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The image you've uploaded is titled **8.EE.C.7a ADDITIONAL PRACTICE**, and it contains algebraic equations and related questions. I'll go through each of them, providing solutions and explanations.

### 1) **What is the value of x in the equation**  
   \( 7 = 5x - 33x \)

#### Solution:
Simplify the equation:  
\[
7 = 5x - 33x
\]
Combine like terms:  
\[
7 = -28x
\]
Divide both sides by \(-28\):
\[
x = \frac{7}{-28} = -\frac{1}{4}
\]

Thus, the value of \(x\) is \(-\frac{1}{4}\).

---

### 2) **What value of x makes the equation \( 2 + 9x = 11 - 4b \) true?**

There seems to be a typo, as this equation involves both \(x\) and \(b\). For solving \(x\), I would assume the equation should be \(2 + 9x = 11 - 4x\). Let's solve it under that assumption.

#### Solution:
\[
2 + 9x = 11 - 4x
\]
Add \(4x\) to both sides:
\[
2 + 13x = 11
\]
Subtract 2 from both sides:
\[
13x = 9
\]
Divide both sides by 13:
\[
x = \frac{9}{13}
\]

Thus, \(x = \frac{9}{13}\).

---

### 3) **Complete the equation below for A and B so that there are infinitely many solutions:**
\[
7 - 5 = Ax + B
\]

For there to be infinitely many solutions, the equation must be consistent for all values of \(x\). This typically means that both sides of the equation must be identical. So, let's set \(A = 0\) and \(B = 2\) (since \(7 - 5 = 2\)).

Therefore, the completed equation is:
\[
7 - 5 = 0x + 2
\]

---

### 4) **How many solutions does the equation \( 3 - x = 3 \) have? If one, provide the solution.**

#### Solution:
Solve for \(x\):
\[
3 - x = 3
\]
Subtract 3 from both sides:
\[
-x = 0
\]
Multiply by \(-1\):
\[
x = 0
\]

Thus, there is **one solution**, and \(x = 0\).

---

### 5) **How many solutions does the equation \( 3x + 1 = x + 1 \) have? If one, provide the solution.**

#### Solution:
Simplify the equation:
\[
3x + 1 = x + 1
\]
Subtract \(x\) and 1 from both sides:
\[
2x = 0
\]
Divide by 2:
\[
x = 0
\]

Thus, there is **one solution**, and \(x = 0\).

---

### 6) **How many solutions does the equation \( 2x + 1 = 3 - 2x \) have? If one, provide the solution.**

#### Solution:
Add \(2x\) to both sides:
\[
4x + 1 = 3
\]
Subtract 1 from both sides:
\[
4x = 2
\]
Divide by 4:
\[
x = \frac{1}{2}
\]

Thus, there is **one solution**, and \(x = \frac{1}{2}\).

---

Would you like further details or have any questions? Here are some additional questions related to these problems:

1. How would you solve an equation with variables on both sides and a constant?
2. Can you explain the process of determining how many solutions an equation has?
3. What does it mean for an equation to have no solution or infinitely many solutions?
4. How do you verify the solution of an equation once you've found it?
5. How do different operations (like multiplying both sides by a constant) affect the solution of an equation?

**Tip:** When solving equations, always check your solution by substituting it back into the original equation to ensure both sides are equal.

The uploaded worksheet contains several algebraic problems related to solving equations and finding the number of solutions.

This worksheet contains algebra problems focusing on solving linear equations, determining the number of solutions, and completing equations for infinite solutions. Suitable for 8th-grade students.

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