We are given two sets of equations and asked to solve them. Set 1: $x^2 - y^2 = 3$ $x - y = 1$ Set 2: $x = y + 1$ $x = y^2 + 3$

1. Problem Description

We are given two sets of equations and asked to solve them.

Set 1:

$x^2 - y^2 = 3$

$x - y = 1$

Set 2:

$x = y + 1$

$x = y^2 + 3$

2. Solution Steps

Set 1:

We can factor the first equation using the difference of squares formula:

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

Since $x^2 - y^2 = 3$ and $x-y = 1$, we have:

$(1)(x+y) = 3$

$x+y = 3$

We now have two equations:

$x - y = 1$

$x + y = 3$

Adding the two equations, we get:

$2x = 4$

$x = 2$

Substituting $x=2$ into $x-y=1$:

$2 - y = 1$

$y = 1$

So the solution for Set 1 is $x=2, y=1$.

Set 2:

We are given:

$x = y + 1$

$x = y^2 + 3$

Since both equations are equal to $x$, we can set them equal to each other:

$y+1 = y^2 + 3$

$0 = y^2 - y + 2$

We can use the quadratic formula to solve for $y$:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a=1, b=-1, c=2$.

$y = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(2)}}{2(1)}$

$y = \frac{1 \pm \sqrt{1 - 8}}{2}$

$y = \frac{1 \pm \sqrt{-7}}{2}$

$y = \frac{1 \pm i\sqrt{7}}{2}$

So the solutions for $y$ are complex numbers.

Now, substitute $y$ back into $x = y + 1$ to find the corresponding $x$ values:

$x = \frac{1 + i\sqrt{7}}{2} + 1 = \frac{3 + i\sqrt{7}}{2}$

$x = \frac{1 - i\sqrt{7}}{2} + 1 = \frac{3 - i\sqrt{7}}{2}$

So the solutions for Set 2 are $x = \frac{3 + i\sqrt{7}}{2}, y = \frac{1 + i\sqrt{7}}{2}$ and $x = \frac{3 - i\sqrt{7}}{2}, y = \frac{1 - i\sqrt{7}}{2}$.

3. Final Answer

Set 1: $x = 2, y = 1$

Set 2: $x = \frac{3 + i\sqrt{7}}{2}, y = \frac{1 + i\sqrt{7}}{2}$ and $x = \frac{3 - i\sqrt{7}}{2}, y = \frac{1 - i\sqrt{7}}{2}$