Question

Help please! Solve for x

Answer 1

can you factor the denominator of the first term?

Answer 2

like this\[x^2-2x+1 = (x-\dots)(x-\dots)\]

Answer 3

no...

Answer 4

then ...'s have to multiply to give +1, and add to give -2

Answer 5

Do you mean that I have to add and subtract numbers from both sides so I can cancel them out?

Answer 6

not yet, we have to factor the denominators first

Answer 7

what are the factors of 1?

Answer 8

1

Answer 9

right,

now we and multiply in the two ...'s together to get +1

so we could have
+1 x +1 = +1,
or
-1 x -1 = +1

Answer 10

if we add -1 and -1 we get -2 which is what we want.

Answer 11

We get:

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

Answer 12

We want -2 so we can get rid of -2x, right?

Answer 13

yeah.

If we expand it back out
\[(x-1)(x-1) = x(x-1)-1(x-1)\\
\qquad\qquad\qquad=x^2-x-x+1\\
\qquad\qquad\qquad=x^2-2x+1\\
\]
so we know we have factored correctly

Answer 14

Now we want to factor the denominator of the second term,
something like this
\[1-x^2 = (\dots)(\dots)\]

Answer 15

this is a difference of squares

Answer 16

\[a^2-b^2 = (a-b)(a+b)\]

if you expand it back out , you can check the middle term cancels away

Answer 17

Can you factor
\[1-x^2 = \]

Answer 18

(−x+1)(x+1)

Answer 19

cool,

Answer 20

so you have
\[\begin{align}
\frac{3}{x^2-2x+1}+\frac2{1-x^2}
&=\frac1{1+x}\\
\frac{3}{(x-1)(x-1)}+\frac2{(-x+1)(x+1)}
&=\frac1{1+x}\\
\frac{3}{(x-1)(x-1)}-\frac2{(x-1)(x+1)}
&=\frac1{1+x}\\
\end{align}\]
right?

Answer 21

Right.

Answer 22

Now, we want to get rid of all those denominators by multiplying.

multiply both sides of the equation by \((x-1)(x-1)(x+1)\)
(but don't expand any brackets)

Answer 23

3 - 2/x-1 = I have trouble with the right side....

Answer 24

\[\begin{align}
\frac{3}{(x-1)(x-1)}+\frac2{(x-1)(x+1)}
&=\frac1{1+x}\\[2ex]
\frac{3(x−1)(x−1)(x+1) }{(x-1)(x-1)}-\frac{2(x−1)(x−1)(x+1) }{(x-1)(x+1)}
&=\frac{(x−1)(x−1)(x+1) }{x+1}\\[2ex]
\frac{3\cancel{(x−1)(x−1)}(x+1)}{\cancel{(x-1)(x-1)}}-\frac{2(x−1)\cancel{(x−1)(x+1)}}{\cancel{(x-1)(x+1)}}
&=\frac{(x−1)(x−1)\cancel{(x+1)}}{\cancel{x+1}}\\[2ex]
3(x+1)-2(x-1)&=(x-1)(x-1)
\end{align}\]

Answer 25

Ohhhh....right XD

Answer 26

now i guess you can expand the brackets and solve from here

Answer 27

what do you get when you expand?

Answer 28

yep! That's all I need help with! Thanks a bunch!! (sorry for the late reply)

Answer 29

@UnkleRhaukus

Answer 30

do you need help

Answer 31

@LYRA_Z you still need help?

Answer 32

Nope, not anymore, thanks!