The problem is to solve the following system of equations for $x$ and $y$: $x - y = 1$ $4 > x^2 - y^2 = 3$

1. Problem Description

The problem is to solve the following system of equations for $x$ and $y$:

$x - y = 1$

$4 > x^2 - y^2 = 3$

2. Solution Steps

First, consider the first equation:

$x - y = 1$

Second, consider the inequality:

$3 \le x^2 - y^2 < 4$

We can rewrite the difference of squares as a product:

$x^2 - y^2 = (x-y)(x+y)$

Since $x-y = 1$, we can substitute this into the inequality:

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

So, $3 \le x+y < 4$

We now have the system of equations:

$x - y = 1$

$3 \le x + y < 4$

We can solve the linear system using substitution or elimination. Let us solve for $x$ in the first equation:

$x = y+1$

Substitute this into the inequality:

$3 \le (y+1) + y < 4$

$3 \le 2y+1 < 4$

Subtract 1 from all sides of the inequality:

$2 \le 2y < 3$

Divide all sides by 2:

$1 \le y < \frac{3}{2}$

$1 \le y < 1.5$

Now, substitute $x = y+1$ into $x+y$ expression:

$x + y = (y+1) + y = 2y + 1$

Since $3 \le 2y + 1 < 4$,

$2 \le 2y < 3$

$1 \le y < \frac{3}{2}$

Now, let's find $x$. Since $x = y+1$,

$x = y+1$. Therefore, $x$ is bounded as follows:

$1 + 1 \le x < \frac{3}{2} + 1$

$2 \le x < \frac{5}{2}$

$2 \le x < 2.5$

So, we have $1 \le y < \frac{3}{2}$ and $2 \le x < \frac{5}{2}$.

3. Final Answer

$2 \le x < \frac{5}{2}$ and $1 \le y < \frac{3}{2}$

or

$2 \le x < 2.5$ and $1 \le y < 1.5$