Question

Determine what kind of solution the following equations have and why. (2 points) 1) 3z+12+2=8+4+z has 2) 4 (z+1)=4z+1 has Item 1 Item 2 Item 3 Item 4 Item 5 Item 6 Item 7 Item 8 :: One solution because when you solve for x, the result is x = 0 :: One solution because when you solve for x, the result is x = 6 No solution because you end up with a false statement - no value of x will make the equation true :: Infinite solutions because you end up with a true statement - any value of x will make the equation true

Answer 1

give me only the answers

Answer 2

Let's analyze each equation systematically to determine the type of solution they have.

1) **Equation: \( 3z + 12 + 2 = 8 + 4 + z \)**

First, simplify both sides:

Left-hand side:
\( 3z + 12 + 2 = 3z + 14 \)

Right-hand side:
\( 8 + 4 + z = 12 + z \)

Now the equation looks like:
\[ 3z + 14 = 12 + z \]

Next, let's isolate \( z \). Subtract \( z \) from both sides:
\[ 3z - z + 14 = 12 \]
\[ 2z + 14 = 12 \]

Now, subtract 14 from both sides:
\[ 2z = 12 - 14 \]
\[ 2z = -2 \]

Finally, divide by 2:
\[ z = -1 \]

Since we found a specific value for \( z\), this means this equation has **one solution**.

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2) **Equation: \( 4(z + 1) = 4z + 1 \)**

First, expand the left-hand side:
\[ 4(z + 1) = 4z + 4 \]

So now the equation looks like:
\[ 4z + 4 = 4z + 1 \]

Next, we subtract \( 4z \) from both sides:
\[ 4 = 1 \]

This results in a **false statement**. Since there is no value of \( z \) that can make \( 4 = 1 \) true, this equation has **no solution**.

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### Summary:

- **Item 1:** One solution because when you solve for \( z \), the result is \( z = -1 \).
- **Item 2:** No solution because you end up with a false statement - no value of \( z \) will make the equation true.

Answer 3

1) One solution.
2) No solution.