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#1 h Apr 17, 2025, 12:30 PM

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Solve equations: $\begin{cases} x^4+x^3y+x^2y^2=7x+9 \\ x(y-x+1)=3 \end{cases}$

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#2 h Apr 18, 2025, 5:38 AM

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We have given the system of equations

$(1) \;\; x^4 + x^3y + x^2y^2 = 7x + 9$ ,

$(2) \;\; x(y - x + 1) = 3$ .

Clearly $x \neq 0$ . Futhermore equations (1) and (2) are equivalent to

${\textstyle (3) \;\; x^2 + xy + y^2 + \frac{49}{36} = (\frac{3}{x} + \frac{7}{6})^2}$ ,

${\textstyle (4) \;\; y - x + \frac{13}{6} = \frac{3}{x} + \frac{7}{6}}$ .

By combining equations (3) and (4) we obtain

${\textstyle x^2 + xy + y^2 + \frac{49}{36} = (y - x + \frac{13}{6})^2}$ ,

yielding

${\textstyle x^2 + xy + y^2 + \frac{49}{36} = x^2 - 2xy + y^2 + \frac{13}{3}(y - x) + \frac{169}{36}}$ ,

i.e.

$(5) \;\; 9xy = 13(y - x) + 10$ .

According to equation (2)

$9xy = 9(x^2 - x + 3)$ ,

which combined with equation (5) give us

$13y - 13x + 10 = 9x^2 - 9x + 27$ ,

i.e.

${\textstyle (6) \;\; y = \frac{9x^2 \,+\, 4x \,+\,17}{13}}$ .

Equation (2) van be expressed as

${\textstyle y = \frac{3}{x} + x - 1}$ ,

which according to equation (6) means

${\textstyle 9x^2 + 4x + 17 = 13(\frac{3}{x} + x - 1)}$ ,

or alternatively

$3x^3 - 3x^2 + 10x - 13 = 0$ ,

which only non-rational real solution is $x \approx 1,2086$ , yielding $y \approx 2,6908$ by equation (7).

Conclusion: The system of equations (1)-(2) has only one solution, namely $(x,y) \approx (1.2086,2.6908)$ .

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#3 h Apr 18, 2025, 6:01 AM

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Solar Plexsus wrote:

You could have just noticed $xy=x^2-x+3$ and replaced $xy$ in the first eq to get $3x^3-3x^2+10x-13=0$

This post has been edited 1 time. Last edited by cazanova19921, Apr 18, 2025, 6:01 AM

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