Question

{(x+y)^2-6(x+y)+5=0 {(x-y)^2-2(x-y)-3=0​

Answer 1

Объяснение:

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Answer 2

It looks like you've provided two quadratic equations involving variables x and y. The equations are:

1. $(x+y)^2 - 6(x+y) + 5 = 0$
2. $(x-y)^2 - 2(x-y) - 3 = 0$

To solve these equations, we can follow these steps:

Equation 1: $(x+y)^2 - 6(x+y) + 5 = 0$

Let's expand and simplify: $x^2 + 2xy + y^2 - 6x - 6y + 5 = 0$

Combine like terms: $x^2 + 2xy + y^2 - 6x - 6y + 5 = 0$

This equation doesn't seem to factor easily, so we can solve for one variable in terms of the other and substitute to simplify further.

Solve for y in terms of x: $y = \frac{-x^2 + 6x - 5}{2x - 6}$

Equation 2: $(x-y)^2 - 2(x-y) - 3 = 0$

Let's expand and simplify: $x^2 - 2xy + y^2 - 2x + 2y - 3 = 0$

Combine like terms: $x^2 - 2xy + y^2 - 2x + 2y - 3 = 0$

Again, this equation doesn't seem to factor easily.

Solve for y in terms of x: $y = \frac{x^2 - 2x + 3}{2x - 2}$

Keep in mind that these equations might have complex solutions or might not have a real solution at all, depending on the values of x and y that satisfy them. If you have specific values for x and y or if you are looking for a specific type of solution, please provide more context so I can assist you further.