To solve the system of equations by substitution, we start with the two equations:

1. $4x + 5y = 2$ (Equation 1)
2. $4x - 20y = -3$ (Equation 2)

Step 1: Solve one equation for one variable.

Let's solve Equation 1 for $x$:

$4x = 2 - 5y$

Dividing both sides by 4 gives:

$x = \frac{2 - 5y}{4}$

Step 2: Substitute $x$ into the other equation.

Now, substitute $x$ into Equation 2:

$4\left(\frac{2 - 5y}{4}\right) - 20y = -3$

This simplifies to:

$2 - 5y - 20y = -3$

Combine like terms:

$2 - 25y = -3$

Step 3: Solve for $y$.

Subtract 2 from both sides:

$-25y = -3 - 2$

$-25y = -5$

Divide by -25:

$y = \frac{-5}{-25} = \frac{1}{5}$

Step 4: Substitute $y$ back to find $x$.

Now substitute $y = \frac{1}{5}$ back into the equation for $x$:

$x = \frac{2 - 5\left(\frac{1}{5}\right)}{4}$

$x = \frac{2 - 1}{4} = \frac{1}{4}$

Final Solution:

The solution to the system of equations is:

$x = \frac{1}{4}, \quad y = \frac{1}{5}$

Summary:

The solution is $(x, y) = \left(\frac{1}{4}, \frac{1}{5}\right)$.

Would you like more details or have any questions? Here are some related questions to consider:

1. How can we verify the solution by substituting back into the original equations?
2. What would be the graphical representation of these equations?
3. How can we solve this system using the elimination method?
4. What happens if we change one of the equations slightly?
5. Can this system of equations have no solutions or infinitely many solutions?

Tip: Always check your solutions by substituting them back into the original equations to confirm their validity!