"Simple solution to system of equationsx2+y+z=0y2+x+z=02z+3y+3x=0" was answered by our community

Sydney Stanley

Answered question

2022-04-25

Simple solution to system of equations

x^{2} + y + z = 0

y^{2} + x + z = 0

2 z + 3 y + 3 x = 0

Elliot Roy

Beginner2022-04-26Added 14 answers

Step 1
1)

x^{2} + y + z = 0

y^{2} + x + z = 0

2 z + 3 y + 3 x = 0

From (3), we have

z = - \frac{3}{2} (x + y)

Replace in (1) to obtain

y = 2 x^{2} - 3 x

This gives

z = 3 (x - 1) x

Replace now in (2) to obtain

4 x^{4} - 12 x^{3} + 6 x^{2} + 4 x = 2 (x - 2) x (2 x^{2} - 2 x - 1)

So the roots

x_{1} = 0 x_{2} = 2 x_{3, 4} = \frac{1}{2} (1 \pm \sqrt{3})

Spencer Murillo

Beginner2022-04-27Added 9 answers

Step 1
To begin with, add the first and second equations together.
Then subtract the third equation from them.
Hence we get that

x^{2} + y^{2} - 2 x - 2 y = 0 L o n g \leftrightarrow {(x - 1)}^{2} + {(y - 1)}^{2} = 2

Then we can make the change of variable

x = 1 + \sqrt{2} \cos (θ)

and

y = 1 + \sqrt{2} \sin (θ)

Consequently,

z = - 3 - \frac{3 \sqrt{2} \cos (θ) + 3 \sqrt{2} \sin (θ)}{2}

Now solve for

θ

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