Sample page from a 10-question algebra quiz focusing on linear equations and solution types.

Let's solve each problem step by step:

---

Problem 1:

Equation: \( 2x + 5 = 5x + 5 - 3x \)
Question: How many solutions does the equation have?

#### Solution:
1. Simplify the right-hand side of the equation:
\[
5x + 5 - 3x = (5x - 3x) + 5 = 2x + 5
\]
So the equation becomes:
\[
2x + 5 = 2x + 5
\]

2. Subtract \( 2x \) from both sides:
\[
2x + 5 - 2x = 2x + 5 - 2x
\]
This simplifies to:
\[
5 = 5
\]

3. The statement \( 5 = 5 \) is always true, which means the equation is an identity. Therefore, it has infinitely many solutions.

#### Answer:
\[
\boxed{\text{B}}
\]

---

Problem 2:

Equation: \( 2(x - 3) = \_\_ x - 4 \)
Question: What number can be placed in front of \( x \) to create an equation with no solutions?

#### Solution:
1. Expand the left-hand side:
\[
2(x - 3) = 2x - 6
\]
So the equation becomes:
\[
2x - 6 = \_\_ x - 4
\]

2. Let the unknown coefficient of \( x \) on the right-hand side be \( k \). The equation is:
\[
2x - 6 = kx - 4
\]

3. To have no solutions, the coefficients of \( x \) must be equal, but the constant terms must be different. Set the coefficients of \( x \) equal:
\[
2 = k
\]

4. Substitute \( k = 2 \) into the equation:
\[
2x - 6 = 2x - 4
\]

5. Subtract \( 2x \) from both sides:
\[
-6 = -4
\]
This is a contradiction, so the equation has no solutions when \( k = 2 \).

#### Answer:
\[
\boxed{\text{D}}
\]

---

Problem 3:

Equation: \( 5x - 4 = \_\_ x + 3 \)
Question: Which numbers could be used to create an equation with one solution?

#### Solution:
1. Let the unknown coefficient of \( x \) on the right-hand side be \( k \). The equation is:
\[
5x - 4 = kx + 3
\]

2. To have one solution, the coefficients of \( x \) must be different. Rearrange the equation:
\[
5x - kx = 3 + 4
\]
\[
(5 - k)x = 7
\]

3. For the equation to have one solution, \( 5 - k \neq 0 \). This means:
\[
k \neq 5
\]

4. Check the given options:
- \( k = -5 \): \( 5 - (-5) = 10 \neq 0 \) (valid)
- \( k = 4 \): \( 5 - 4 = 1 \neq 0 \) (valid)
- \( k = 3 \): \( 5 - 3 = 2 \neq 0 \) (valid)
- \( k = 5 \): \( 5 - 5 = 0 \) (invalid)

#### Answer:
\[
\boxed{\text{A, B, C}}
\]

---

Problem 4:

Equation: \( 16 = \frac{k}{11} \)
Question: How should you show your work to isolate the variable \( k \)?

#### Solution:
1. The equation is:
\[
16 = \frac{k}{11}
\]

2. To isolate \( k \), multiply both sides by 11:
\[
16 \times 11 = \frac{k}{11} \times 11
\]
\[
176 = k
\]

#### Answer:
\[
\boxed{\text{C}}
\]

---

Problem 5:

Equation: \( 6z + 3 - 4z = 9 \)
Question: Solve for \( z \).

#### Solution:
1. Combine like terms on the left-hand side:
\[
6z - 4z + 3 = 9
\]
\[
2z + 3 = 9
\]

2. Subtract 3 from both sides:
\[
2z + 3 - 3 = 9 - 3
\]
\[
2z = 6
\]

3. Divide both sides by 2:
\[
\frac{2z}{2} = \frac{6}{2}
\]
\[
z = 3
\]

#### Answer:
\[
\boxed{\text{C}}
\]

---

Problem 6:

Equation: \( r - 11 = 5 \)
Question: Solve for \( r \).

#### Solution:
1. Add 11 to both sides:
\[
r - 11 + 11 = 5 + 11
\]
\[
r = 16
\]

#### Answer:
\[
\boxed{\text{B}}
\]

---

Final Answers:

1. \(\boxed{\text{B}}\)
2. \(\boxed{\text{D}}\)
3. \(\boxed{\text{A, B, C}}\)
4. \(\boxed{\text{C}}\)
5. \(\boxed{\text{C}}\)
6. \(\boxed{\text{B}}\)