Question

Determine what kind of solution the following equations have and why
3x + 12 + x = 8 + 4 + x
4 ( x + 1 ) = 4x + 1

Answer 1

Let's solve each equation step-by-step to determine the kind of solutions they have.

### 1. First Equation:

\[ 3x + 12 + x = 8 + 4 + x \]

First, simplify both sides:

- Left side: \(3x + x + 12 = 4x + 12\)
- Right side: \(8 + 4 + x = 12 + x\)

Now, we can write the equation as:

\[ 4x + 12 = 12 + x \]

Next, we will isolate \(x\) by moving all terms involving \(x\) to one side and constant terms to the other:

Subtract \(x\) from both sides:

\[ 4x - x + 12 = 12 \]

This simplifies to:

\[ 3x + 12 = 12 \]

Now, subtract 12 from both sides:

\[ 3x = 0 \]

Now, divide both sides by 3:

\[ x = 0 \]

This equation has a **unique solution**: \(x = 0\).

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### 2. Second Equation:

\[ 4(x + 1) = 4x + 1 \]

First, expand the left side:

\[ 4x + 4 = 4x + 1 \]

Next, we can try to simplify this equation. We will subtract \(4x\) from both sides:

\[ 4 = 1 \]

This results in a statement that is false.

The equation simplifies to 4 = 1 which is not true. Therefore, this system of equations has no solutions.

### Summary:

1. The first equation \(3x + 12 + x = 8 + 4 + x\) has a **unique solution**: \(x = 0\).
2. The second equation \(4(x + 1) = 4x + 1\) has **no solutions** because it leads to a contradiction (4 = 1).