Solve using elimination method: 6x - 5y = 0; 2x - y = 4.

Steps to Solve

1. Write down the equations
   Identify the given system of linear equations. For example:
   $$
   2x + 3y = 6 \quad (1)
   4x - y = 5 \quad (2)
   $$

2. Choose a variable to eliminate
   Decide which variable you want to eliminate. Let's eliminate $y$ here. We can do this by modifying the equations so that the coefficients of $y$ in both equations are equal.

3. Multiply the equations if necessary
   Multiply equation (2) by $3$ to align the coefficients of $y$:
   $$
   3(4x - y) = 3(5) \implies 12x - 3y = 15 \quad (3)
   $$
   So the new equations are:
   $$
   2x + 3y = 6 \quad (1)
   12x - 3y = 15 \quad (3)
   $$

4. Add the equations
   Now, add equations (1) and (3) together to eliminate $y$:
   $$
   (2x + 3y) + (12x - 3y) = 6 + 15
   $$
   This simplifies to:
   $$
   14x = 21
   $$

5. Solve for the remaining variable
   Now, divide both sides by $14$ to find $x$:
   $$
   x = \frac{21}{14} = \frac{3}{2}
   $$

6. Substitute back to find the other variable
   Substitute $x$ back into one of the original equations to find $y$. We use equation (1):
   $$
   2\left(\frac{3}{2}\right) + 3y = 6
   $$
   This simplifies to:
   $$
   3 + 3y = 6
   $$
   Now, solve for $y$:
   $$
   3y = 6 - 3
   $$
   $$
   3y = 3 \implies y = 1
   $$

7. Write the final solution
   The solution of the system is:
   $$
   x = \frac{3}{2}, \quad y = 1
   $$