Question

1)x + y = 4,xy = 4;​

Answer 1

Ответ: х1=2;  х2=2;   у1=2;  у2=2.

Объяснение:

Это система. Решаем методом подстановки, так как другого не указано.

x + y = 4,

xy = 4;​

х=4-у;

(4-y)y=4;

4y-y²=4;

-y²+4y-4=0;  [*(-1)]

y²-4y+4=0;

По теореме  Виета

у1+у2=4;

у1*у2=4;

у1=2; у2=2.

х=4-у=4-2=2.

Answer 2

To solve the system of equations $x + y = 4$ and $xy = 4$, you can use substitution or elimination. Let's use substitution:

1. Solve one of the equations for one variable. In this case, let's solve $x + y = 4$ for $x$:

   $x = 4 - y$

2. Substitute this expression for $x$ into the second equation $xy = 4$:

   $(4 - y)y = 4$

3. Expand and solve for $y$:

   $4y - y^2 = 4$

4. Rearrange this equation into a quadratic equation:

   $y^2 - 4y + 4 = 0$

5. Factor the quadratic:

   $(y - 2)(y - 2) = 0$

6. Set each factor equal to zero:

   $y - 2 = 0$

7. Solve for $y$:

   $y = 2$

8. Now that you have found $y$, substitute it back into the equation $x = 4 - y$ to find $x$:

   $x = 4 - 2$

   $x = 2$

So, the solution to the system of equations $x + y = 4$ and $xy = 4$ is $x = 2$ and $y = 2$.