Question

Решите систему уравнений { 2x-y+3=0 { y-x^2=0

Answer 1

2х-у+3=0
а=2; б=1; с=3
D=b^2-4ac=1^2-4*2*3=1-24=23
x1=6
x2=5,5

вроде так

Answer 2

Solution to the System of Equations

To solve the system of equations:

2x - y + 3 = 0
y - x^2 = 0

we can use various methods such as substitution, elimination, or matrix methods. Let's solve it using the substitution method.

Substitution Method

1. Solve one equation for one variable in terms of the other variable. Let's solve the second equation for y:

   y = x^2
   

2. Substitute the expression for y in terms of x into the other equation. Substituting `y = x^2` into the first equation, we get:

   2x - (x^2) + 3 = 0
   

3. Rearrange the equation to solve for x. Rearranging the equation, we have:

   x^2 - 2x + 3 = 0
   

4. Solve the quadratic equation for x. Using the quadratic formula or factoring, we find that the solutions for x are complex numbers. Therefore, there are no real solutions for x.

5. Substitute the value of x back into one of the original equations to solve for y. Since we don't have a real solution for x, we cannot find the corresponding value of y.

Therefore, the system of equations does not have any real solutions.

Note: The above solution is based on the search results and the information provided. However, please double-check the calculations and verify the solution independently.