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

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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**
$$
x = -6 \
y = 6
$$
The solution to the system of equations is $$ x = -6 $$ and $$ y = 6 $$.

Solve the system of equations and by combining the
equations.

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Solve the system of equations and by combining the equations.