To solve the system of equations using the elimination method, we start with the two equations:
- \( 2x + 3y = 30 \) (Equation 1)
- \( -12x + 8y = -24 \) (Equation 2)
We'll manipulate these equations to eliminate one of the variables. Let's eliminate \(x\).
First, we can multiply Equation 1 by 6 to make the coefficient of \(x\) the same in both equations. This gives us:
\[ 6(2x + 3y) = 6(30) \] \[ 12x + 18y = 180 \quad (Equation 3) \]
Now we have:
- \( 12x + 18y = 180 \) (Equation 3)
- \( -12x + 8y = -24 \) (Equation 2)
Next, we add Equation 3 and Equation 2 together:
\[ (12x + 18y) + (-12x + 8y) = 180 + (-24) \]
This simplifies to:
\[ 0x + 26y = 156 \]
So we have:
\[ 26y = 156 \] Dividing both sides by 26 gives:
\[ y = 6 \]
Now that we have \(y\), we can substitute it back into one of the original equations to solve for \(x\). We'll use Equation 1:
\[ 2x + 3(6) = 30 \] \[ 2x + 18 = 30 \] Subtracting 18 from both sides gives:
\[ 2x = 12 \] Dividing both sides by 2 gives:
\[ x = 6 \]
So the solution to the system of equations is:
\[ x = 6, \quad y = 6 \]
Thus, the solution is \( (x, y) = (6, 6) \).