Solve the following system of equations, or show that no real solution exists. graphically confirm your answer. 3x²+3y²=6 ; x-y=3

No real solution exists.

Step-by-step 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!

b.my answer is b.

brainliest

is should be y = 5x

step-by-step explanation:

correct me if im wrong