Let's solve each of these equations step by step.

### 1) \( 9x + 6x = -1 \)

First, combine the like terms on the left side:
\[
9x + 6x = 15x
\]
So the equation becomes:
\[
15x = -1
\]

Now, solve for \( x \) by dividing both sides by 15:
\[
x = \frac{-1}{15}
\]

Thus, the solution for the first equation is:
\[
x = \frac{-1}{15}
\]

### 2) \( \frac{6}{x} - 2x = 9 \)

To solve this equation, first eliminate the fraction by multiplying every term by \( x \) (assuming \( x \neq 0 \)):
\[
6 - 2x^2 = 9x
\]

Now, rearrange the equation to form a standard quadratic equation:
\[
2x^2 + 9x - 6 = 0
\]

Next, solve this quadratic equation using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 2 \), \( b = 9 \), and \( c = -6 \).

Substituting these values into the formula:
\[
x = \frac{-9 \pm \sqrt{9^2 - 4(2)(-6)}}{4}
\]
\[
x = \frac{-9 \pm \sqrt{81 + 48}}{4}
\]
\[
x = \frac{-9 \pm \sqrt{129}}{4}
\]

Therefore, the solutions for the second equation are:
\[
x = \frac{-9 + \sqrt{129}}{4} \quad \text{or} \quad x = \frac{-9 - \sqrt{129}}{4}
\]

### Summary
- The solution for the first equation is \( x = \frac{-1}{15} \).
- The solutions for the second equation are \( x = \frac{-9 + \sqrt{129}}{4} \) or \( x = \frac{-9 - \sqrt{129}}{4} \).

Would you like any further details or have any questions?

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**Related Questions:**
1. How do you simplify and solve linear equations with one variable?
2. What are the steps to solve quadratic equations using the quadratic formula?
3. How do you factorize quadratic equations to find their solutions?
4. What is the discriminant, and how does it determine the nature of the roots?
5. Can the quadratic formula be derived from completing the square?

**Tip:** When solving quadratic equations, always check the discriminant (\(b^2 - 4ac\)) first to determine if the solutions are real and distinct, real and equal, or complex.

Learn how to solve two equations involving linear and quadratic forms with detailed step-by-step solutions. Covering concepts such as linear equation solving and quadratic formula applications. Suitable for high school students.

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