Answer :

To find the solutions for the system of equations:

[tex]\[ \left\{\begin{array}{r} x^2 + y^2 = 4 \\ x - y = 1 \end{array}\right. \][/tex]

we will solve this step-by-step:

1. Start with the second equation:

[tex]\[ x - y = 1 \][/tex]

2. Express \( x \) in terms of \( y \):

[tex]\[ x = y + 1 \][/tex]

3. Substitute \( x \) from the second equation (into the first equation):

[tex]\[ (y + 1)^2 + y^2 = 4 \][/tex]

4. Expand and simplify the equation:

[tex]\[ (y + 1)^2 + y^2 = 4 \\ (y^2 + 2y + 1) + y^2 = 4 \\ 2y^2 + 2y + 1 = 4 \][/tex]

5. Move all terms to one side to set up a quadratic equation:

[tex]\[ 2y^2 + 2y + 1 - 4 = 0 \\ 2y^2 + 2y - 3 = 0 \][/tex]

6. Solve the quadratic equation using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) :

[tex]\[ a = 2, \; b = 2, \; c = -3 \][/tex]

[tex]\[ y = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} \\ y = \frac{-2 \pm \sqrt{4 + 24}}{4} \\ y = \frac{-2 \pm \sqrt{28}}{4} \\ y = \frac{-2 \pm 2\sqrt{7}}{4} \\ y = \frac{-1 \pm \sqrt{7}}{2} \][/tex]

So we have two possible values for \( y \):

[tex]\[ y_1 = \frac{-1 - \sqrt{7}}{2} \quad \text{and} \quad y_2 = \frac{-1 + \sqrt{7}}{2} \][/tex]

7. Substitute these values back into \( x = y + 1 \) to find the corresponding \( x \) values:

For \( y_1 = \frac{-1 - \sqrt{7}}{2} \):

[tex]\[ x_1 = \left( \frac{-1 - \sqrt{7}}{2} \right) + 1 = \frac{-1 - \sqrt{7} + 2}{2} = \frac{1 - \sqrt{7}}{2} \][/tex]

For \( y_2 = \frac{-1 + \sqrt{7}}{2} \):

[tex]\[ x_2 = \left( \frac{-1 + \sqrt{7}}{2} \right) + 1 = \frac{-1 + \sqrt{7} + 2}{2} = \frac{1 + \sqrt{7}}{2} \][/tex]

Thus, we have the solutions:

[tex]\[ \left( \frac{1 - \sqrt{7}}{2}, \frac{-1 - \sqrt{7}}{2} \right) \quad \text{and} \quad \left( \frac{1 + \sqrt{7}}{2}, \frac{-1 + \sqrt{7}}{2} \right) \][/tex]

Which graph represents the solutions:

To visualize the solutions, you would plot the points \(\left( \frac{1 - \sqrt{7}}{2}, \frac{-1 - \sqrt{7}}{2} \right)\) and \(\left( \frac{1 + \sqrt{7}}{2}, \frac{-1 + \sqrt{7}}{2} \right)\) on the coordinate plane along with the circle \(x^2 + y^2 = 4\) and the line \(x - y = 1\).

These points will lie on the intersection of the circle with radius 2 centered at the origin and the line [tex]\(x - y = 1\)[/tex]. The graphical representation will show where the line intersects the circle, indicating the solutions to the system.