Solve the system of equations: $x^2 + y^2 = 5$ $y = 3x - 5$

1. Problem Description

Solve the system of equations:

$x^2 + y^2 = 5$

$y = 3x - 5$

2. Solution Steps

Substitute $y$ from the second equation into the first equation:

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

Expand the expression:

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

Combine like terms:

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

Subtract 5 from both sides:

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

Divide the equation by 10:

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

Factor the quadratic equation:

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

Solve for $x$:

$x - 1 = 0$ or $x - 2 = 0$

$x = 1$ or $x = 2$

Now, find the corresponding $y$ values using the equation $y = 3x - 5$:

If $x = 1$, then $y = 3(1) - 5 = 3 - 5 = -2$

If $x = 2$, then $y = 3(2) - 5 = 6 - 5 = 1$

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

3. Final Answer

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