X +y =6x2- y=14[tex]x + y = 6 \\ x2 - y = 14[/tex]

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вопрос

Solving these two equations simultaneously:

From the first equation, we can rewrite it as:

x = 6 - y

Substitute x into the second equation:

(6 - y)^2 - y = 14
36 - 12y + y^2 - y = 14
y^2 - 13y + 22 = 0

Now we can solve for y by using the quadratic formula:

y = [13 ± √(13^2 - 4122)] / 2
y = [13 ± √(169 - 88)] / 2
y = [13 ± √81] / 2
y = [13 ± 9] / 2

So, y = (13 + 9) / 2 = 11 or y = (13 - 9) / 2 = 2

Now, substitute the values of y back into x = 6 - y to find the corresponding values of x:

For y = 11:
x = 6 - 11 = -5

For y = 2:
x = 6 - 2 = 4

Therefore, the solutions to the system of equations are:
x = -5, y = 11
OR
x = 4, y = 2