Question

Algebraic fractions please help. stuck half way

Answer 1

mari try posting your question please

Answer 2

\[\frac{ x+y }{ x-y }+\frac{ xy+x^2-y }{ (y-x)(y+x) }\]

Answer 3

do you know how assume two fractions ?

Answer 4

do you know what mean common denominator ?

Answer 5

Yes. so do something like this?

\[\frac{ x+y }{ x-y }\times \frac{ (y-x)(y+x) }{ (y-x)(y+x) }\]

Answer 6

that is to the first term

Answer 7

no sorry

Answer 8

Not sure about assuming two fractions @jhonyy9

Answer 9

ok.like an example

1 2
--- + ---- = ?
2 3

Answer 10

\[\mathrm{Find\:the\:LCD\:for}\:\frac{x+y}{x-y}+\frac{x^2+xy-y}{\left(y-x\right)\left(x+y\right)}:\quad \left(x-y\right)\left(x+y\right)\]
\[\mathrm{Adjust\:Fractions\:based\:on\:the\:LCD}\]
\[=\frac{\left(x+y\right)\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}-\frac{\left(xy+x^2-y\right)}{\left(x-y\right)\left(x+y\right)}\]
\[\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}\]
\[=\frac{\left(x+y\right)\left(x+y\right)-\left(x^2+xy-y\right)}{\left(x-y\right)\left(x+y\right)}\]
\[\left(x+y\right)\left(x+y\right)-\left(x^2+xy-y\right)=\left(x+y\right)^2-\left(x^2+xy-y\right)\]
\[=\frac{\left(x+y\right)^2-\left(x^2+xy-y\right)}{\left(x-y\right)\left(x+y\right)}\]
\[\mathrm{Expand}\:\left(x+y\right)^2-\left(x^2+xy-y\right):\quad xy+y^2+y\]
\[=\frac{xy+y^2+y}{\left(x-y\right)\left(x+y\right)}\]

Answer 11

@DecentNabeel can you please tell me how you went about getting the LCD

Answer 12

@DecentNabeel nice job - GREAT !!! - ty

Answer 13

@marigirl
\[\mathrm{Find\:the\:least\:common\:denominator\:}\]
\[\left(x-y\right)\left(x+y\right)\]

Answer 14

@jhonyy9 thanks :)

Answer 15

yw.