X^2+xy-y^2=11 X-2y=1
To solve the system of equations:
- \( x^2 + xy - y^2 = 11 \)
- \( x - 2y = 1 \)
we can start by expressing \(x\) in terms of \(y\) from the second equation:
\[ x = 2y + 1 \]
Now we can substitute this expression for \(x\) into the first equation.
Substituting \(x\) gives us:
\[ (2y + 1)^2 + (2y + 1)y - y^2 = 11 \]
Expanding the terms:
\[ (4y^2 + 4y + 1) + (2y^2 + y) - y^2 = 11 \]
Now combine like terms:
\[ 4y^2 + 2y^2 - y^2 + 4y + y + 1 = 11 \]
This simplifies to:
\[ 5y^2 + 5y + 1 = 11 \]
Subtracting 11 from both sides gives:
\[ 5y^2 + 5y + 1 - 11 = 0 \]
which simplifies to:
\[ 5y^2 + 5y - 10 = 0 \]
To simplify, divide the entire equation by 5:
\[ y^2 + y - 2 = 0 \]
Now we can factor this quadratic equation:
\[ (y + 2)(y - 1) = 0 \]
Setting each factor to zero gives us the solutions for \(y\):
\[ y + 2 = 0 \quad \Rightarrow \quad y = -2 \]
\[ y - 1 = 0 \quad \Rightarrow \quad y = 1 \]
Now we will find the corresponding \(x\) values for each \(y\):
- For \(y = -2\):
\[ x = 2(-2) + 1 = -4 + 1 = -3 \]
Thus, one solution is \((-3, -2)\).
- For \(y = 1\):
\[ x = 2(1) + 1 = 2 + 1 = 3 \]
Thus, the second solution is \((3, 1)\).
Finally, the solutions to the system of equations are:
\[ (-3, -2) \quad \text{and} \quad (3, 1) \]
