Question

plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz hellllllllllllllllllllllllppppppppppppppp meeeeeeeeeeeeeeeeeeeeeeeeeee. this is my third time asking for help. please show all your work so i undersatnd how you did it alng with the answer

Answer 1

The solutions are
`x=0 \text{ or } x=1`.

I had no extraneous solutions (solutions that were contradictory).

Explanation:

`(x)/(x-2)+(x-1)/(x+1)=-1`

First step: Notice the domain of the equation.

What values can
`x` definitely not take on.

`x \\eq 2 \text{ or } -1` since that would make the denominators of
`x-2` and
`x+1` zero.

That is
`x-2=0` when
`x=2` (added 2 on both sides).

That is
`x+1=0` when
`x=-1` (subtracted 1 on both sides).

So neither one of those can be solutions to the equation.

Second step: Multiply both sides by the least common multiple of the denominators, or a common multiple which happens to be the product of the denominators.

So I'm multiplying both sides by
`(x-2)(x+1)`.

This gives me:

`(x-2)(x+1)(x)/(x-2)+(x-2)(x+1)(x-1)/(x+1)=(x-2)(x+1)(-1)`

`(x+1)(x)+(x-2)(x-1)=-1(x-2)(x+1)`

Third step: We are going to do some distributive property here.

`(x^2+x)+(x^2-2x-1x+2)=-1(x^2-2x+1x-2)`

`(x^2+x)+(x^2-2x-1x+2)=-x^2+2x-1x+2`

Fourth step: Combine all like terms on left hand side and then do the right hand side.

`2x^2-2x+2=-x^2+x+2`

Fifth step: Move everything to one side so one of the sides is just equal to 0.

Subtract 2 on both sides:

`2x^2-2x+2-2=-x^2+x+2-2`

Simplify:

`2x^2-2x=-x^2+x`

Subtract
`x` on both sides:

`2x^2-2x-x=-x^2+x-x`

Simplify:

`2x^2-3x=-x^2`

Add
`x^2` on both sides:

`2x^2-3x+x^2=-x^2+x^2`

Simplify:

`3x^2-3x=0`

Sixth step: I'm going to factor and set any factors equal to right hand side,0. I will then solve those equations.

Factoring the left hand side:

I notice the left hand side has terms that each contain a factor of
`3x` so I will factor that common factor out:

`3x(x-1)=0`

This gives us either:
`3x=0` or
`x-1=0`

The first equation can be solved by dividing both sides by 3 giving us
`x=0`.

The second equation can be solved by adding 1 on both sides giving us
`x=1`.

These do not contradict the
`x`'s we said
`x` could definitely not be so these do appear to be our solutions if we have made no mistake.

Just to verify I'm going to add one more step.

Seventh step: Let's verify our answer.

`x=0`?

`(x)/(x-2)+(x-1)/(x+1)=-1` with
`x=0`:

`(0)/(0-2)+(0-1)/(0+1)=-1`

`(0)/(-2)+(-1)/(1)=-1`

`0+-1=-1`

`-1=-1` is a true equation so we have verified
`x=0` is a solution.

`x=1`?

`(x)/(x-2)+(x-1)/(x+1)=-1` with
`x=1`:

`(1)/(1-2)+(1-1)/(1+1)=-1`

`(1)/(-1)+(0)/(1)=-1`

`-1+0=-1`

`-1=-1` is a true equation so we have verified
`x=1` is a solution.