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?
