Question

У²-xy=2 2у²+3xy=14 (Ето система уровнений)

Answer 1

Умножим первое уравнение на 3 и прибавим к третьему:
5y^2=20
Y^2=4
y=2 или  y=-2
xy=2   x=1  или x=-1
Ответ(-1,-2) или (1,2)

Answer 2

To solve the system of equations:

1. $u^2 - xy = 2$
2. $2y^2 + 3xy = 14$

We can use substitution or elimination method. Let's use the elimination method:

Step 1: Multiply the first equation by 3 and the second equation by 1 to make the coefficients of $xy$ terms the same:

1. $3(u^2 - xy) = 3(2) \Rightarrow 3u^2 - 3xy = 6$
2. $2y^2 + 3xy = 14$

Step 2: Add the modified first equation to the second equation to eliminate the $xy$ term:

$3u^2 - 3xy + 2y^2 + 3xy = 6 + 14$

Simplify the equation:

$3u^2 + 2y^2 = 20$

Step 3: Now, let's solve each equation separately.

From the first equation: $u^2 - xy = 2$ ➜ $u^2 = 2 + xy$

From the equation we derived in Step 2: $3u^2 + 2y^2 = 20$ ➜ $u^2 = \frac{{20 - 2y^2}}{3}$

Since both expressions equal $u^2$, they must be equal to each other:

$2 + xy = \frac{{20 - 2y^2}}{3}$

Step 4: Solve for $x$ in terms of $y$:

$3(2 + xy) = 20 - 2y^2$ $6 + 3xy = 20 - 2y^2$ $3xy = 14 - 2y^2$ $xy = \frac{{14 - 2y^2}}{3}$

Step 5: Substitute the value of $xy$ from Step 4 into the first equation to solve for $u$:

$u^2 - \frac{{14 - 2y^2}}{3} = 2$ $u^2 = 2 + \frac{{14 - 2y^2}}{3}$ $u^2 = \frac{{6 + 14 - 2y^2}}{3}$ $u^2 = \frac{{20 - 2y^2}}{3}$

We've already derived this expression for $u^2$ in Step 3. This means that $u^2 = \frac{{20 - 2y^2}}{3}$.

Step 6: Now, we have a system of two equations:

1. $xy = \frac{{14 - 2y^2}}{3}$
2. $u^2 = \frac{{20 - 2y^2}}{3}$

Step 7: Since we have two equations and two variables, we can solve for $u$ and $y$:

From the first equation, we can express $x$ in terms of $y$:

$x = \frac{{14 - 2y^2}}{3y}$

Now, we have the values of $x$ and $u$ in terms of $y$:

$x = \frac{{14 - 2y^2}}{3y}$

$u^2 = \frac{{20 - 2y^2}}{3}$