X+y=4x^2-2xy+y^2=64

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вопрос

To solve the system of equations x + y = 4 and x^2 - 2xy + y^2 = 64, we can first substitute y = 4 - x into the second equation:

x^2 - 2x(4 - x) + (4 - x)^2 = 64
x^2 - 8x + 2x^2 - 8x + 16 = 64
3x^2 - 16x + 16 = 64
3x^2 - 16x - 48 = 0
(3x + 8)(x - 6) = 0

This gives us two possible solutions for x: x = -8/3 or x = 6.

Substitute each of these x values back into the first equation x + y = 4:

For x = -8/3, -8/3 + y = 4
y = 4 + 8/3
y = 12/3 + 8/3
y = 20/3

So, one solution is x = -8/3 and y = 20/3.

For x = 6, 6 + y = 4
y = 4 - 6
y = -2

The other solution is x = 6 and y = -2.

Therefore, the two solutions to the system of equations are (-8/3, 20/3) and (6, -2).