Solve the equation. $$ 2=\frac{7}{2 x-1}+\frac{-3}{(2 x-1)^{2}} $$

Question

UpStudy AI · Accepted Answer

To solve the equation:

$$
2 = \frac{7}{2x - 1} + \frac{-3}{(2x - 1)^2}
$$

**Step 1: Combine the Fractions**

First, combine the terms on the right side:

$$
2 = \frac{7(2x - 1) - 3}{(2x - 1)^2}
$$

Simplify the numerator:

$$
2 = \frac{14x - 7 - 3}{(2x - 1)^2} = \frac{14x - 10}{(2x - 1)^2}
$$

**Step 2: Clear the Denominator**

Multiply both sides by $$(2x - 1)^2$$ to eliminate the denominator:

$$
2(2x - 1)^2 = 14x - 10
$$

**Step 3: Expand and Simplify**

Expand $$(2x - 1)^2$$:

$$
(2x - 1)^2 = 4x^2 - 4x + 1
$$

Substitute back into the equation:

$$
2(4x^2 - 4x + 1) = 14x - 10
$$

Expand the left side:

$$
8x^2 - 8x + 2 = 14x - 10
$$

Move all terms to one side:

$$
8x^2 - 22x + 12 = 0
$$

**Step 4: Solve the Quadratic Equation**

Divide the entire equation by 2 for simplicity:

$$
4x^2 - 11x + 6 = 0
$$

Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$, where $$a = 4$$, $$b = -11$$, and $$c = 6$$:

$$
x = \frac{11 \pm \sqrt{121 - 96}}{8} = \frac{11 \pm \sqrt{25}}{8} = \frac{11 \pm 5}{8}
$$

This gives two solutions:

$$
x = \frac{16}{8} = 2 \quad \text{and} \quad x = \frac{6}{8} = \frac{3}{4}
$$

**Step 5: Verify the Solutions**

Ensure that neither solution makes the denominator zero:

- For $$x = 2$$: $$2(2) - 1 = 3 \neq 0$$
- For $$x = \frac{3}{4}$$: $$2\left(\frac{3}{4}\right) - 1 = \frac{1}{2} \neq 0$$

Both solutions are valid.

**Final Answer:**

Both real solutions are $$x = 2$$ and $$x = \dfrac{3}{4}$$.
The solutions are $$x = 2$$ and $$x = \frac{3}{4}$$.

Solve the equation. \[ 2=\frac{7}{2 x-1}+\frac{-3}{(2 x-1)^{2}} \]