Solve the system of equations \( 2 x+3 y=6 \) and \( -3 x-5 y=-12 \) by combining the equations. \[ \begin{array}{l}(2 x+3 y=6) \\ (-3 x-5 y=-12) \\ 2 x+3 y=6 \\ -3 x-5 y=-12\end{array} \] \( \begin{array}{l}\text { o } x+0 y=\end{array} \)

To solve the system of equations using the **elimination method** (combining the equations to eliminate one of the variables), follow these steps: \[ \begin{cases} 2x + 3y = 6 \quad \text{(Equation 1)} \\ -3x - 5y = -12 \quad \text{(Equation 2)} \end{cases} \] ### Step 1: Align the Equations First, write both equations clearly: \[ \begin{align*} 2x + 3y &= 6 \quad \text{(Equation 1)} \\ -3x - 5y &= -12 \quad \text{(Equation 2)} \end{align*} \] ### Step 2: Eliminate One Variable We'll eliminate \( y \) by making the coefficients of \( y \) opposites. To do this, find the least common multiple (LCM) of the coefficients of \( y \): - Coefficient of \( y \) in Equation 1: \( 3 \) - Coefficient of \( y \) in Equation 2: \( -5 \) The LCM of \( 3 \) and \( 5 \) is \( 15 \). **Multiply Equation 1 by 5:** \[ 5 \times (2x + 3y) = 5 \times 6 \\ 10x + 15y = 30 \quad \text{(Equation 3)} \] **Multiply Equation 2 by 3:** \[ 3 \times (-3x - 5y) = 3 \times (-12) \\ -9x - 15y = -36 \quad \text{(Equation 4)} \] ### Step 3: Add the Modified Equations Now, add Equation 3 and Equation 4 to eliminate \( y \): \[ \begin{align*} 10x + 15y &= 30 \quad \text{(Equation 3)} \\ -9x - 15y &= -36 \quad \text{(Equation 4)} \\ \hline (10x - 9x) + (15y - 15y) &= 30 - 36 \\ x &= -6 \end{align*} \] ### Step 4: Solve for the Second Variable Now that we have \( x = -6 \), substitute this value back into one of the original equations to find \( y \). Let's use Equation 1: \[ 2x + 3y = 6 \\ 2(-6) + 3y = 6 \\ -12 + 3y = 6 \\ 3y = 18 \\ y = 6 \] ### **Solution** \[ x = -6 \\ y = 6 \]