X-2y=1 xy+y=12

To solve the system of equations, we have:

1. \( x - 2y = 1 \) (Equation 1)
2. \( xy + y = 12 \) (Equation 2)

First, we can express \( x \) from Equation 1:

\[ x = 2y + 1 \]

Now we can substitute this expression for \( x \) into Equation 2:

\[ (2y + 1)y + y = 12 \]

This simplifies to:

\[ 2y^2 + y + y = 12 \]

Combining the \( y \) terms gives us:

\[ 2y^2 + 2y = 12 \]

We can simplify this by dividing the entire equation by 2:

\[ y^2 + y - 6 = 0 \]

Next, we will factor the quadratic equation:

\[ (y + 3)(y - 2) = 0 \]

Setting each factor to zero gives us the possible values for \( y \):

1. \( y + 3 = 0 \) → \( y = -3 \)
2. \( y - 2 = 0 \) → \( y = 2 \)

Now we will find the corresponding \( x \) values for each \( y \):

1. For \( y = -3 \): \[ x = 2(-3) + 1 = -6 + 1 = -5 \] So, one solution is \( (x, y) = (-5, -3) \).

2. For \( y = 2 \): \[ x = 2(2) + 1 = 4 + 1 = 5 \] So, another solution is \( (x, y) = (5, 2) \).

Therefore, the system of equations has two solutions:

1. \( (x, y) = (-5, -3) \)
2. \( (x, y) = (5, 2) \)