

For which k , x=-2 value is one of the roots of the equation?

the equation is 3/2x-1+5/k=9x/2x+1




Let's solve the equation:

\[
\frac{3}{2x - 1} + \frac{5}{k} = \frac{9x}{2x + 1}
\]

We want to find the value of \( k \) for which \( x = -2 \) is a root.

### Step 1: Substitute \( x = -2 \) into the equation.

Substituting \( x = -2 \):

\[
\frac{3}{2(-2) - 1} + \frac{5}{k} = \frac{9(-2)}{2(-2) + 1}
\]

Simplify each part:

- \( 2(-2) - 1 = -4 - 1 = -5 \)
- \( 2(-2) + 1 = -4 + 1 = -3 \)
- \( 9(-2) = -18 \)

So, the equation becomes:

\[
\frac{3}{-5} + \frac{5}{k} = \frac{-18}{-3}
\]

### Step 2: Simplify the equation.

\[
-\frac{3}{5} + \frac{5}{k} = 6
\]

### Step 3: Solve for \( k \).

To isolate \( \frac{5}{k} \), add \( \frac{3}{5} \) to both sides:

\[
\frac{5}{k} = 6 + \frac{3}{5}
\]

Convert 6 to a fraction with a denominator of 5:

\[
\frac{5}{k} = \frac{30}{5} + \frac{3}{5} = \frac{33}{5}
\]

Now, solve for \( k \):

\[
\frac{5}{k} = \frac{33}{5}
\]

To find \( k \), take the reciprocal of both sides:

\[
k = \frac{5 \times 5}{33} = \frac{25}{33}
\]

### Final Answer

The value of \( k \) for which \( x = -2 \) is a root of the equation is \( k = \frac{25}{33} \).

Would you like more details or have any questions?

### Related Questions:
1. What happens if \( x = 2 \) is substituted into the equation?
2. How would the solution change if the equation were slightly different, e.g., with a different constant on the right side?
3. How can we verify that \( k = \frac{25}{33} \) is indeed the correct solution?
4. What if the equation had additional terms, how would that affect finding \( k \)?
5. Can you describe how the function behaves around \( x = -2 \)?

**Tip:** When solving equations with fractions, it's often helpful to clear the fractions by multiplying through by the least common denominator (LCD) early in the process.

Learn how to find the value of k for which x = -2 is a root of the equation involving fractions. Detailed step-by-step solution with equation manipulation and fraction operations explained. Suitable for high school students.

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