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OpenStudy (anonymous):

1/(x+a)=[2/(x-1)]-[1/(x-a)] solve for x

OpenStudy (anonymous):

show steps please

myininaya (myininaya):

\[\frac{1}{x+a}=\frac{2}{x-1}-\frac{1}{x-a}\]

OpenStudy (anonymous):

let me know when you get \[x=a^2\]

myininaya (myininaya):

\[(x+a)(x-a)(x-1)\frac{1}{x+a}=(x+a)(x-a)(x-1)\frac{2}{x-1}-(x+a)(x-a)(x-1)\frac{1}{x-a}\]

myininaya (myininaya):

\[(x-a)(x-1)=(x+a)(x-a)(2)-(x+a)(x-1)\]

myininaya (myininaya):

\[(x-a)(x-1)=(x+a)(2(x-a)-(x-1))\]

myininaya (myininaya):

\[(x-a)(x-1)=(x+a)(2x-2a-x+1)\]

myininaya (myininaya):

\[(x-a)(x-1)=(x+a)(x-2a+1)\]

myininaya (myininaya):

\[x^2-x-ax+1=x(x-2a+1)+a(x-2a+1)\]

myininaya (myininaya):

\[x^2-x-ax+1=x^2-2ax+x+ax-2a^2+a\]

myininaya (myininaya):

put like terms together

myininaya (myininaya):

move everything to one side

myininaya (myininaya):

well actually the x^2's cancel so don't

OpenStudy (anonymous):

okay I got it from there thank you so much!!!

myininaya (myininaya):

make sure \[x \neq -a, x \neq 1, x \neq a \]

OpenStudy (anonymous):

ok

myininaya (myininaya):

let me know if you need further assistance :)

OpenStudy (anonymous):

Im lost how you went from (x+a)(2x-2a-x+1) to (x+a) (x-2a+1) where did the 2's and x go to

myininaya (myininaya):

2x-x=x

myininaya (myininaya):

this is was the last step i got to for him \[x^2-x-ax+1=x^2-2ax+x+ax-2a^2+a \]

myininaya (myininaya):

next step is to: put x terms on one side and everything without x on the other side

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