(x2+y2–1)3-x2y3=0

The teacher will be very surprised to see your correct solution 😉

The solution set is obviously symmetric with respect to the $y$-axis. Therefore we may assume $x\geq 0$. In the domain $\{(x,y)\in {\mathbb R}^2\ |\ x\geq0\}$ the equation is equivalent with\n$$x^2+ y^2 -1=x^{2/3} y\ ,$$\nwhich can easily be solved for $y$:\n$$y={1\over2}\bigl(x^{2/3}\pm\sqrt{x^{4/3}+4(1-x^2)}\bigr)\ .$$\nNow plot this, taking both branches of the square root into account. You might have to numerically solve the equation $x^{4/3}+4(1-x^2)=0$ in order to get the exact $x$-interval.

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(x2 + y2 - 1)*3 - x2*y3 = 0

$$- x_{2} y_{3} + \left(x_{2} + y_{2} - 1\right) 3 = 0$$

Detail solution

Given the linear equation:

Expand brackets in the left part

Looking for similar summands in the left part:

Move free summands (without y3)

from left part to right part, we given:

$$- x_{2} y_{3} + 3 x_{2} + 3 y_{2} = 3$$

Move the summands with the other variables

from left part to right part, we given:

$$- x_{2} y_{3} + 3 y_{2} = - 3 x_{2} + 3$$

Divide both parts of the equation by (3*y2 - x2*y3)/y3

We get the answer: y3 = 3*(-1 + x2 + y2)/x2

(x2+y2-1)*3-x2*y3 = 0

Expand brackets in the left part

x2*3+y2*3-1*3-x2*y3 = 0

Looking for similar summands in the left part:

-3 + 3*x2 + 3*y2 - x2*y3 = 0

Move free summands (without y3)

from left part to right part, we given:

$$- x_{2} y_{3} + 3 x_{2} + 3 y_{2} = 3$$

Move the summands with the other variables

from left part to right part, we given:

$$- x_{2} y_{3} + 3 y_{2} = - 3 x_{2} + 3$$

Divide both parts of the equation by (3*y2 - x2*y3)/y3

y3 = 3 - 3*x2 / ((3*y2 - x2*y3)/y3)

We get the answer: y3 = 3*(-1 + x2 + y2)/x2

The solution of the parametric equation

Given the equation with a parameter:

$$- x_{2} y_{3} + 3 x_{2} + 3 y_{2} - 3 = 0$$

The coefficient at y3 is equal to

$$- x_{2}$$

then possible cases for x2 :

$$x_{2} < 0$$

$$x_{2} = 0$$

Consider all cases in more detail:

With

$$x_{2} < 0$$

the equation

$$3 y_{2} + y_{3} - 6 = 0$$

its solution

$$y_{3} = - 3 y_{2} + 6$$

With

$$x_{2} = 0$$

the equation

$$3 y_{2} - 3 = 0$$

its solution

$$- x_{2} y_{3} + 3 x_{2} + 3 y_{2} - 3 = 0$$

The coefficient at y3 is equal to

$$- x_{2}$$

then possible cases for x2 :

$$x_{2} < 0$$

$$x_{2} = 0$$

Consider all cases in more detail:

With

$$x_{2} < 0$$

the equation

$$3 y_{2} + y_{3} - 6 = 0$$

its solution

$$y_{3} = - 3 y_{2} + 6$$

With

$$x_{2} = 0$$

the equation

$$3 y_{2} - 3 = 0$$

its solution

The graph

Rapid solution
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3*(-1 + x2 + y2) y3_1 = ---------------- x2

$$y_{3 1} = \frac{3 \left(x_{2} + y_{2} - 1\right)}{x_{2}}$$

Sum and product of roots
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sum

3*(-1 + x2 + y2) ---------------- x2

$$\left(\frac{3 \left(x_{2} + y_{2} - 1\right)}{x_{2}}\right)$$

=

3*(-1 + x2 + y2) ---------------- x2

$$\frac{3 \left(x_{2} + y_{2} - 1\right)}{x_{2}}$$

product

3*(-1 + x2 + y2) ---------------- x2

$$\left(\frac{3 \left(x_{2} + y_{2} - 1\right)}{x_{2}}\right)$$

=

3*(-1 + x2 + y2) ---------------- x2

$$\frac{3 \left(x_{2} + y_{2} - 1\right)}{x_{2}}$$