At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
And ended up with 3(x-1) 2x-1 But I'm very unsteady with these sorts of things. Check my work?
first, factor the denominator of the first rational function. From that you'll know the common denominator of both the functions.
I'm trying to solve it now, and I'll check if we got the same thing
Factored, isn't it (2x-1)(x-3)? Then the x-3s would cancel out.
nope, it should be (2x+1)(x-3)
Oops! Sorry. that's what I meant; typo. The buttons are right next to each other lol
so it will look like this: |dw:1433537147204:dw| then since they already have the same denominator, it will be like this: |dw:1433537285450:dw|
Would the numerator end up as x+1, or am I taking things to literally?
hmm let's see.. 3x-2-(2x+1) --> remove the brackets first 3x-2-2x-1 --> apply the operations with like terms, what is 3x-2x? and what is -2-1?
Oh! 1x-3 ( or just x-3), then, not 1. (I'm so awful at math....Thanks for the help..)
^_^ you're not done yet ! so the numerator will be x-3, and you also have x-3 in the denominator, so it means, they will cancel out and you'll end up with..?
1 over 2x+1?
Thank you sooooo much.... You don't realise how much good you've just done me!!!!