To solve the system of equations using substitution, we can follow these steps:

Given equations:

1. $y = -6x + 5$
2. $-2x + y = 5$

Step 1: Substitute the expression for $y$ from the first equation into the second equation.

Since $y = -6x + 5$, we can replace $y$ in the second equation:

$-2x + (-6x + 5) = 5$

Simplify the equation:

$-2x - 6x + 5 = 5$

$-8x + 5 = 5$

Step 2: Solve for $x$.

Subtract 5 from both sides:

$-8x = 0$

Divide by $-8$:

$x = 0$

Step 3: Substitute $x = 0$ into the first equation to find $y$.

$y = -6(0) + 5$

$y = 5$

So, the solution is:

$x = 0, \quad y = 5$

Would you like further details on this solution, or do you have any questions? Here are some related problems you might find interesting:

1. Solve the system using the elimination method: $3x + 2y = 12$ and $x - y = 3$.
2. Find the intersection point of the lines represented by the equations $2x - 3y = 6$ and $4x + y = 7$.
3. Solve by substitution: $y = 3x - 4$ and $5x - y = 11$.
4. Determine if the system $x + y = 1$ and $2x + 2y = 2$ has one solution, no solution, or infinitely many solutions.
5. Use substitution to solve the system: $2y = 4x - 6$ and $3x - y = 7$.

Tip: When solving systems of equations by substitution, always start by isolating one variable in one of the equations to make the substitution easier.