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Mathematics 53 Online
OpenStudy (anonymous):

(2/x) + (2/x-1) - (2/x-2) simplify with work

OpenStudy (matheducatormcg):

If you're just simplifying, then you need to find a LCD and combine the fractions. Do you know the LCD?

OpenStudy (anonymous):

nope, i was only given the problem as shown above. im having a super hard time with this class lol

OpenStudy (matheducatormcg):

Well, common denominators are numbers that can be divide by all the denominators. Since you have all expressions in your denominators, you have to use the old multiple them together to find the common denominator. The LCD here would be x(x-1)(x-2).

OpenStudy (anonymous):

ok, so if i have the LCD, what do i do with it now? I honestly wasnt looking for an answer, just how to solve it

OpenStudy (matheducatormcg):

so now, multiply each fraction in the numerator and denominator by the part of the LCD that is not present in the fraction. \[\left( \frac{(x-1)(x-2)}{(x-1)(x-2)} \right)\frac{2}{x}+\left( \frac{(x)(x-2)}{(x)(x-2)} \right)\frac{2}{x-1}-\left( \frac{(x)(x-1)}{(x)(x-1)} \right)\frac{2}{x-2}\]

OpenStudy (anonymous):

wait, so am i supposed to foil what ever is in the parenthesis?

OpenStudy (matheducatormcg):

not in the denominator because your combining fractions which means the denominators stay the same. but the numerator does need to be foiled in each fraction and then combine like terms in the numerator.

OpenStudy (matheducatormcg):

your answer should have \[x(x-1)(x-2)\] in the denominator. Don't foil this part because understanding what makes the denominator zero is very important.

OpenStudy (anonymous):

ok, so after i foil in the first parenthesis i get x^2-3x+2/(x-1)(x-2). what do i do with this?

OpenStudy (matheducatormcg):

there should be an x in the denominator and now the 2 can distribute into the \[x^2-3x+2\]

OpenStudy (matheducatormcg):

your first fraction should be \[\frac{2x^2-6x+4}{x(x-1)(x-2)}\]

OpenStudy (matheducatormcg):

again, each fraction will have the same denominator. FOIL the numerators and combine like terms.

OpenStudy (anonymous):

ok thanks :)

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