(8)/(x-2) + (2)/(x-3) ??? help please its pre-cal

Question

anonymous · Answer

Hey there! I think I can help you out here! :]

So in order to add two fractions, you have to have a common denominator!

So take 8/(x-2) and multiply it by (x-3)/(x-3). This is the same thing as multiplying it by ONE! Multiply 2/(x-3) by (x-2)/(x-2). This is also multiplying by ONE! Then, your denominators will be the same.

anonymous · Answer

can you be a tad morespecific i dont get what you mean :s sorry

anonymous · Answer

That's fine! So if you were to add 3/4 and 5/3, you would have to make the denominator an 12 on the bottom of both fractions to add them, right?

anonymous · Answer

right

anonymous · Answer

You're essentially doing the same with your problem, except that instead of finding a number they have in common, you'll multiply both by a number equal to 1 that will give them the same denominator.  

The denominator for both needs to be (x-2)(x-3). So you'll multiply the 8/(x-2) by (x-3)/(x-3) and the 2/(x-3) by (x-2)/(x-2). So you'll end up with 8(x-3)/(x-2)(x-3) + 2(x-2)/(x-2)(x-3). Once they have the same denominator, you can drop it and simply add the numerators.

anonymous · Answer

so it would basically be$$8(x-3)+2(x-2)\over(x-3)(x-2)$$ or am i being stupid?

anonymous · Answer

Correct! Can you simplify it from there?

anonymous · Answer

$$10x-20\over (x-3)(x-2) $$ ?