Question

1. 6.8 - 4.2b = 5.6b - 3
2. 1/3 + 2/3m = 2/3m - 2/3
3. -2(5.25 + 6.2x) = 4(-3.1x + 2.68)
4. 1/2r + 6 = 3 - 2r

Answer 1

Is #1 an identity or no solution?

Answer 2

Let's solve each equation step-by-step.

### Equation 1:
\[ 6.8 - 4.2b = 5.6b - 3 \]

1. Move all terms involving \( b \) to one side and constant terms to the other side:
\[ 6.8 + 3 = 5.6b + 4.2b \]
\[ 9.8 = 9.8b \]

2. Divide both sides by 9.8:
\[ b = 1 \]

### Equation 2:
\[ \frac{1}{3} + \frac{2}{3}m = \frac{2}{3}m - \frac{2}{3} \]

1. Move \( \frac{2}{3}m \) to the left side:
\[ \frac{1}{3} + \frac{2}{3}m - \frac{2}{3}m = -\frac{2}{3} \]
\[ \frac{1}{3} = -\frac{2}{3} \] (invalid, since \(\frac{1}{3} \neq -\frac{2}{3}\))

Thus, there are no solutions for this equation.

### Equation 3:
\[ -2(5.25 + 6.2x) = 4(-3.1x + 2.68) \]

1. Distribute on both sides:
\[ -2 \times 5.25 - 2 \times 6.2x = 4 \times -3.1x + 4 \times 2.68 \]
\[ -10.5 - 12.4x = -12.4x + 10.72 \]

2. Add \( 12.4x \) to both sides:
\[ -10.5 = 10.72 \]

This is also an invalid statement, implying that there are no solutions.

### Equation 4:
\[ \frac{1}{2}r + 6 = 3 - 2r \]

1. Move all terms involving \( r \) to one side and constants to the other side:
\[ \frac{1}{2}r + 2r = 3 - 6 \]
\[ \frac{1}{2}r + 4/2r = -3 \]
\[ \frac{5}{2}r = -3 \]

2. Multiply both sides by \( \frac{2}{5} \):
\[ r = -\frac{3 \times 2}{5} \]
\[ r = -\frac{6}{5} \]
\[ r = -1.2 \]

### Summary of Solutions:
1. \( b = 1 \)
2. No solution.
3. No solution.
4. \( r = -1.2 \)

Answer 3

To determine whether the first equation \(6.8 - 4.2b = 5.6b - 3\) is an identity or has no solution, we analyze the steps we took to solve it.

### Original Equation:
\[ 6.8 - 4.2b = 5.6b - 3 \]

### Combining Like Terms:
1. Move all terms involving \( b \) to one side:
\[ 6.8 + 3 = 5.6b + 4.2b \]
\[ 9.8 = 9.8b \]

2. Divide both sides by 9.8:
\[ b = 1 \]

### Analysis:
1. The equation simplifies to \( 9.8 = 9.8b \), which means there is a specific value for \( b\) (in this case, \( b = 1 \)) that satisfies the equation.
2. Since we found a solution for \( b \), the equation is **not** an identity (which would hold for all values of \( b \)) and is **not** an empty set (which would indicate no solutions).

### Conclusion:
The equation has a specific solution:

**\( b = 1 \)**

It is not an identity (like \( 0 = 0 \)), and it is not a no-solution scenario (like \( 0 = 5 \)).