) eliminate the denominators since they are the same, you are left with x^2 -2x = 3

3) bring the 3 to the other side

x^2 -2x -3 = 0

4) factor

(x+ 1)(x-3)= 0

5) solve for x

x+1= 0  x-3=0

x= -1     x=3