Question

X2 + y2 = 41 Y - x = -1​

Answer 1

Ответ:

1 какое-то не понятно но я сделал его авторое вообще нет решения

Answer 2

The given system of equations is:

1. $x^2 + y^2 = 41$
2. $y - x = -1$

To solve this system, we can use substitution or elimination. Let's use the substitution method.

From equation 2, we can express $y$ in terms of $x$:

$y = x - 1$

Now, substitute this into equation 1:

$x^2 + (x - 1)^2 = 41$

Expand and simplify:

$x^2 + x^2 - 2x + 1 = 41$

$2x^2 - 2x - 40 = 0$

Now, divide the entire equation by 2:

$x^2 - x - 20 = 0$

Now, we can factor this quadratic equation:

$(x - 5)(x + 4) = 0$

So, $x = 5$ or $x = -4$.

Now, let's find the corresponding values of $y$ for each $x$:

1. When $x = 5$: $y = x - 1 = 5 - 1 = 4$

2. When $x = -4$: $y = x - 1 = -4 - 1 = -5$

So, the solutions for $x$ and $y$ are:

1. $x = 5, y = 4$
2. $x = -4, y = -5$