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 \( (x, y) = (1, -1) \).