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Mathematics 10 Online
OpenStudy (anonymous):

Please explain how I would simplify this

OpenStudy (anonymous):

OpenStudy (anonymous):

Do you know what the greatest common factor is to combine those fractions?

OpenStudy (anonymous):

Sorry for not responding. GCF for the numerator or denominator?

OpenStudy (anonymous):

Denominator.

OpenStudy (anonymous):

2?

OpenStudy (anonymous):

sorry.... i didn't mean GCF... what I meant to say is the Least Common Denominator, LCD. the LCD of each of those terms is x(x+1)(x+2)

OpenStudy (anonymous):

how´d you find that?

OpenStudy (anonymous):

it's the same way you'd add/subtract regular fractions.... you need to find the LCD.. for example, to add \(\large \frac{1}{2}+\frac{1}{3} \), the LCD is \(\large 2\cdot3 \) or 6.

OpenStudy (anonymous):

oh, ok

OpenStudy (anonymous):

so: \(\large \frac{1}{2}+\frac{1}{3}=\frac{1}{2}\cdot\frac{3}{3}+\frac{1}{3}\cdot\frac{2}{2} =\frac{3}{6}+\frac{2}{6} = \frac{3+2}{6}=\frac{5}{6}\) this is basically what you'd do to your original problem

OpenStudy (anonymous):

so you wanna work it out with me?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

so for the first term, \(\large \frac{3}{x} \), to get the denominator to be x(x+1)(x+2), you need to multiply both top and bottom by (x+1)(x+2): \(\large \frac{3}{x}=\frac{3}{x}\cdot \frac{(x+1)(x+2)}{(x+1)(x+2)}=\frac{3(x+1)(x+2)}{x(x+1)(x+2)} \) that's it for the first term...

OpenStudy (anonymous):

ohh... no one here.... :(

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