Question

Please help with Addition and Subtraction of algebraic fractions!

Answer 1

First step is to get a common denominator. Can you factor the denominators of each of the fractions?

Answer 2

You can factor the denominator of the first fraction to get (x+5)(x+3)

Answer 3

Okay, good. So as it has no factors which don't appear in the other fractions denominators, you can multiply the other fractions top and bottom by the missing factors in their denominators. The \[\frac{3x}{x+5} = \frac{3x}{x+5}*\frac{x+3}{x+3}\]for example, and the result will have a common denominator with the first fraction. Do the same for the other fraction, and then you can combine the numerators over that denominator.

Answer 4

In general, the common denominator will be the product of all of the unique factors you find in the denominators of all of the fractions you are trying to combine.

Answer 5

So now it looks like this:

Answer 6

To add fractions with different denominators, we must learn how to construct the Lowest Common Multiple of a series of terms. The Lowest Common Multiple (LCM) of a series of terms is the smallest product that contains every factor from every term.

For example, consider this series of three terms:
pq pr ps

We will now construct their LCM -- factor by factor.To begin, it will have the factors of the first term:

LCM = pq

Moving on to the second term, the LCM must have the factors pr. But it already has the factor p -- therefore, we need add only the factor r:

LCM = pqr

Finally, moving on to the last term, the LCM must contain the factors ps. But again it has the factor p, so we need add only the factor s:

LCM = pqrs.

That product is the Lowest Common Multiple of pq, pr, ps. It is the smallest product that contains each of them as factors.

\[\frac{ x + 6 }{ x^2 + 8x + 15 } + \frac{ 3x }{ x + 5 } - \frac{ x - 3 }{ x + 3 }\] We need to find the common denominator. Start by factoring your denominators

\[\frac{ x + 6 }{(x+5)(x+3) } +\frac{ 3x }{ x + 5 } - \frac{ x - 3 }{ x + 3 }\]

\[\frac{ x + 6 }{(x+5)(x+3) } +\frac{ 3x(x+3) }{ x + 5 } - \frac{ x - 3(x+5) }{ x + 3 }\]

Now, simply solve for the numerator by adding like terms and putting it al over the common denominator.

Answer 7

Yes. Now expand the numerators, add 'em up (or subtract, as appropriate), and you've got your result. It may be possible to factor the result and cancel common terms.

Answer 8

OK but this is actually the part I was having trouble with because either I am incorrectly adding and subtracting or I just am missing a step.

Answer 9

Last expression in @compassionate's writeup should be

\[\frac{ x + 6 }{(x+5)(x+3) } +\frac{ 3x(x+3) }{ (x + 5)(x+3) } - \frac{( x - 3)(x+5) }{( x + 3)(x+5) }\]

Answer 10

Whoops! Missed that. My keyboard is really something right now.

Answer 11

@MathHater82 what is \[3x(x+3)\]after applying the distributive property?
what is \[(x-3)(x+5)\]after applying the distributive property?

Answer 12

Basically, once you have a common denominator, you can add and subtract like usual

Answer 13

For the first one \[3x^2+9x\]
For the second one \[x^2+2x-15\]

Answer 14

Yes. Add those plus the numerator of the first fraction, and what do you get?

Answer 15

shouldn't the numerator of the last fraction be subtraction?

Answer 16

Yes, it should

Answer 17

right, earlier I did say add or subtract as appropriate :-)

Answer 18

So i should add x+6 and 3x^2+9x and then subtract x^2+2x-15?

Answer 19

yes, \[\frac{(x+6)+(3x^2+9x)-(x^2+2x-15)}{(x+5)(x+3)}\]

Answer 20

this would be what it looks like?

Answer 21

Yes, that is right. Can you factor the numerator at all?

Answer 22

If you can, and you find a common factor in the numerator and denominator, you can simplify some more. If not, you are done, aside from deciding whether to write the denominator in factored form as you have it, or expanded as it originally appeared in the problem statement.

Answer 23

I don't think I can factor the numerator but maybe I'm wrong.

Answer 24

No, it cannot be factored any more. But always worth checking at this point to see if you can get a simpler answer...

Answer 25

OK thanks!

Answer 26

any questions about what we did? sometimes you'll end up with something like

\[\frac{x}{a*b} + \frac{y}{a*c} +\frac{z}{b * c}\]or some variation where no fraction has all of the factors in the denominator, unlike this problem. Proceed in the same way, you'll just end up with a common denominator that is more complicated than any of the starting ones...just tedious algebra, not rocket surgery :-)

Answer 27

Haha you mean rocket science?

Answer 28

or brain surgery

Answer 29

commingling two common expressions for difficult activities, "it's not rocket science", and "it's not brain surgery"

Answer 30

Yeah well thanks anyway I don't have anymore questions.