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Mathematics 11 Online
OpenStudy (anonymous):

Subtract:

OpenStudy (anonymous):

|dw:1393392757815:dw|

OpenStudy (luigi0210):

Get common denominators.

OpenStudy (anonymous):

this might sound dumb but... how exactly do I do that??

OpenStudy (luigi0210):

Apply the basic fraction rules you learned.

OpenStudy (anonymous):

Yeah that's what I am confused about..... Can you explain how?

OpenStudy (anonymous):

Here use this formula :) \[\large \frac{ a }{ b } + \frac{ c }{ d } = \frac{ ad + bc }{ bd }\]

OpenStudy (luigi0210):

^Yea, pretty much that

OpenStudy (anonymous):

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

OpenStudy (anonymous):

8x+4/ (x-2)(x+2) ??

OpenStudy (luigi0210):

Think of it this way, say we have: \[\LARGE \frac{1}{3}+\frac{2}{5}\] How could we add them? Get common denominators: \[\LARGE \frac{1}{3}*\frac{5}{5}+\frac{2}{5}*\frac{3}{3}\] Now we can add them: \[\LARGE \frac{5}{15}+\frac{6}{15}\] Same with yours,,

OpenStudy (luigi0210):

Thank you @AshleyAlley<3 T_T

OpenStudy (anonymous):

Okay so was my answer correct? I get it now

OpenStudy (luigi0210):

Yea, it's right.

OpenStudy (anonymous):

Oh, wait your original question was to subtract them not add, ahhhh my bad >.<

OpenStudy (anonymous):

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

OpenStudy (anonymous):

you're welcome @Luigi0210 xD

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