https://file.aiquickdraw.com/mathbot/file/172571464004184bk5alw.jpg

The equation in the image is:

\[
\frac{2x}{x+1} + \frac{5}{2x} = 2
\]

Let's solve this step by step.

### Step 1: Eliminate the fractions by multiplying the entire equation by the least common denominator (LCD).
The denominators are \(x+1\) and \(2x\), so the LCD is \((x+1)(2x)\).

Multiply both sides of the equation by \((x+1)(2x)\):

\[
(x+1)(2x) \left(\frac{2x}{x+1} + \frac{5}{2x}\right) = 2(x+1)(2x)
\]

### Step 2: Simplify each term.
- The first term simplifies as:
  \[
  (x+1)(2x) \cdot \frac{2x}{x+1} = 2x \cdot 2x = 4x^2
  \]
- The second term simplifies as:
  \[
  (x+1)(2x) \cdot \frac{5}{2x} = (x+1) \cdot 5 = 5(x+1)
  \]
- The right-hand side simplifies as:
  \[
  2(x+1)(2x) = 4x(x+1) = 4x^2 + 4x
  \]

So the equation becomes:

\[
4x^2 + 5(x+1) = 4x^2 + 4x
\]

### Step 3: Expand and simplify.
Expand the second term:

\[
4x^2 + 5x + 5 = 4x^2 + 4x
\]

### Step 4: Subtract \(4x^2\) from both sides.

\[
5x + 5 = 4x
\]

### Step 5: Solve for \(x\).
Subtract \(4x\) from both sides:

\[
x + 5 = 0
\]

Subtract 5 from both sides:

\[
x = -5
\]

### Final answer:
The solution is \(x = -5\).

Would you like further details or explanations?

Here are some follow-up questions to consider:
1. What is the least common denominator, and why is it important in solving this equation?
2. How do you eliminate fractions in an equation like this?
3. Why did we subtract \(4x^2\) from both sides in step 4?
4. What happens if the value of \(x\) makes a denominator zero? Does that affect the solution?
5. Can this method be used for all rational equations?

**Tip:** Always check your solutions in the original equation to ensure they do not make any denominator zero, which would make the solution invalid.

Solve the equation: \frac{2x}{x+1} + \frac{5}{2x} = 2

Learn how to solve the rational equation \frac{2x}{x+1} + \frac{5}{2x} = 2 step-by-step. This problem covers key algebraic concepts such as rational equations, cross multiplication, and linear equations. Suitable for students in Grades 9-11.

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