y=x-1, x^2-2y=26.

To solve the system of equations given by:

1. \( y = x - 1 \)
2. \( x^2 - 2y = 26 \)

we can substitute the expression for \( y \) from the first equation into the second equation.

Substituting \( y \) into the second equation:

\[ x^2 - 2(x - 1) = 26 \]

Now, simplify the equation:

\[ x^2 - 2x + 2 = 26 \]

Subtract 26 from both sides:

\[ x^2 - 2x + 2 - 26 = 0 \]

This simplifies to:

\[ x^2 - 2x - 24 = 0 \]

Next, we will factor the quadratic equation:

\[ (x - 6)(x + 4) = 0 \]

Setting each factor to zero gives us:

1. \( x - 6 = 0 \) ⟹ \( x = 6 \)
2. \( x + 4 = 0 \) ⟹ \( x = -4 \)

Now, we can find the corresponding \( y \) values for each \( x \):

1. For \( x = 6 \):

\[ y = 6 - 1 = 5 \]

2. For \( x = -4 \):

\[ y = -4 - 1 = -5 \]

Thus, the solutions to the system of equations are:

1. \( (6, 5) \)
2. \( (-4, -5) \)

So, the pairs \((x, y)\) are:

\[ (6, 5) \quad \text{and} \quad (-4, -5) \]