We are given the system of equations:

1. $y = 6 - 2x$
2. $3x + 2y = 10$

We will solve this system using substitution.

Step 1: Substitute $y = 6 - 2x$ into the second equation.

The second equation is $3x + 2y = 10$, and since $y = 6 - 2x$, we substitute this into the second equation:

$3x + 2(6 - 2x) = 10$

Step 2: Simplify the equation.

Distribute the 2 to both terms inside the parentheses:

$3x + 12 - 4x = 10$

Combine like terms:

$-x + 12 = 10$

Step 3: Solve for $x$.

Subtract 12 from both sides:

$-x = -2$

Multiply both sides by -1:

$x = 2$

Step 4: Substitute $x = 2$ into the first equation to find $y$.

Now that we know $x = 2$, substitute this value into the first equation $y = 6 - 2x$:

$y = 6 - 2(2)$ $y = 6 - 4$ $y = 2$

Final Solution:

The solution to the system of equations is:

$x = 2, \quad y = 2$

Would you like a more detailed explanation, or do you have other questions?

Here are 5 related questions to explore further:

1. How would you solve this system using the elimination method?
2. Can this system of equations have infinitely many solutions?
3. How would you graph these two equations?
4. How does the substitution method compare to the elimination method for solving systems of equations?
5. What happens if we substitute a different value into the first equation?

Tip: When solving systems by substitution, it's helpful to solve for one variable in one equation first, then substitute that expression into the other equation to eliminate one variable at a time.