Question

{2x2 – 3у2=-19,{ху = -6;​ Смотреть ответ shmyrevAleksandr shmyrevAleksandr Ответ: 2x²-3y²=-19 xy=-6              x=-6/y 2(-6/y)²-3y²=-19 72/y²-3y²=-19 72-3y⁴=-19y² -3y⁴+19y²+72=0 y²=t -3t²+19t+72=0 D=19²-4*(-3)*72=361+864=1225=35² t=(-19-35)/2*(-3)=9        t=(-19+35)/2*(-3)=16/-6=-8/3 y²=9 y=3              y=-3 x=-6/3=-2    x=-6/(-3)=2 Ответ: (-2;3)  (2;-3) Объяснение:

Answer 1

Looks like you've provided a solution for a system of equations involving two equations:

1. $2x^2 - 3y^2 = -19$
2. $xy = -6$

Your solution process seems correct, and you've correctly found the solutions for $x$ and $y$:

For equation (2), $xy = -6$, you've solved for $x$: $x = -\frac{6}{y}$

You substituted this value of $x$ into equation (1), $2x^2 - 3y^2 = -19$, simplified, and eventually arrived at the quadratic equation: $-3t^2 + 19t + 72 = 0$

By solving this quadratic equation, you found two possible values for $t$: $t = 9$ and $t = -\frac{8}{3}$.

Since $t = y^2$, you found two possible values for $y$: $y = 3$ and $y = -3$.

Substituting these values of $y$ back into $x = -\frac{6}{y}$, you found the corresponding values for $x$: $x = -2$ and $x = 2$.

So, the solutions for the system of equations are $(-2, 3)$ and $(2, -3)$, which you've correctly identified.

Great job on solving the system and explaining the steps! If you have any further questions or topics you'd like to discuss, feel free to ask.