Question

For each linear equation, indicate whether the equation has no solution, one solution. or infinitely many solutions. This problem requires you to show your work.(3 points)Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.6x+2−4x=2(x+2)3(x+3)=7+3x+6−x8(x+2)=5x+16+3x

Answer 1

Let's go through each of the linear equations provided and determine whether they have no solution, one solution, or infinitely many solutions by solving them step by step.

### 1. Equation: \( 6x + 2 - 4x = 2(x + 2) \)

**Step 1**: Simplify both sides of the equation.

- Left Side: \( 6x - 4x + 2 = 2x + 2 \)
- Right Side: \( 2(x + 2) = 2x + 4 \)

Thus, the equation simplifies to:
\[ 2x + 2 = 2x + 4 \]

**Step 2**: Isolate the variable.
Subtract \( 2x \) from both sides:
\[ 2 = 4 \]

**Conclusion**: Since \( 2 = 4 \) is a false statement, the equation has **no solution**.

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### 2. Equation: \( 3(x + 3) = 7 + 3x + 6 - x \)

**Step 1**: Simplify both sides.

- Left Side: \( 3(x + 3) = 3x + 9 \)
- Right Side: \( 7 + 3x + 6 - x = 3x + 13 \)

Thus, the equation simplifies to:
\[ 3x + 9 = 3x + 13 \]

**Step 2**: Isolate the variable.
Subtract \( 3x \) from both sides:
\[ 9 = 13 \]

**Conclusion**: Since \( 9 = 13 \) is a false statement, the equation has **no solution**.

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### 3. Equation: \( 8(x + 2) = 5x + 16 + 3x \)

**Step 1**: Simplify both sides.
- Left Side: \( 8(x + 2) = 8x + 16 \)
- Right Side: \( 5x + 16 + 3x = 8x + 16 \)

Thus, the equation simplifies to:
\[ 8x + 16 = 8x + 16 \]

**Step 2**: Isolate the variable.
Subtract \( 8x \) from both sides:
\[ 16 = 16 \]

**Conclusion**: Since \( 16 = 16 \) is a true statement, the equation has **infinitely many solutions**.

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### Summary of Results:
1. **Equation 1**: No solution.
2. **Equation 2**: No solution.
3. **Equation 3**: Infinitely many solutions.