2x+y=0 1/x + 9/y-1=0

To solve the system of equations:

1. \( 2x + y = 0 \)
2. \( \frac{1}{x} + \frac{9}{y} - 1 = 0 \)

First, we can solve the first equation for \( y \):

\[ y = -2x \]

Next, we substitute \( y \) into the second equation:

\[ \frac{1}{x} + \frac{9}{-2x} - 1 = 0 \]

This simplifies to:

\[ \frac{1}{x} - \frac{9}{2x} - 1 = 0 \]

Combining the fraction terms:

\[ \frac{1 - \frac{9}{2}}{x} - 1 = 0 \]

\[ \frac{2 - 9}{2x} - 1 = 0 \]

\[ \frac{-7}{2x} - 1 = 0 \]

Now we can isolate the fraction:

\[ \frac{-7}{2x} = 1 \]

Cross-multiplying gives:

\[ -7 = 2x \]

Solving for \( x \):

\[ x = -\frac{7}{2} \]

Now, substitute \( x \) back into the equation for \( y \):

\[ y = -2\left(-\frac{7}{2}\right) = 7 \]

So the solution to the system of equations is:

\[ x = -\frac{7}{2}, \quad y = 7 \]

In summary, the solution to the original equations is:

\[ (x, y) = \left(-\frac{7}{2}, 7\right) \]