Can someone please help me with my algebra?

Question

I'm so desperate and I've been trying to look for good sites and have been looking over my notes for hours. I need help adding and subtracting rational expressions.
They are:
.x.......x
---- + ---
3.......5

.........6..............4
--------------- - -------
x^2 - 3x +2.....x - 2

........4...............1
--------------- + -------
c^2 + 4c+3.....c+3

........3...........9
--------- + -----------
2x + 6......a^2 - 2

....3.........4
------- + ------
2d^2......3d

..4........6
----- + -----
x+3....x-2

3t.....3t
--- + ---
8......8

..4.........8
------ - ------
k+3.....k+3

..2x........3
------- - ------
x^2-1...x+1

Please atleast show me how to do some of them, i'm begginyou from my soul. P.S. ignore the dots.

I'm sorry the dots are to make it more readable as fractions. And i cannot find a website for this. SOmeone pleas help me i have ADD and no one will giv me medication.

Answer 1

for the first one,


to do it you must multiply the numerator and denominator for each fraction by some form of one so that the denominator is the same for each, so
x/3 * 5/5 = 5x/15
and
x/5 * 3/3 = 3x/15

And then you can finally add the numbers because they have the same denominator.
5x/15 + 3x/15 = 8x/15.

Hope that makes sense!!! i don't have mch time as i have homework too..

Answer 2

First, you went to a lot of work trying to format that stuff into meaningful fractions. It's appreciated, but perhaps a little confusing at first.



(x/5)+(x/3)
Don't get freaked out when you see letters or symbols. 3 is just a symbol too.
You're trying to add fractions with different denominators.
The trick, just as if the numerator were a number, is to get the denominators the same. If you know how to find the lowest common denominator, that's what you need to do. If you don't know how to do that, you can just multiply the denominators together.
(3)(5) = 15
so you new denominator will be 15.
The problem is how many 15ths are x/3 and x/5?
Here's how. Divide the new denominator by the old denominator and multiply the numerator by that value.

Let's do x/3
Divide the new denominator (15) by the old denominator(3) 15/3=5.
Now multiply the numerator (x) by the value we just calculated (5). We get
x/3 = 5x/15

Doing the same with x/5
15/5=3
x/5=3x/15

Now add 5x/15 + 3x/15 = 8x/15
=================
Next one:
This one is a little less trivial.
6/(X² - 3x + 2) + 4/(x-2)

Just like the last one, we need to get this over a common denominator.
Fortunately X² - 3x + 2 =(x-2)(x-1) So x-2 is a factor in both denominators. So the new denominator will be X² - 3x + 2

6/(X² - 3x + 2) is already done.

How many (X² - 3x + 2)ths is 4/(x-2)
First divide the new denominator (X² - 3x + 2) by (x-2) and you get (x-1) because (x-1)(x-2)=X² - 3x + 2
Now multiply the denominator by x-1
4/(x-2)=4(x-1)/(X² - 3x + 2) = (4x-4)/(X² - 3x + 2)
Now we can add
6/(X² - 3x + 2) + (4x - 4)/(X² - 3x + 2)=
(6 + 4x -4)/(X² - 3x + 2)=
(4x + 2)/(X² - 3x + 2)=

I'm not certain, but I suspect your teacher likes stuff factored so
2(2x + 1)/[(x-2)(x-1)]

Believe me, the rest are done exactly the same way. Some are not as "convenient" as the one I just did, but the process is always the same.

Step 1. Find the lowest common denominator, if you can, If you can't just use the product of the two denominators.
Step 2. For each fraction, divide the new denominator you developed in step 1, by the old denominator.
Step 3. For each fraction multiply the numerator by the value you came up in step 2 for that fraction.
Now, all the denominators are the same.
Step 4. Add the new numerators.
Step 5. Factor everything you can.
Step 6. Reduce the fractions if possible. For example (5x²-10x)/(x² - 4) factors into
[(5x)(x-2)]/[(x+2)(x-2)]
That's the same as [(5x)/(x+2)][(x-2)/(x-2)] and since (x-2)/(x-2)=1. you would wind up with 5x/(x+2)
If you didn't factor everything, you might have missed that.

Answer 3

How about I solve the first three for you? The trick to these problems is to make the denominator's the same, and then add or subtract them together. For the first problem, follow these steps:
X/3 + X/5
5X/15 + 3X/15
8X/15

For the second problem, follow these steps:
6/(X^2 - 3X + 2) - 4/(X - 2)
6/((X - 2)(X - 1)) - 4/(X - 2)
6/((X - 2)(X - 1)) - 4(X - 1)/((X - 2)(X - 1))
(6 - 4(X - 1))/((X - 2)(X - 1))
(10 - 4X)/((X - 2)(X - 1))

For the third problem, follow these steps:
4/(C^2 + 4C + 3) + 1/(C + 3)
4/((C + 1)(C + 3)) + 1/(C + 3)
4/((C + 1)(C + 3)) + 1(C + 1)/((C + 1)(C + 3))
(4 + (C + 1))/((C + 1)(C + 3))
(5 + C)/((C + 1)(C + 3))

Answer 4

That's too many to do at bedtime, but here's a couple (without the dots):

x/3 + x/5 =
(5/5)(x/3) + (3/3)(x/5) =
5x/15 + 3x/15 =
8x/15

you multiply each fraction by 1 (same thing top and bottom), in the form of whatever it takes to end up with the same denominators.

6 / [ x² - 3x + 2] - 4 / [ x - 2 ] =
6 / [ (x-2)(x-1)] - 4(x-1) / [(x-2)(x-1)] =
[ 6 - 4x + 4] / [ (x-2)(x-1)] =
(10 - 4x) / [ (x-2)(x-1)]

and one with the dots so you can see the added factors that make the common denominators:

...4........6
------ + ------ =
x+3 ..... x-2

...4(x-2) ........6(x+3)
------------- + ------------- =
(x+3)(x-2).....(x-2)(x+3)

4x - 8 + 6x + 18
----------------------- =
....(x-2)(x+3)

10x + 10
-------------
(x-2)(x+3)

Answer 5

They're similar to regular fractions, you just need to find a common denominator and work from there. Some of them can be difficult since they're polynomials, but if you just factor them out, you should be able to find the answer.

For example, your very last one: x^2 -1 = (x+1)(x-1)

so your equation would be [(2x)/(x-1)(x+1)] - [{3(x-1)}/{(x+1)(x-1)]
= (-x+3)/(x^2 -1)

I hope this helps.

Answer 6

I can tell u
Its easy if you pay attention. IN algebra only two like variables can be added or subtracted with each other ex.
2x + 2x will be 4x
But u cannot add 2x + 2y because the variable is not like.
However in multiplication or division like or unlike is not there you can just multiply or divide any variable with the other ex.
2x *2x is 4x square(2) and u can even multiply 2x * 2y the answer will be 2xy
Best Of Luck!!11

Answer 7

i have no idea but i can still help by telling u that u can type in google the name of the kind of math problems you are doing and it will bring up a bunch of websites and you click one and it should show you how to do them......
try several sites from the google search until you find one that's shows you in a way you can understand.....
:O)

Answer 8

Just what are you asking? As far as I can tell you are mumbling a the keyboard. What the heck are the periods for?