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