Solve the following equation for $$ x $$ $$ -\frac{3}{x-1}+\frac{5}{x-3}=-\frac{1}{x^{2}-4 x+3} $$

Question

UpStudy AI · Accepted Answer

To solve the equation:

$$
-\frac{3}{x-1} + \frac{5}{x-3} = -\frac{1}{x^{2} - 4x + 3}
$$

**Step 1: Factor the Denominator on the Right Side**

The denominator on the right side can be factored:

$$
x^{2} - 4x + 3 = (x - 1)(x - 3)
$$

So, the equation becomes:

$$
-\frac{3}{x-1} + \frac{5}{x-3} = -\frac{1}{(x-1)(x-3)}
$$

**Step 2: Multiply Both Sides by $$(x-1)(x-3)$$ to Eliminate the Denominators**

Assuming $$x \neq 1$$ and $$x \neq 3$$ (to avoid division by zero), multiply each term by $$(x-1)(x-3)$$:

$$
-3(x - 3) + 5(x - 1) = -1
$$

**Step 3: Expand and Simplify**

Expand the left side:

$$
-3x + 9 + 5x - 5 = -1
$$

Combine like terms:

$$
( -3x + 5x ) + ( 9 - 5 ) = -1
$$

$$
2x + 4 = -1
$$

**Step 4: Solve for $$x$$**

Subtract 4 from both sides:

$$
2x = -5
$$

Divide by 2:

$$
x = -\frac{5}{2}
$$

**Step 5: Verify the Solution**

Ensure that $$x = -\frac{5}{2}$$ does not make any denominator zero:

$$
x \neq 1, \quad x \neq 3
$$

Since $$-\frac{5}{2}$$ is neither 1 nor 3, it is a valid solution.

**Final Answer:**

$$
x = -\frac{5}{2}
$$
$$ x = -\frac{5}{2} $$

Solve the following equation for \( x \) \[ -\frac{3}{x-1}+\frac{5}{x-3}=-\frac{1}{x^{2}-4 x+3} \]