Question

Determine what kind of solution the following equations have and why.

3x+12+x=8+4+x has_____________________
4(x+1)=4x+1 has_____________________

Answer 1

Let's analyze each equation one at a time.

### Equation 1:
\( 3x + 12 + x = 8 + 4 + x \)

First, combine like terms on both sides:

**Left side:**
\( 3x + x + 12 = 4x + 12 \)

**Right side:**
\( 8 + 4 + x = 12 + x \)

Now, the equation becomes:
\[ 4x + 12 = 12 + x \]

Next, let's isolate \( x \) by moving \( x \) to the left side and \( 12 \) to the right side:
\[ 4x - x = 12 - 12 \]
\[ 3x = 0 \]

Now, divide both sides by \( 3 \):
\[ x = 0 \]

Since we found a single value for \( x \), the equation has **exactly one solution**.

### Equation 2:
\( 4(x + 1) = 4x + 1 \)

First, distribute on the left side:
\[ 4x + 4 = 4x + 1 \]

Now, we can see that the \( 4x \) terms on both sides are the same. So, the equation simplifies to:
\[ 4 = 1 \]

This statement is false. Since there’s no value of \( x \) that can make this statement true, the equation has **no solutions**.

### Summary:
1. \( 3x + 12 + x = 8 + 4 + x \) has **exactly one solution**.
2. \( 4(x + 1) = 4x + 1 \) has **no solutions**.