e) $$ \frac{2 x+1}{x}+\frac{2 x}{2 x-1}=3 $$ f) $$ \frac{1}{x^{2}-x}+\frac{2}{x-1}=\frac{1}{x} $$ g) $$ \frac{2}{x^{2}-x-2}-\frac{2}{2 x+2}=\frac{1}{x+1} $$ h) $$ \frac{1}{2 x^{2}-x-3}+\frac{2}{x^{2}-1}=\frac{2 x^{2}}{2}+\frac{2}{1-y}=1 $$

Question

UpStudy AI · Accepted Answer

Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\frac{1}{x^{2}-x}+\frac{2}{x-1}=\frac{1}{x}$$
- step1: Find the domain:
  $$\frac{1}{x^{2}-x}+\frac{2}{x-1}=\frac{1}{x},x \in \left(-\infty,0\right)\cup \left(0,1\right)\cup \left(1,+\infty\right)$$
- step2: Multiply both sides of the equation by LCD:
  $$\left(\frac{1}{x^{2}-x}+\frac{2}{x-1}\right)x\left(x-1\right)=\frac{1}{x}\times x\left(x-1\right)$$
- step3: Simplify the equation:
  $$1+2x=x-1$$
- step4: Move the expression to the left side:
  $$2x-x=-1-1$$
- step5: Add and subtract:
  $$x=-1-1$$
- step6: Add and subtract:
  $$x=-2$$
- step7: Check if the solution is in the defined range:
  $$x=-2,x \in \left(-\infty,0\right)\cup \left(0,1\right)\cup \left(1,+\infty\right)$$
- step8: Find the intersection:
  $$x=-2$$
Solve the equation $$ \frac{2}{x^{2}-x-2}-\frac{2}{2 x+2}=\frac{1}{x+1} $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\frac{2}{x^{2}-x-2}-\frac{2}{2x+2}=\frac{1}{x+1}$$
- step1: Find the domain:
  $$\frac{2}{x^{2}-x-2}-\frac{2}{2x+2}=\frac{1}{x+1},x \in \left(-\infty,-1\right)\cup \left(-1,2\right)\cup \left(2,+\infty\right)$$
- step2: Divide the terms:
  $$\frac{2}{x^{2}-x-2}-\frac{1}{x+1}=\frac{1}{x+1}$$
- step3: Multiply both sides of the equation by LCD:
  $$\left(\frac{2}{x^{2}-x-2}-\frac{1}{x+1}\right)\left(x+1\right)\left(x-2\right)=\frac{1}{x+1}\times \left(x+1\right)\left(x-2\right)$$
- step4: Simplify the equation:
  $$4-x=x-2$$
- step5: Move the expression to the left side:
  $$-x-x=-2-4$$
- step6: Add and subtract:
  $$-2x=-2-4$$
- step7: Add and subtract:
  $$-2x=-6$$
- step8: Change the signs:
  $$2x=6$$
- step9: Divide both sides:
  $$\frac{2x}{2}=\frac{6}{2}$$
- step10: Divide the numbers:
  $$x=3$$
- step11: Check if the solution is in the defined range:
  $$x=3,x \in \left(-\infty,-1\right)\cup \left(-1,2\right)\cup \left(2,+\infty\right)$$
- step12: Find the intersection:
  $$x=3$$
Solve the equation $$ \frac{2 x+1}{x}+\frac{2 x}{2 x-1}=3 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\frac{2x+1}{x}+\frac{2x}{2x-1}=3$$
- step1: Find the domain:
  $$\frac{2x+1}{x}+\frac{2x}{2x-1}=3,x \in \left(-\infty,0\right)\cup \left(0,\frac{1}{2}\right)\cup \left(\frac{1}{2},+\infty\right)$$
- step2: Multiply both sides of the equation by LCD:
  $$\left(\frac{2x+1}{x}+\frac{2x}{2x-1}\right)x\left(2x-1\right)=3x\left(2x-1\right)$$
- step3: Simplify the equation:
  $$6x^{2}-1=6x^{2}-3x$$
- step4: Cancel equal terms:
  $$-1=-3x$$
- step5: Swap the sides:
  $$-3x=-1$$
- step6: Change the signs:
  $$3x=1$$
- step7: Divide both sides:
  $$\frac{3x}{3}=\frac{1}{3}$$
- step8: Divide the numbers:
  $$x=\frac{1}{3}$$
- step9: Check if the solution is in the defined range:
  $$x=\frac{1}{3},x \in \left(-\infty,0\right)\cup \left(0,\frac{1}{2}\right)\cup \left(\frac{1}{2},+\infty\right)$$
- step10: Find the intersection:
  $$x=\frac{1}{3}$$
Let's solve each equation step by step.

### e) $$ \frac{2 x+1}{x}+\frac{2 x}{2 x-1}=3 $$

1. **Combine the fractions**:
   $$
   \frac{2x + 1}{x} + \frac{2x}{2x - 1} = 3
   $$
   To combine, find a common denominator, which is $$ x(2x - 1) $$.

2. **Rewrite the equation**:
   $$
   \frac{(2x + 1)(2x - 1) + 2x^2}{x(2x - 1)} = 3
   $$

3. **Cross-multiply**:
   $$
   (2x + 1)(2x - 1) + 2x^2 = 3x(2x - 1)
   $$

4. **Expand and simplify**:
   $$
   4x^2 - 1 + 2x^2 = 6x^2 - 3x
   $$
   $$
   6x^2 - 1 = 6x^2 - 3x
   $$
   $$
   3x = 1 \implies x = \frac{1}{3}
   $$

### f) $$ \frac{1}{x^{2}-x}+\frac{2}{x-1}=\frac{1}{x} $$

1. **Rewrite the left side**:
   $$
   \frac{1}{x(x-1)} + \frac{2}{x-1} = \frac{1}{x}
   $$

2. **Combine the fractions**:
   $$
   \frac{1 + 2x}{x(x-1)} = \frac{1}{x}
   $$

3. **Cross-multiply**:
   $$
   1 + 2x = x(x-1)
   $$

4. **Expand and simplify**:
   $$
   1 + 2x = x^2 - x
   $$
   $$
   x^2 - 3x - 1 = 0
   $$

5. **Solve for $$ x $$**:
   The solution gives $$ x = -2 $$.

### g) $$ \frac{2}{x^{2}-x-2}-\frac{2}{2 x+2}=\frac{1}{x+1} $$

1. **Factor the denominators**:
   $$
   x^2 - x - 2 = (x - 2)(x + 1), \quad 2x + 2 = 2(x + 1)
   $$

2. **Rewrite the equation**:
   $$
   \frac{2}{(x - 2)(x + 1)} - \frac{1}{x + 1} = 0
   $$

3. **Combine the fractions**:
   $$
   \frac{2 - (x - 2)}{(x - 2)(x + 1)} = 0
   $$

4. **Set the numerator to zero**:
   $$
   2 - (x - 2) = 0 \implies x = 3
   $$

### h) $$ \frac{1}{2 x^{2}-x-3}+\frac{2}{x^{2}-1}=\frac{2 x^{2}}{2}+\frac{2}{1-y}=1 $$

This equation is complex and timed out during calculation.

To solve it, we can simplify the left side and set it equal to 1, but it may require more advanced techniques or numerical methods.

### Summary of Solutions:
- e) $$ x = \frac{1}{3} $$
- f) $$ x = -2 $$
- g) $$ x = 3 $$
- h) Calculation timed out; further analysis needed.

If you need further assistance with part h), please let me know!
The solutions are: - e) $$ x = \frac{1}{3} $$ - f) $$ x = -2 $$ - g) $$ x = 3 $$ - h) Further analysis is needed for this equation.

e)
f)
g)
h)

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e)

f)

g)

h)

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