To solve the system of equations using the elimination method, we start with the given equations:

$5x + 3y = -6 \quad \text{(Equation 1)}$ $3x - 2y = 4 \quad \text{(Equation 2)}$

Step 1: Make the coefficients of one variable the same

We'll eliminate $x$ by making the coefficients of $x$ the same in both equations. We can do this by multiplying both equations by suitable numbers:

• Multiply Equation 1 by 3: $3(5x + 3y) = 3(-6)$ $15x + 9y = -18 \quad \text{(Equation 3)}$

• Multiply Equation 2 by 5: $5(3x - 2y) = 5(4)$ $15x - 10y = 20 \quad \text{(Equation 4)}$

Step 2: Subtract the equations to eliminate $x$

Now, subtract Equation 4 from Equation 3:

$(15x + 9y) - (15x - 10y) = -18 - 20$

Simplifying, we get:

$15x + 9y - 15x + 10y = -38$

$19y = -38$

Step 3: Solve for $y$

Divide both sides by 19:

$y = \frac{-38}{19}$

$y = -2$

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

We'll substitute $y = -2$ into Equation 1:

$5x + 3(-2) = -6$

Simplify:

$5x - 6 = -6$

Add 6 to both sides:

$5x = 0$

Divide by 5:

$x = 0$

Solution

The solution to the system of equations is $x = 0$ and $y = -2$.

Would you like further details or have any questions? Here are some related questions you might consider:

1. How does the elimination method differ from the substitution method?
2. What are the advantages of using the elimination method?
3. How can you solve a system of equations using matrices?
4. What if both variables were eliminated during the elimination process?
5. How do you determine if a system of equations has no solution or infinitely many solutions?

Tip: When solving systems of equations, always double-check your steps by substituting your solution back into the original equations.