Let's solve the system of equations:

1. \( -3x + 5y = 12 \)
2. \( 6x + 4y = 18 \)

We can solve this system using either substitution or elimination. I'll use the **elimination method** here.

### Step 1: Eliminate one of the variables

To eliminate \( x \), we'll first make the coefficients of \( x \) the same in both equations. We can achieve this by multiplying the first equation by 2.

Multiply the first equation by 2:
\[
2(-3x + 5y) = 2(12)
\]
This gives:
\[
-6x + 10y = 24
\]

Now we have the system:
1. \( -6x + 10y = 24 \)
2. \( 6x + 4y = 18 \)

### Step 2: Add the two equations

Now, add the two equations to eliminate \( x \):
\[
(-6x + 10y) + (6x + 4y) = 24 + 18
\]
This simplifies to:
\[
14y = 42
\]

### Step 3: Solve for \( y \)

Now solve for \( y \):
\[
y = \frac{42}{14} = 3
\]

### Step 4: Substitute \( y = 3 \) back into one of the original equations

Substitute \( y = 3 \) into the first equation \( -3x + 5y = 12 \):
\[
-3x + 5(3) = 12
\]
\[
-3x + 15 = 12
\]

Subtract 15 from both sides:
\[
-3x = 12 - 15
\]
\[
-3x = -3
\]

### Step 5: Solve for \( x \)
\[
x = \frac{-3}{-3} = 1
\]

### Solution
The solution to the system is:
\[
x = 1, \quad y = 3
\]

Would you like further clarification or details?

Here are 5 related questions:
1. How do you solve a system of equations using substitution instead of elimination?
2. What are the conditions under which a system of equations has no solution?
3. Can a system of linear equations have infinitely many solutions? When?
4. How do you graphically represent the solution to a system of linear equations?
5. What happens if both equations in a system are multiples of each other?

**Tip:** When using elimination, always check if multiplying the equations by constants can simplify the process of eliminating variables.

Learn how to solve a system of linear equations -3x + 5y = 12 and 6x + 4y = 18 using the elimination method. This problem involves algebraic techniques to eliminate variables and find solutions. Suitable for students in Grades 9-12.

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