We are given a system of two equations: $x^2 + y^2 = 5$ $y = 3x - 5$ We need to find the solutions $(x, y)$ that satisfy both equations.

1. Problem Description

We are given a system of two equations:

$x^2 + y^2 = 5$

$y = 3x - 5$

We need to find the solutions $(x, y)$ that satisfy both equations.

2. Solution Steps

We will use the substitution method to solve this system of equations.

Since $y = 3x - 5$, we can substitute this expression for $y$ into the first equation:

$x^2 + (3x - 5)^2 = 5$

Expanding the square, we get:

$x^2 + (9x^2 - 30x + 25) = 5$

Combining like terms, we have:

$10x^2 - 30x + 25 = 5$

Subtracting 5 from both sides gives:

$10x^2 - 30x + 20 = 0$

We can divide the entire equation by 10 to simplify:

$x^2 - 3x + 2 = 0$

Now, we can factor the quadratic equation:

$(x - 1)(x - 2) = 0$

So, the solutions for $x$ are $x = 1$ and $x = 2$.

For $x = 1$, we have $y = 3(1) - 5 = 3 - 5 = -2$.

For $x = 2$, we have $y = 3(2) - 5 = 6 - 5 = 1$.

Therefore, the solutions are $(1, -2)$ and $(2, 1)$.

3. Final Answer

The solutions are $(1, -2)$ and $(2, 1)$.