You are given a system of linear equations:

1. $3x + 2y = 0$
2. $y = 2x - 1$

Let's solve this system by substitution since $y = 2x - 1$ is already isolated.

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

Substitute $2x - 1$ for $y$ in the equation $3x + 2y = 0$:

$3x + 2(2x - 1) = 0$

Step 2: Simplify the equation.

Expand the terms inside the parentheses:

$3x + 4x - 2 = 0$

Combine like terms:

$7x - 2 = 0$

Step 3: Solve for $x$.

Add 2 to both sides:

$7x = 2$

Now, divide by 7:

$x = \frac{2}{7}$

Step 4: Solve for $y$.

Now that we have $x = \frac{2}{7}$, substitute this value into the second equation $y = 2x - 1$:

$y = 2\left(\frac{2}{7}\right) - 1 = \frac{4}{7} - 1$

Convert 1 to a fraction:

$y = \frac{4}{7} - \frac{7}{7} = \frac{-3}{7}$

Final Solution:

The solution to the system of equations is: $x = \frac{2}{7}, \quad y = \frac{-3}{7}$

Would you like further details or have any questions?

Here are 5 related questions for you to explore:

1. How can you graphically represent the system of equations?
2. What happens if the equations represent parallel lines?
3. Can this method be used to solve non-linear systems?
4. How does the substitution method compare to the elimination method?
5. What would change if the second equation was $y = 3x - 1$?

Tip: When solving systems of equations, always check if substitution or elimination is more straightforward based on how the equations are given.