Question

Решить систему x +y = -11 если xy = 28​

Answer 1

Х=-11-у

ху=28

(-11-у)у=28

-11у-у^2-28=0

у^2+11у+28=0

Д=121-4*28=9

у1=-11+9/2=-1

у2=-11-9/2=-10

у1=-1

х1=-11-(-1)=-10

у2=-10

х2=-11-(-10) =-1

Answer 2

To solve the system of equations:

1. x + y = -11
2. xy = 28

We can use the method of substitution or elimination. Let's solve it using the substitution method.

From equation 1, we can rewrite it as x = -11 - y.

Substituting this value of x into equation 2:

(-11 - y)y = 28

Expanding and rearranging the equation:

-y^2 - 11y = 28

Rearranging again and setting it equal to zero:

y^2 + 11y + 28 = 0

Now, we can factorize the quadratic equation:

(y + 4)(y + 7) = 0

This gives us two possible solutions:

1. y + 4 = 0, which gives y = -4
2. y + 7 = 0, which gives y = -7

Now, substitute these values back into equation 1 to find the corresponding values of x:

For y = -4: x + (-4) = -11 x - 4 = -11 x = -11 + 4 x = -7

For y = -7: x + (-7) = -11 x - 7 = -11 x = -11 + 7 x = -4

Therefore, the solutions to the system of equations are x = -7, y = -4 and x = -4, y = -7.