Consider the following system of equations:

To solve the system of equations using substitution, we start with the given equations:

1. \(-x + y = 5\) (Eq. 1)
2. \(y = (x - 3)^2 + 2\) (Eq. 2)

Step 1: Solve Eq. 1 for \(y\)

From Eq. 1: \[ y = x + 5 \]

Step 2: Substitute \(y\) in Eq. 2

Now we substitute \(y\) from Eq. 1 into Eq. 2: \[ x + 5 = (x - 3)^2 + 2 \]

Step 3: Expand and simplify the equation

Expanding the right side: \[ x + 5 = (x^2 - 6x + 9) + 2 \] \[ x + 5 = x^2 - 6x + 11 \]

Step 4: Rearranging the equation

Move all terms to one side to set the equation to zero: \[ 0 = x^2 - 6x + 11 - x - 5 \] \[ 0 = x^2 - 7x + 6 \]

Step 5: Factor the quadratic equation

Now we factor \(x^2 - 7x + 6\): \[ 0 = (x - 1)(x - 6) \]

Step 6: Solve for \(x\)

Setting each factor to zero gives: \[ x - 1 = 0 \implies x = 1 \] \[ x - 6 = 0 \implies x = 6 \]

Step 7: Substitute \(x\) back to find \(y\)

Now we will find \(y\) for both \(x\) values using \(y = x + 5\):

1. For \(x = 1\): \[ y = 1 + 5 = 6 \]

2. For \(x = 6\): \[ y = 6 + 5 = 11 \]

Step 8: Solutions

The solutions to the system of equations are:

1. \((1, 6)\)
2. \((6, 11)\)

Therefore, the final answer is: \[ \text{The solutions are } (1, 6) \text{ and } (6, 11). \]