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\begin{cases} x+y+\dfrac{x^2}{y^2}=7\\ \dfrac{(x-y)x^2}{y^2}=12 \end{cases} I don't have any idea to solve this. I tried to subtract, add and multiply the given equations, but nothing help me to find $x$ or $y$.

Can someone help me to solve this system? Thanks for attention.

Mohsen Shahriari
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Cavalo
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    Can always do substitution. Use the first to solve for $x$ in terms of $y$. $x^2 + y^2x + (y^3 - 7) =0$ so $x=\frac {-y^2\pm\sqrt{y^4 - 4y^2}}2$. Then plug those into the second and solve for $y$. – fleablood Jul 23 '20 at 21:03
  • Well, if the minus sign in the second equation is changed to plus, then $x=2, y=1$ is a solution. Or if you change the 12 to a 4. – B. Goddard Jul 23 '20 at 21:04
  • Yes, I tried this, but the second equuation will be so big! And I coudn't solve it :| – Cavalo Jul 23 '20 at 21:07
  • Which is the source of the problem? (Before we are searching for the roots of an equation of the shape $y^5 - 11y^4 + 79y^3 - 33y^2 + 84y + 72=0$, we should have the feeling that there is no sign typo in the equations.) – dan_fulea Jul 23 '20 at 21:26
  • The real solution, obtained using numerical methods, is approximately $x = 1.386320503, ; y = -0.5580341319$. – Robert Israel Jul 23 '20 at 23:26

1 Answers1

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The second equation tells you that $$\frac{x^2}{y^2}=\frac{12}{x-y}$$

(The possibility $x-y=0$ is ruled out, because this would violate the second equation.)

If we substitute this in the first equation, we get $$x+y+\frac{12}{x-y}=7$$

which implies

$$(x+y)(x-y)+12=7(x-y)$$

$$12=7(x-y)-(x-y)(x+y)$$

$$12=(x-y)[7-(x+y)]$$

I was trying to find integer solutions. $12 = 6.2 \textrm{ or } 2.6 \textrm{ or } 12.1 \textrm{ or } 1.12 \textrm{ or } 4.3 \textrm{ or } 3.4$ (6 different cases).

The first case results in $x=5.5$ and $y=-0.5$. (This is very easy)

The second case results in $x=1.5$ and $y=-0.5$.

The third case results in $x=9$ and $y=-3$.

The fourth case results in $x=-2$ and $y=-3$.

The fifth case results in $x=4$ and $y=0$ (but $y \neq 0)$.

The sixth case results in $x=3$ and $y=0$ (but $y \neq 0)$.

We can do the same with the negative integer factors of 12. That would give us extra solutions.

  • Something is not adding up, try, for example $x = 9, y = -3$ in the original equations. – Moo Jul 23 '20 at 21:54
  • That is true, Moo. Thanks. I can't find a reason. –  Jul 23 '20 at 22:20
  • "I was trying to find integer solutions" That's naive . If one equation, which has infinite solutions, has an integer solution, that in no way means that integer solution will be compatible with the second equation.... after all. If you has $x+ y = 26$ and $32x + 59y=-7$ if you figuer that $x+y=26$ has a nice solution $x=y=13$ that won't do us any good for the second equation. $3213+5913 \ne -7$ – fleablood Jul 23 '20 at 22:20
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    Solve it as a quadratic. – fleablood Jul 23 '20 at 22:23