Solve this system of equations.

Question

2x + 6y = −6,

4x − 3y = −12

What is the solution to the system of equations?

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Answer 1

To solve the system of equations:

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

We can use either the substitution method or the elimination method. Here, I will use the elimination method.

First, let's simplify Equation 1 by dividing all terms by 2:

\[ x + 3y = -3 \quad \text{(Equation 1 simplified)} \]

Now we have:

1. \( x + 3y = -3 \) (Equation 3)
2. \( 4x - 3y = -12 \) (Equation 2)

Next, let's add Equation 3 and Equation 2 to eliminate \( y \). First, we can multiply Equation 3 by 3 so that the coefficients of \( y \) will cancel out:

\[ 3(x + 3y) = 3(-3) \] \[ 3x + 9y = -9 \quad \text{(Equation 4)} \]

Now we can write the two equations we have:

1. \( 3x + 9y = -9 \) (Equation 4)
2. \( 4x - 3y = -12 \) (Equation 2)

Now let's eliminate \( y \) by multiplying Equation 2 by 3:

\[ 3(4x - 3y) = 3(-12) \] \[ 12x - 9y = -36 \quad \text{(Equation 5)} \]

Now we have:

1. \( 3x + 9y = -9 \) (Equation 4)
2. \( 12x - 9y = -36 \) (Equation 5)

Next, we can add Equation 4 and Equation 5:

\[ (3x + 9y) + (12x - 9y) = -9 + (-36) \]

This simplifies to:

\[ 15x = -45 \]

Now, divide both sides by 15:

\[ x = -3 \]

Now that we have \( x \), we can substitute it back into Equation 3 to solve for \( y \):

\[ -3 + 3y = -3 \] \[ 3y = -3 + 3 \] \[ 3y = 0 \] \[ y = 0 \]

Thus, the solution to the system of equations is:

\[ (x, y) = (-3, 0) \]

So, the final answer is:

\[ \boxed{(-3, 0)} \]