Question

What is the simplified form of x plus 1 over x squared plus x minus 6 divided by x squared plus 5x plus 4 over x plus 3 ?

Answer 1

If I write it in the equation editor, it is simpler already ;)\[\frac{ x+1 }{ x^2 }+\frac{x-6}{x^2}+\frac{5x+4}{x+3}\]

You can start with the two leftmost fractions; they have the same denominator, so can be written as one fraction.

Answer 2

Which makes it \[\frac{ 2x-5 }{ x ^{2} }+\frac{ 5x+4 }{ x+3 }\] right?

Answer 3

Yes! Now you have write the fractions with a new denominator, that is the same for both.
Do you know how to do that?

Answer 4

Hint: use x²(x+3) as the new denominator.

Answer 5

\[\frac{ 2x-5 }{ x ^{3} +3x^{2} }+\frac{ 5x+4 }{ x ^{3}+3x ^{2} }\] right?

Answer 6

Right, although I would prefer to leave the denominator as x²(x+3).

Now write it as one fraction and simplify the numerator.

Answer 7

\[\frac{ 7x-1 }{ x ^{2}(x+3) }\] right?

Answer 8

Oops, I only looked at the denominator. YOu can't just change them without changing the numerator as well:\[\frac{ (2x-5)(x+3) }{ x^2(x+3) }+\frac{ x^2(5x+4) }{ x^2(x+3) }\]

Answer 9

Denominator of first fraction has been multiplied with x+3, so do that in the numerator as well.

Second one: both multiplied with x²

Answer 10

Have to go now, back in half an hour...

Answer 11

Okay, so then it's \[\frac{ 2x ^{3}+x-15 }{ x ^{2}(x+3) }+\frac{ 5x^{3}+4x^{2} }{ x^{2}(x+3) }\]Right?

Answer 12

Okay.

Answer 13

Hey that's right! Now write it as one fraction.

(now I really must go, sorry..)

Answer 14

Okay, see you later!

Answer 15

\[\frac{ 7x^{3}+4x^{2}+x-15 }{ x^{2}(x+3) }\]right?

Answer 16

@HarvestMan: You have got it right!

Answer 17

Awesome!! Can it be simplified any further?

Answer 18

The only answers for my homework are
1 over the quantity x plus 3 times the quantity x plus 4
1 over the quantity x plus 3 times the quantity x minus 2
1 over the quantity x plus 4 times the quantity x minus 2
or 1 over the quantity x plus 3 times the quantity x plus 1

Answer 19

No, because this could only be done if you'd factor the numerator. If there were common factors, the would cancel.
But: x is not a common factor, and x-3 neither (to see why, set x=3 in the numerator - it won't come out at 0), so there is nothing to be done about it...

Answer 20

If these are your only options, then have we worked with the right fractions from the beginning?

Answer 21

Yes, we have. But, can we simplify our answer any further?

Answer 22

oh

Answer 23

Hold on, I'll check everything from the beginning....

Answer 24

Okay

Answer 25

When I do the last step again, I get:

Answer 26

\[\frac{ (2x-5)(x+3) +x^2(5x+4)}{ x^2(x+3) }=\frac{ 2x^2+6x-5x-15+5x^3+4x^2 }{ x^2(x+3) }=\]\[\frac{ 5x^3+6x^2+x-15 }{ x^2(x+3) }\]Cannot simplify this further...

Answer 27

Hm.. which one of the answers do you think would be the closest to this?

Answer 28

I'm guessing answer 1 is the closest

Answer 29

I've got to go, talk to you later!

Answer 30

OK, wouldn't have a clue, I'm afraid...

Answer 31

I'll just go with answer #1