Question

X²+y²+xy=7 x+y=3 2*x*y-?

Answer 1

$x^{2}+xy+y^{2}=7$

x+y=3

Поднесем до квадрата:

$(x+y)^{2}=9$

$x^{2}+2xy+y^{2}=9$

Отнимем обе части уравнений:

xy=9-7=2

2xy=4

Ответ:4

Answer 2

{x²+y²+ху=7
{x+y=3
2*x*y-?
Второе уравнение системы возведём в квадрат
{x²+y²+ху=7
{х^2+2ху+у^2=9
х^2+2ху+у^2-(х^2+у^2+ху)=9-7
х^2+2ху+у^2-х^2-у^2-ху=2
ху=2
2*ху=2*2
2ху=4
Ответ: 2ху=4

Answer 3

To solve the given system of equations and find the value of $2xy$, we'll use substitution or elimination. Let's start by solving for $x$ and $y$ using the second equation ($x + y = 3$).

From the second equation, we can express $x$ in terms of $y$ (or vice versa):

$x = 3 - y$.

Now substitute this into the first equation ($x^2 + y^2 + xy = 7$):

$(3 - y)^2 + y^2 + (3 - y)y = 7$.

Expand and solve for $y$:

$9 - 6y + y^2 + y^2 + 3y - y^2 = 7$.

Simplify:

$9 - 6y + 3y = 7$.

$9 - 3y = 7$.

$-3y = -2$.

$y = \frac{2}{3}$.

Now, plug this value of $y$ back into the second equation to solve for $x$:

$x = 3 - \frac{2}{3} = \frac{7}{3}$.

Now that we have $x$ and $y$, we can find $2xy$:

$2xy = 2 \times \frac{7}{3} \times \frac{2}{3} = \frac{28}{9}$.