Question

Решите систему уравнений методом подстановки: xy = - 2, x + y = 1

Answer 1

$\left \{ {{xy= -2} \atop {x+y =1}} \right.$

x=1-y;

(1-y)*y= -2;

y-y²= -2

y²-y-2=0

D=1+8=9

y1=$\frac{1+3}{2}=2$

y2=$\frac{1-3}{2}= -1$

x1= -1

x2 = 2

Ответ: (-1;2) или (2;-1)

Answer 2

To solve the system of equations using the method of substitution, we need to isolate one of the variables in one equation and then substitute its value into the other equation. Let's solve the system step by step:

1. Start with the given equations: xy = -2 ...(Equation 1) x + y = 1 ...(Equation 2)

2. Isolate one variable in terms of the other. We'll isolate y from Equation 2: y = 1 - x ...(Equation 3)

3. Substitute the value of y from Equation 3 into Equation 1: x(1 - x) = -2

4. Simplify and solve for x: x - x^2 = -2 x^2 - x + 2 = 0

Now, we can use the quadratic formula to find the values of x:

The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by: x = (-b ± √(b^2 - 4ac)) / 2a

In our case, a = 1, b = -1, and c = 2: x = (1 ± √((-1)^2 - 412)) / 2*1 x = (1 ± √(1 - 8)) / 2 x = (1 ± √(-7)) / 2

Since the square root of a negative number is not a real number, this means that the system has no real solutions. The graphs of the two equations represent two different curves that do not intersect in the real plane, which is why there is no common solution for x and y.