Question

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

Answer 1

show steps please

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

put like terms together

Answer 12

move everything to one side

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

ok

Answer 17

let me know if you need further assistance :)

Answer 18

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

Answer 19

2x-x=x

Answer 20

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

Answer 21

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