Question

Х+у=2 Х²+2у=19 Решите систему уравнений

Answer 1

Y=2-x
$x^{2}$+2(2-x)=19
y=2-x
$x^{2}$-2x-15=0
D=b*b -4ac=4+60=64
x=(2+8)/2=5 y=2-5=-3
x=(2-8)/2=-3 y=2+3=5

Answer 2

Problem Analysis

We are given a system of equations: 1. x + y = 2 2. x^2 + 2y = 19

We need to find the values of x and y that satisfy both equations.

Solution

To solve this system of equations, we can use the substitution method or the elimination method. Let's use the substitution method.

From equation 1, we can express x in terms of y: 1. x = 2 - y

Substituting this value of x into equation 2, we get: 2. (2 - y)^2 + 2y = 19

Expanding equation 2, we have: 3. 4 - 4y + y^2 + 2y = 19

Combining like terms, we get: 4. y^2 - 2y - 15 = 0

Now, we can solve this quadratic equation for y. Factoring equation 4, we have: 5. (y - 5)(y + 3) = 0

Setting each factor equal to zero, we get two possible values for y: 6. y - 5 = 0 or y + 3 = 0

Solving these equations, we find: 7. y = 5 or y = -3

Now, we can substitute these values of y back into equation 1 to find the corresponding values of x.

For y = 5: Substituting y = 5 into equation 1, we get: 8. x + 5 = 2 Solving for x, we find: 9. x = -3

For y = -3: Substituting y = -3 into equation 1, we get: 10. x - 3 = 2 Solving for x, we find: 11. x = 5

Therefore, the solutions to the system of equations are: (x, y) = (-3, 5) and (5, -3).

Conclusion

The solutions to the given system of equations are (x, y) = (-3, 5) and (5, -3).