Sydney Stanley

2022-04-25

Simple solution to system of equations

${x}^{2}+y+z=0$

${y}^{2}+x+z=0$

$2z+3y+3x=0$

Elliot Roy

Beginner2022-04-26Added 14 answers

Step 1

1)${x}^{2}+y+z=0$

2)${y}^{2}+x+z=0$

3)$2z+3y+3x=0$

From (3), we have

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

Replace in (1) to obtain

$y=2{x}^{2}-3x$

This gives

$z=3(x-1)x$

Replace now in (2) to obtain

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

So the roots

${x}_{1}=0{\textstyle \phantom{\rule{2em}{0ex}}}{x}_{2}=2{\textstyle \phantom{\rule{2em}{0ex}}}{x}_{3,4}=\frac{1}{2}(1\pm \sqrt{3})$

1)

2)

3)

From (3), we have

Replace in (1) to obtain

This gives

Replace now in (2) to obtain

So the roots

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

${x}^{2}+{y}^{2}-2x-2y=0Long\leftrightarrow {(x-1)}^{2}+{(y-1)}^{2}=2$

Then we can make the change of variable

$x=1+\sqrt{2}\mathrm{cos}\left(\theta \right)$ and $y=1+\sqrt{2}\mathrm{sin}\left(\theta \right)$

Consequently,

$z=-3-\frac{3\sqrt{2}\mathrm{cos}\left(\theta \right)+3\sqrt{2}\mathrm{sin}\left(\theta \right)}{2}$

Now solve for$\theta$

To begin with, add the first and second equations together.

Then subtract the third equation from them.

Hence we get that

Then we can make the change of variable

Consequently,

Now solve for

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