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

Answer:

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

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

Step-by-step explanation:

$\frac{x}{x-2}+\frac{x-1}{x+1}=-1$

First step: Notice the domain of the equation.

What values can $x$ definitely not take on.

$x \neq 2 \text{ or } -1$ 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:

$(x-2)(x+1)\frac{x}{x-2}+(x-2)(x+1)\frac{x-1}{x+1}=(x-2)(x+1)(-1)$

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

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.

$x=0$?

$\frac{x}{x-2}+\frac{x-1}{x+1}=-1$ with $x=0$:

$\frac{0}{0-2}+\frac{0-1}{0+1}=-1$

$\frac{0}{-2}+\frac{-1}{1}=-1$

$0+-1=-1$

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

$x=1$?

$\frac{x}{x-2}+\frac{x-1}{x+1}=-1$ with $x=1$:

$\frac{1}{1-2}+\frac{1-1}{1+1}=-1$

$\frac{1}{-1}+\frac{0}{1}=-1$

$-1+0=-1$

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