x-y=3 xy=-20
To solve the system of equations
- \( x - y = 3 \) (Equation 1)
- \( xy = -20 \) (Equation 2)
we can start by expressing one variable in terms of the other using Equation 1.
From Equation 1, we can express \( x \) in terms of \( y \):
\[ x = y + 3 \]
Next, we can substitute this expression for \( x \) into Equation 2:
\[ (y + 3)y = -20 \]
Expanding this, we get:
\[ y^2 + 3y + 20 = 0 \]
Now, we can rearrange it into standard quadratic form:
\[ y^2 + 3y + 20 = 0 \]
Next, we will use the quadratic formula to find the roots of this equation, where \( a = 1 \), \( b = 3 \), and \( c = 20 \):
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Calculating the discriminant:
\[ b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot 20 = 9 - 80 = -71 \]
Since the discriminant is negative, there are no real solutions for \( y \), and consequently, no real solutions for \( x \).
Therefore, the system of equations has no real solution. The solutions would be complex. If you need, I can help find the complex solutions as well. Would you like to proceed with that?
