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Mathematics 10 Online
OpenStudy (matimaticas):

please help (x/x + 3) + (x + 2/x + 5)

OpenStudy (anonymous):

First you need to create a common denominator. Do you know how?

OpenStudy (matimaticas):

would that be like x + 15

OpenStudy (anonymous):

No It would be (x+3)(x+5)

OpenStudy (matimaticas):

oh ._. lol then what?

OpenStudy (anonymous):

You would need to multiply that common denominator towards our numerators.

OpenStudy (anonymous):

Like so, \[\frac{ x(x+5) }{ (x+3)(x+5) }\]

OpenStudy (anonymous):

Yes like @Johnbc said whatever you multiply the denominator by you must also do to the numerator

OpenStudy (anonymous):

Because the only term missing was (x+5) to get our common denominator we only had to multiply by that and to the other term you can already tell it is only missing (x+3) so multiply the other term by that and add them over your new common denominator.

OpenStudy (matimaticas):

so what would it look like when the two numerators are added

OpenStudy (anonymous):

\[\frac{ x(x+5) + (x+2)(x+3) }{ (x+3)(x+5) }\]

OpenStudy (anonymous):

You know have to perform the multiplication to get your new terms and then combine like terms.

OpenStudy (matimaticas):

\[2x^2 + 10x + 6 \ \]

OpenStudy (anonymous):

Correct that is the numerator

OpenStudy (matimaticas):

and then you have to factor?

OpenStudy (anonymous):

Well you also have a denominator.

OpenStudy (anonymous):

\[\frac{ 2x^2+10x+6 }{ x^2+8x+15 }\]

OpenStudy (matimaticas):

ok cool, ^ is that simplified?

OpenStudy (anonymous):

Nope.

OpenStudy (anonymous):

You can cancel stuff out and simplify further now.

OpenStudy (anonymous):

If you leave the denominator factored and perform the operation you said it is actually easier.

OpenStudy (anonymous):

\[\frac{ 2(x^2+5x+3) }{ (x+3)(x+5) }\]

OpenStudy (matimaticas):

ok now I think that is totally factored

OpenStudy (anonymous):

Yeah that would be close to a factored answer

OpenStudy (matimaticas):

thanks so much for holding my hand through this lol!

OpenStudy (anonymous):

My pleasure

OpenStudy (anonymous):

If you are asked to perform the division you can actually get a more simplified value.|dw:1405371694798:dw|

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