Answer:
No real solution exists.
Explanation:
Hi there!
We have the following system of equations:
3x² + 3y² = 6
x-y = 3
Let´s take the second equation and rewrite it:
x-y = 3
-y = 3 -x
y = x - 3
Now, let´s replace y in the first equation and solve for x:
3x² + 3(x-3)² = 6
divide by 3 both sides of the equation
x² + (x-3)² = 2
x² + (x-3)(x-3) = 2
x² + x² -6x + 9 = 2
2x² - 6x + 9 = 2
subtract 2 from both sides
2x² - 6x +7 = 0
Now, let´s apply the quadratic formula to solve this quadratic equation:
(-b±√(b² - 4ac))/2a
a = 2
b = -6
c = 7
(6±√(36 - 4·2·7))/ 4
Since the term inside the root is negative (36 - 4·2·7 = -20), the quadratic equation has no real solution.
Then, the system has no real solution.
Graphically, you can see that both curves don´t intersect.
Have a nice day!