Solve `3(x+4) + 1 lt 2(3x+1) + 15`

Answer

To solve the inequality \( 3(x + 4) + 1 < 2(3x + 1) + 15 \), we will follow these steps:

### Step 1: Expand both sides of the inequality
First, we will distribute the terms inside the parentheses.

\[ 3(x + 4) + 1 < 2(3x + 1) + 15 \]

Expanding both sides:

\[ 3x + 12 + 1 < 6x + 2 + 15 \]

### Step 2: Combine like terms
Now, we will combine the constants on both sides.

\[ 3x + 13 < 6x + 17 \]

### Step 3: Rearrange the inequality
Next, we will move all terms involving \(x\) to one side and constant terms to the other side.

Subtract \(3x\) from both sides:

\[ 13 < 6x - 3x + 17 \]

This simplifies to:

\[ 13 < 3x + 17 \]

Now, subtract 17 from both sides:

\[ 13 - 17 < 3x \]

This simplifies to:

\[ -4 < 3x \]

### Step 4: Divide by 3
Now, we will divide both sides by 3 to solve for \(x\):

\[ \frac{-4}{3} < x \]

### Step 5: Write the final solution
This can be rewritten as:

\[ x > -\frac{4}{3} \]

In interval notation, the solution is:

\[ x \in \left(-\frac{4}{3}, \infty\right) \]