Add :
`a - 3b + 3, 2a + 5 - 3c, 6c - 15 + 6b`

Answer

To solve the problem of adding the algebraic expressions \(a - 3b + 3\), \(2a + 5 - 3c\), and \(6c - 15 + 6b\), we will follow these steps:

### Step 1: Write down the expressions
We start with the three expressions:
1. \(a - 3b + 3\)
2. \(2a + 5 - 3c\)
3. \(6c - 15 + 6b\)

### Step 2: Combine all expressions
We will combine all the expressions into one:
\[ (a - 3b + 3) + (2a + 5 - 3c) + (6c - 15 + 6b) \]

### Step 3: Rearrange the terms
Next, we will rearrange the terms by grouping similar types:
- **a terms**: \(a + 2a\)
- **b terms**: \(-3b + 6b\)
- **c terms**: \(-3c + 6c\)
- **constant terms**: \(3 + 5 - 15\)

So we can rewrite it as:
\[ (a + 2a) + (-3b + 6b) + (-3c + 6c) + (3 + 5 - 15) \]

### Step 4: Simplify each group
Now we will simplify each group:
- For the **a terms**: \(a + 2a = 3a\)
- For the **b terms**: \(-3b + 6b = 3b\)
- For the **c terms**: \(-3c + 6c = 3c\)
- For the **constant terms**: \(3 + 5 - 15 = 8 - 15 = -7\)

### Step 5: Combine the simplified terms
Now we can combine all the simplified terms:
\[ 3a + 3b + 3c - 7 \]

### Final Answer
Thus, the final result of adding the three expressions is:
\[ 3a + 3b + 3c - 7 \]

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