Solve the following equations :
`(1)/(x-1)-(1)/(x)=(1)/(x+3)-(1)/(x+4)`

Options

A. `-1(1)/(6)`
B. `-1(1)/(2)`
C. `-1(1)/(3)`
D. `-1(7)/(19)`

Answer

To solve the equation

\[ \frac{1}{x-1} - \frac{1}{x} = \frac{1}{x+3} - \frac{1}{x+4} \]

we will follow these steps:

### Step 1: Find the LCM of the fractions on both sides.

**Left-Hand Side (LHS):**
The LCM of \(x\) and \(x-1\) is \(x(x-1)\).

**Right-Hand Side (RHS):**
The LCM of \(x+3\) and \(x+4\) is \((x+3)(x+4)\).

### Step 2: Rewrite the equation using the LCM.

Rewriting the LHS:

\[ \frac{1 \cdot x}{x(x-1)} - \frac{1 \cdot (x-1)}{x(x-1)} = \frac{x - (x-1)}{x(x-1)} = \frac{1}{x(x-1)} \]

Rewriting the RHS:

\[ \frac{1 \cdot (x+4)}{(x+3)(x+4)} - \frac{1 \cdot (x+3)}{(x+3)(x+4)} = \frac{(x+4) - (x+3)}{(x+3)(x+4)} = \frac{1}{(x+3)(x+4)} \]

### Step 3: Set the rewritten LHS equal to the rewritten RHS.

Now we have:

\[ \frac{1}{x(x-1)} = \frac{1}{(x+3)(x+4)} \]

### Step 4: Cross-multiply to eliminate the fractions.

Cross-multiplying gives us:

\[ 1 \cdot (x+3)(x+4) = 1 \cdot x(x-1) \]

This simplifies to:

\[ (x+3)(x+4) = x(x-1) \]

### Step 5: Expand both sides.

Expanding the LHS:

\[ x^2 + 4x + 3x + 12 = x^2 + 7x + 12 \]

Expanding the RHS:

\[ x^2 - x \]

### Step 6: Set the equation to zero.

Now we set the equation:

\[ x^2 + 7x + 12 = x^2 - x \]

Subtract \(x^2\) from both sides:

\[ 7x + 12 = -x \]

### Step 7: Combine like terms.

Adding \(x\) to both sides gives:

\[ 7x + x + 12 = 0 \implies 8x + 12 = 0 \]

### Step 8: Solve for \(x\).

Subtracting 12 from both sides:

\[ 8x = -12 \]

Dividing by 8:

\[ x = -\frac{12}{8} = -\frac{3}{2} \]

### Final Answer:

The solution to the equation is

\[ x = -\frac{3}{2} \]

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