We can solve the system of equations by elimination.

The system of equations is:

1. $3x + 4y = -3$
2. $2x + 6y = 8$

Step 1: Multiply the equations

We want to eliminate one variable, so we need to manipulate the equations. First, multiply both equations by factors that make the coefficients of $x$ or $y$ the same.

To eliminate $x$, we can multiply:

• The first equation by 2: $2(3x + 4y) = 2(-3)$
• The second equation by 3: $3(2x + 6y) = 3(8)$

This gives:

1. $6x + 8y = -6$
2. $6x + 18y = 24$

Step 2: Subtract the equations

Now, subtract equation 1 from equation 2 to eliminate $x$:

$(6x + 18y) - (6x + 8y) = 24 - (-6)$

This simplifies to:

$10y = 30$

Step 3: Solve for $y$

$y = \frac{30}{10} = 3$

Step 4: Substitute $y$ back into one of the original equations

Now, substitute $y = 3$ into one of the original equations (let's use the first one):

$3x + 4(3) = -3$ $3x + 12 = -3$ $3x = -3 - 12$ $3x = -15$ $x = \frac{-15}{3} = -5$

Final Solution:

The solution is: $x = -5, \quad y = 3$

Would you like further details on this process or have any questions?

Here are 5 related questions to expand on this:

1. Can you solve a similar system of equations with different coefficients using elimination?
2. How would you approach this problem if the coefficients of $y$ were the same initially?
3. What happens if both equations are multiples of each other? How would you solve such a system?
4. Can you use substitution instead of elimination for this system? What would be the steps?
5. How can you verify that the solution is correct?

Tip: Always double-check your final answer by plugging the values of $x$ and $y$ back into the original equations to verify their correctness.