X+y=1y-z=-2x²+y²+z²=21

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To solve this system of equations, we can start by rearranging the first two equations to solve for y in terms of x and z.

From the first equation, we have:

y = 1 - x

Substitute this value of y into the second equation:

1 - x - z = -2

Rearranging, we get:

x + z = 3

Now, substitute the value of y into the third equation:

x² + (1 - x)² + z² = 21
x² + 1 - 2x + x² + z² = 21
2x² - 2x + z² = 20

Now we have a system of two equations:

x + z = 3
2x² - 2x + z² = 20

From the first equation, we can rearrange to get z in terms of x:

z = 3 - x

Now substitute this into the second equation:

2x² - 2x + (3 - x)² = 20
2x² - 2x + 9 - 6x + x² = 20
3x² - 8x - 11 = 0

Now we have a quadratic equation that we can solve for x. Using the quadratic formula:

x = (8 ± √(8² - 43(-11))) / 6
x = (8 ± √(64 + 132)) / 6
x = (8 ± √196) / 6
x = (8 ± 14) / 6

This gives us two possible values for x:

x = (8 + 14) / 6 = 22 / 6 = 11 / 3
or
x = (8 - 14) / 6 = -6 / 6 = -1

For x = 11/3, we can find the corresponding z and y values:

z = 3 - 11/3 = 0
y = 1 - 11/3 = -8/3

Thus, one solution is x = 11/3, y = -8/3, z = 0.

For x = -1, we can find the corresponding z and y values:

z = 3 - (-1) = 4
y = 1 - (-1) = 2

So the other solution is x = -1, y = 2, z = 4.

Therefore, the solutions to the system of equations are (11/3, -8/3, 0) and (-1, 2, 4).