Question

Найдите все значения p при которых система x-y=1 y-x=1-2p² не имеет решений

Answer 1

$\left \{ {{x-y=1} \atop {y-x=1-2p^{2} }} \right. \\ \left \{ {{y-x=-1} \atop {y-x=1-2p^{2} }} \right.$

Левые части  уравнений равны,  значит чтобы система имела решения , должны быть равны и  правые части,  т.е.

$1-2p^{2} = -1 \\ 2p^{2} = 2 \\ p^{2} = 1 \\ p = 1; p = -1$

При любых других  значениях р  решений нет.

Ответ:  система не имеет решений при р ≠ +- 1 .

Answer 2

Finding the values of p for which the system x-y=1 and y-x=1-2p² has no solutions

To find the values of p for which the system x-y=1 and y-x=1-2p² has no solutions, we need to solve the system of equations and determine when it becomes inconsistent.

Let's start by solving the system of equations:

Equation 1: x - y = 1 Equation 2: y - x = 1 - 2p²

To eliminate the variable x, we can add Equation 1 and Equation 2:

(x - y) + (y - x) = 1 + (1 - 2p²) 0 = 2 - 2p²

Simplifying the equation, we get:

2p² = 2

Dividing both sides by 2, we find:

p² = 1

Taking the square root of both sides, we have:

p = ±1

Therefore, the values of p for which the system x-y=1 and y-x=1-2p² has no solutions are p = -1 and p = 1.

Answer: The values of p for which the system x-y=1 and y-x=1-2p² has no solutions are p = -1 and p = 1.