Note: Enter your answer and show all the steps you use to solve this problem in the space provided.

Solve the following equation. Show all your work.

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



Answer :

To solve the equation

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

we will follow these steps:

### Step 1: Simplify the equation by combining the fractions

First, we take the least common denominator (LCD) of the fractions [tex]\(\frac{x}{x-2}\)[/tex] and [tex]\(\frac{x-1}{x+1}\)[/tex], which is [tex]\((x-2)(x+1)\)[/tex]. Rewriting the given equation, we get:

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

### Step 2: Expand the numerators

Next, let's expand and simplify the numerator:

[tex]\[ x(x+1) = x^2 + x \][/tex]

[tex]\[ (x-1)(x-2) = x^2 - 3x + 2 \][/tex]

Adding these together:

[tex]\[ x^2 + x + x^2 - 3x + 2 = 2x^2 - 2x + 2 \][/tex]

So our equation now looks like:

[tex]\[ \frac{2x^2 - 2x + 2}{(x-2)(x+1)} = -1 \][/tex]

### Step 3: Clear the denominator by multiplying both sides by [tex]\((x-2)(x+1)\)[/tex]

Multiplying through by the denominator, we get:

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

### Step 4: Expand and simplify the right-hand side

Expanding [tex]\((x-2)(x+1)\)[/tex]:

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

So our equation becomes:

[tex]\[ 2x^2 - 2x + 2 = - (x^2 - x - 2) \][/tex]

Distribute the negative sign on the right-hand side:

[tex]\[ 2x^2 - 2x + 2 = -x^2 + x + 2 \][/tex]

### Step 5: Bring all terms to one side and combine like terms

Bringing all terms to the left-hand side:

[tex]\[ 2x^2 - 2x + 2 + x^2 - x - 2 = 0 \][/tex]

Combining like terms:

[tex]\[ (2x^2 + x^2) + (-2x - x) + (2 - 2) = 0 \][/tex]

[tex]\[ 3x^2 - 3x = 0 \][/tex]

### Step 6: Factor out the common term

Factor out [tex]\(3x\)[/tex]:

[tex]\[ 3x(x - 1) = 0 \][/tex]

### Step 7: Solve for [tex]\(x\)[/tex]

Set each factor to zero:

1. [tex]\(3x = 0\)[/tex]

[tex]\[ x = 0 \][/tex]

2. [tex]\(x - 1 = 0\)[/tex]

[tex]\[ x = 1 \][/tex]

Therefore, the solutions to the equation are:

[tex]\[ x = 0 \quad \text{and} \quad x = 1 \][/tex]