Question

Решите систему уравнений {x+y=3, xy+4=0 методом полстановки

Answer 1

X=3-y - подстановка
y * (3-y) + 4 =0
3y - y^2 + 4 =0
-y^2+3y+4=0
y^2-3y-4=0
D=9+16=25
y1=4   x=-1
y2=-1  x=4

Ответ: (-1;4) и (4;-1)

Answer 2

Solving the System of Equations using Substitution Method

To solve the system of equations {x+y=3, xy+4=0} using the method of substitution, we can start by solving one of the equations for one variable and then substituting that expression into the other equation.

Let's solve the first equation for y: 1. x + y = 3 - Solving for y: y = 3 - x

Now, we can substitute this expression for y into the second equation xy + 4 = 0.

Substituting y = 3 - x into the second equation: x(3 - x) + 4 = 0

Expanding and simplifying the equation: 3x - x^2 + 4 = 0 -x^2 + 3x + 4 = 0

This results in a quadratic equation, which can be solved using the quadratic formula or factoring.

Let's solve for x using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)

Where: - a = -1 - b = 3 - c = 4

Plugging these values into the quadratic formula: x = (-3 ± √(3^2 - 4*(-1)*4)) / (2*(-1)) x = (-3 ± √(9 + 16)) / (-2) x = (-3 ± √25) / (-2) x = (-3 ± 5) / (-2)

This gives us two possible values for x: 1. x = (-3 + 5) / (-2) = 1 2. x = (-3 - 5) / (-2) = 4

Now that we have the values for x, we can substitute them back into the equation y = 3 - x to find the corresponding values for y.

For x = 1: y = 3 - 1 = 2

For x = 4: y = 3 - 4 = -1

So, the solutions to the system of equations are: - When x = 1, y = 2 - When x = 4, y = -1

Therefore, the solution to the system of equations {x+y=3, xy+4=0} using the method of substitution is: x = 1, y = 2 and x = 4, y = -1.