9х-13у=22 2х+3у=-1

To solve the system of equations:

1. \( 9x - 13y = 22 \) (Equation 1)
2. \( 2x + 3y = -1 \) (Equation 2)

We can use the method of substitution or elimination. Here, I will use the elimination method.

First, let's eliminate one of the variables. We can do this by manipulating the equations.

To eliminate \( y \), we can multiply Equation 2 by 13 to match the coefficient of \( y \) in Equation 1:

\[ 13(2x + 3y) = 13(-1) \]

This gives us:

\[ 26x + 39y = -13 \quad \text{(Equation 3)} \]

Now we have:

1. \( 9x - 13y = 22 \) (Equation 1)
2. \( 26x + 39y = -13 \) (Equation 3)

Next, we can manipulate Equation 1 to make it easier to combine with Equation 3. Let's multiply Equation 1 by 3:

\[ 3(9x - 13y) = 3(22) \]

This gives us:

\[ 27x - 39y = 66 \quad \text{(Equation 4)} \]

Now, we have:

1. \( 27x - 39y = 66 \) (Equation 4)
2. \( 26x + 39y = -13 \) (Equation 3)

Now we can add Equation 4 and Equation 3 together:

\[ (27x - 39y) + (26x + 39y) = 66 - 13 \]

This simplifies to:

\[ 53x = 53 \]

Dividing both sides by \( 53 \):

\[ x = 1 \]

Now that we have \( x \), we can substitute it back into one of the original equations to find \( y \). Let’s use Equation 2:

\[ 2(1) + 3y = -1 \]

This simplifies to:

\[ 2 + 3y = -1 \]

Subtracting \( 2 \) from both sides:

\[ 3y = -3 \]

Dividing by \( 3 \):

\[ y = -1 \]

So the solution to the system of equations is:

\[ x = 1, \quad y = -1 \]

In conclusion, the solution is \( (x, y) = (1, -1) \).