Question

4х^2-y=2 2x^2+y=4 решите систему

Answer 1

Складываем уравнения, получаем

6x²=6
y=4x²-2

x²=1
y=4x²-2

x1=1
y1=2

x2= -1
y2=2
Ответ: (1;2) и (-1;2)

Answer 2

{ 4х² - у  = 2
{ 2х² + у  = 4
Способ сложения:
4х²  - у  + 2х²  + у  = 2 + 4
6х²  = 6
х² =  6 : 6
х²  = 1
х²  - 1²  = 0
(х - 1)(х + 1) = 0
х - 1 = 0
х₁ =  1
х + 1 = 0
х₂ =  - 1

2 * 1²  + у  = 4
2 +у  = 4
у = 4 - 2
у₁ = 2

2* (-1)² + у = 4
2 + у = 4
у = 4-2
у₂ = 2

Ответ :  ( 1 ; 2)  ,   (- 1 ;  2) .

Answer 3

To solve the system of equations:

1. 4x^2 - y = 2
2. 2x^2 + y = 4

We can use the method of substitution or elimination. Let's use the elimination method:

Step 1: Add the equations to eliminate y: (4x^2 - y) + (2x^2 + y) = 2 + 4

Step 2: Simplify and combine like terms: 6x^2 = 6

Step 3: Divide both sides by 6 to isolate x^2: x^2 = 1

Step 4: Take the square root of both sides: x = ±√1 x = ±1

Now, we have two potential values for x: x = 1 and x = -1.

Step 5: Substitute the value of x back into one of the original equations to solve for y. Let's use the first equation:

4(1)^2 - y = 2 4 - y = 2 y = 4 - 2 y = 2

So, one solution is x = 1 and y = 2.

Step 6: Now, let's substitute the other value of x:

4(-1)^2 - y = 2 4 - y = 2 y = 4 - 2 y = 2

In this case, we get the same value of y = 2.

Therefore, the solution to the system of equations is x = 1 and y = 2.