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

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

I think the best way to do this is to just multiply out the first two terms, and for the third term, factor the denominator to (x-3)(x+3).
can you multiply it? I dont understand this. it looks like subtraction to me
oh sorry, I didn't see the / in the first term. Ok, for the second term, (x+1)(x+3), multiply this out Then for the third term, (x-9)/(x^2-9), factor the denominator

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

uh oh i typed this out wrong its supp((x-2)/(x-3))-((x+1)/(x+3))+((x-9)/(x^2-9))osed to be
\[\frac{x-2}{x-3} - \frac{x+1}{x+3} + \frac{x-9}{x^2 -9}\] Recognize that \(x^2 - 9 = (x+3)(x-3)\) and put them all under a common denominator of \(x^2-9\)
Over a common denominator rather.
so i need to multiply the first two fractions by (x+3) or x-3?
Multiply the first by (x+3) on top and bottom, and the second by x-3 on top and bottom.
Oh, was the second term also a fraction? Remember when you learned how to add fractions with different denominators? Use the same logic. Multiply the first fraction by (x+3)/(x+3) and the second by (x-3)/(x-3) so that they all have a common denominator. Basically what polpak is saying but it is worth noting that this is just like what you do when you are adding normal fractions with different denominators.
Oh. I may have mistyped it. Was the middle term a fraction? If not it'd be \[\frac{x-2}{x-3} +(x+1)(x+3) +\frac{x-9}{x^2-9}\]
But you still want to get them over a common denominator of \(x^2-9\)
polpak, she posted that they were all fractions a few minutes ago. I missed it too.
you were correct the first time i mistyped it... thank you both for your help! its greatly appreciated
so i came up with 0(x+3)(x-3) is this correct? i somehow dont think it is
im sorry 0/(x-3)(x+3)
I got something else. Did you forget that the second fraction is being subtracted? I almost forgot that.
no i subtracted it but it all seemed to cancel out. what am i doing wrong?
\[\frac{x-2}{x-3} - \frac{x+1}{x+3} + \frac{x-9}{x^2-9} \] \[= \frac{(x-2)(x+3)}{x^2-9} - \frac{(x+1)(x-3)}{x^2-9} + \frac{x-9}{x^2-9} \] \[= \frac{[x^2+x-6] - [x^2-2x -3] + [x-9]}{x^2-9} \] \[= \frac{x^2-x^2+x + 2x + x -6+3 -9}{x^2-9} \] \[= \frac{4x-12}{x^2-9} \] \[= \frac{4(x-3)}{x^2-9} \] \[= \frac{4(x-3)}{(x+3)(x-3)} \] \[= \frac{4}{x+3} \]
Where'd you go wrong?
And hopefully you can follow that mess.
i didnt change the signs in the second fraction AND i can follow it
Thank you so much!
Cool, np =)
4/( x+3)

Not the answer you are looking for?

Search for more explanations.

Ask your own question