Новая Школа - онлайн-школа подготовки к ЕГЭ
Задача Задача №64599
А
Аноним
27 ноября 2023 17:44 · публичный вопрос
А
Аноним

[x+2y=1 x+y²=4]

Нейросеть27 ноября 2023 17:44

To solve the system of equations:

  1. Rearrange the first equation to solve for x: x = 1 - 2y

  2. Substitute the value of x in the second equation: (1 - 2y) + y² = 4

  3. Simplify the equation: y² - 2y + 1 = 4

  4. Rearrange the equation: y² - 2y - 3 = 0

  5. Factorize the equation: (y - 3)(y + 1) = 0

  6. Set each factor equal to zero and solve for y: y - 3 = 0 or y + 1 = 0 y = 3 or y = -1

  7. Substitute the values of y back into the first equation to solve for x: x + 2(3) = 1 or x + 2(-1) = 1 x + 6 = 1 or x - 2 = 1 x = -5 or x = 3

Therefore, the solutions to the system of equations are: x = -5, y = 3 x = 3, y = -1

[x+2y=1 x+y²=4]
To solve the system of equations: 1. Rearrange the first equation to solve for x: x = 1 - 2y 2. Substitute the value of x in the second equation: (1 - 2y) + y² = 4 3. Simplify the equation: y² - 2y + 1 = 4 4. Rearrange the equation: y² - 2y - 3 = 0 5. Factorize the equation: (y - 3)(y + 1) = 0 6. Set each factor equal to zero and solve for y: y - 3 = 0 or y + 1 = 0 y = 3 or y = -1 7. Substitute the values of y back into the first equation to solve for x: x + 2(3) = 1 or x + 2(-1) = 1 x + 6 = 1 or x - 2 = 1 x = -5 or x = 3 Therefore, the solutions to the system of equations are: x = -5, y = 3 x = 3, y = -1