x^2+y^2=10 y=x-2
To solve the system of equations given by \( x^2 + y^2 = 10 \) and \( y = x - 2 \), we can substitute the second equation into the first.
Substitute \( y \) into the first equation: \[ x^2 + (x - 2)^2 = 10 \]
Expand the equation: \[ x^2 + (x^2 - 4x + 4) = 10 \] \[ 2x^2 - 4x + 4 = 10 \]
Rearrange the equation: \[ 2x^2 - 4x + 4 - 10 = 0 \] \[ 2x^2 - 4x - 6 = 0 \]
Divide the entire equation by 2: \[ x^2 - 2x - 3 = 0 \]
Factor the quadratic: \[ (x - 3)(x + 1) = 0 \]
Find the values of \( x \): \[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \] \[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]
Find the corresponding values of \( y \) using \( y = x - 2 \):
For \( x = 3 \): \[ y = 3 - 2 = 1 \] So, one solution is \( (3, 1) \).
For \( x = -1 \): \[ y = -1 - 2 = -3 \] So, another solution is \( (-1, -3) \).
The solutions to the system of equations are: \[ (3, 1) \quad \text{and} \quad (-1, -3) \]
