$$ \begin{array}{cc} ext { System 1 } & ext { System 2 } \ 2 x+3 y=6 & 2 x+3 y=6 \ -2 x-y=-6 & x+y=4\end{array} $$

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$$ \begin{array}{cc}	ext { System 1 } & 	ext { System 2 } \ 2 x+3 y=6 & 2 x+3 y=6 \ -2 x-y=-6 & x+y=4\end{array} $$

UpStudy AI · Accepted Answer

Sure, let's solve both **System 1** and **System 2** step by step.

---

### **System 1:**
$$
\begin{cases}
2x + 3y = 6 \quad \text{(Equation 1)} \
-2x - y = -6 \quad \text{(Equation 2)}
\end{cases}
$$

**Step 1: Add Equation 1 and Equation 2**

$$
\begin{align*}
(2x + 3y) + (-2x - y) &= 6 + (-6) \
2x - 2x + 3y - y &= 0 \
2y &= 0 \
y &= 0
\end{align*}
$$

**Step 2: Substitute $$ y = 0 $$ into Equation 1**

$$
2x + 3(0) = 6 \
2x = 6 \
x = 3
$$

**Solution for System 1:** 
$$
x = 3, \quad y = 0
$$

---

### **System 2:**
$$
\begin{cases}
2x + 3y = 6 \quad \text{(Equation 1)} \
x + y = 4 \quad \text{(Equation 2)}
\end{cases}
$$

**Step 1: Solve Equation 2 for $$ x $$**

$$
x = 4 - y
$$

**Step 2: Substitute $$ x = 4 - y $$ into Equation 1**

$$
2(4 - y) + 3y = 6 \
8 - 2y + 3y = 6 \
8 + y = 6 \
y = 6 - 8 \
y = -2
$$

**Step 3: Substitute $$ y = -2 $$ back into Equation 2 to find $$ x $$**

$$
x = 4 - (-2) \
x = 4 + 2 \
x = 6
$$

**Solution for System 2:** 
$$
x = 6, \quad y = -2
$$

---

### **Summary of Solutions:**

- **System 1:** $$ x = 3 $$, $$ y = 0 $$
- **System 2:** $$ x = 6 $$, $$ y = -2 $$

If you have any further questions or need additional explanations, feel free to ask!
**Solutions:** - **System 1:** $$ x = 3 $$, $$ y = 0 $$ - **System 2:** $$ x = 6 $$, $$ y = -2 $$

\( \begin{array}{cc}\text { System 1 } & \text { System 2 } \\ 2 x+3 y=6 & 2 x+3 y=6 \\ -2 x-y=-6 & x+y=4\end{array} \)

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