Question

Решите систему уравнений 2x+y=7 x^2-y=1 Решите пожалуйста срочно!!!

Answer 1

2х+у=7 (1)
х²-у=1 (2)
у=7-2х (1)
х²-(7-2х)=1 (2)
х²-7+2х-1=0
х²+2х-8=0
D=36
x=-4
x=2
y=7-2 × (-4)=15
y=7-2 × 2=3

Answer 2

Solving the System of Equations

The given system of equations is: - 2x + y = 7 - x^2 - y = 1

To solve this system, we can use the substitution method or the elimination method. Let's use the substitution method to solve for x and y.

First, let's solve the second equation for y: - y = x^2 - 1

Now, we can substitute this expression for y into the first equation: - 2x + (x^2 - 1) = 7 - x^2 + 2x - 1 = 7 - x^2 + 2x - 8 = 0

We can solve this quadratic equation to find the values of x. Let's proceed with the calculations.

Calculating the Values of x and y

Using the quadratic formula, we can find the values of x: - x = (-b ± √(b^2 - 4ac)) / (2a)

Substituting the values a = 1, b = 2, and c = -8 into the quadratic formula, we get: - x = (-2 ± √(2^2 - 4*1*(-8))) / (2*1) - x = (-2 ± √(4 + 32)) / 2 - x = (-2 ± √36) / 2 - x = (-2 ± 6) / 2

So, we have two possible values for x: - x = (-2 + 6) / 2 = 2 - x = (-2 - 6) / 2 = -4

Now, we can find the corresponding values of y using the equation y = x^2 - 1.

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

For x = -4: - y = (-4)^2 - 1 = 15

Therefore, the solutions to the system of equations are: - When x = 2, y = 3 - When x = -4, y = 15

So, the solutions to the system of equations 2x + y = 7 and x^2 - y = 1 are: x = 2, y = 3 x = -4, y = 15