Question

1.29. 1) y - x = 1 ,\\ x+|y|=1​

Answer 1

Ответ:

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Answer 2

To solve the system of equations:

1. $y - x = 1$
2. $x + |y| = 1$

We can consider different cases for the absolute value $|y|$. The absolute value of a number is always non-negative, so we have two cases:

Case 1: $y \geq 0$ In this case, $|y| = y$, and the system becomes:

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

Now, let's solve this system:

From equation (1), we can express $y$ as $y = x + 1$.

Substitute this into equation (2):

$x + (x + 1) = 1$

Combine like terms:

$2x + 1 = 1$

Subtract 1 from both sides:

$2x = 0$

Divide by 2:

$x = 0$

Now that we have the value of $x$, we can find $y$ using equation (1):

$y - 0 = 1$

$y = 1$

So, in Case 1, the solution is $x = 0$ and $y = 1$.

Case 2: $y < 0$ In this case, $|y| = -y$, and the system becomes:

1. $y - x = 1$
2. $x - y = 1$

Now, let's solve this system:

From equation (1), we can express $y$ as $y = x + 1$.

Substitute this into equation (2):

$x - (x + 1) = 1$

Simplify:

$x - x - 1 = 1$

Simplify further:

$-1 = 1$

This case leads to a contradiction. There is no solution when $y < 0$.

So, the system of equations has one solution in Case 1, which is $x = 0$ and $y = 1$, and no solution in Case 2. Therefore, the overall solution is $x = 0$ and $y = 1$.