Practice Problems
1. Match each equation with an equivalent equation. Some of the answer choices are not used.
e. 7x = 1
2. Mai says that equations A and B have the same solution.
- Equation A: - 3(x + 7) = 24
Equation B: x + 7 = -8
Which statement explains why this is true?
a. Adding 3 to both sides of Equation A gives x + 7 = -8.
b. Applying the distributive property to Equation A gives x + 7 = -8.
c. Subtracting 3 from both sides of Equation A gives x + 7 = -8.
d. Dividing both sides of Equation A by -3 gives x + 7 = -8.
3. Is 0 a solution to 2x + 10 = 4x + 10? Explain or show your reasoning.
4. Kiran says that a solution to the equation x + 4 = 20 must also be a solution to the
equation 5(x + 4) = 100.
Write a convincing explanation as to why this is true.

Let's tackle each of these problems one by one:

---

### 1. **Match each equation with an equivalent equation.**

Given equation:  
**e. \( 7x = 1 \)**

To solve for \( x \), divide both sides by 7:  
\[
x = \frac{1}{7}
\]
Look for an equation in the answer choices that also simplifies to \( x = \frac{1}{7} \).

---

### 2. **Mai says that equations A and B have the same solution.**
- **Equation A: \( -3(x + 7) = 24 \)**
- **Equation B: \( x + 7 = -8 \)**

Let's solve **Equation A** step by step:

Start with:
\[
-3(x + 7) = 24
\]

Divide both sides by \(-3\):
\[
x + 7 = \frac{24}{-3} = -8
\]

This gives the same equation as **Equation B: \( x + 7 = -8 \)**, meaning that the correct statement is **d**:  
**"Dividing both sides of Equation A by -3 gives \( x + 7 = -8 \)."**

---

### 3. **Is 0 a solution to \( 2x + 10 = 4x + 10 \)?**

We want to check if \( x = 0 \) satisfies this equation. Substitute \( x = 0 \) into the equation:

\[
2(0) + 10 = 4(0) + 10
\]
\[
10 = 10
\]

Since both sides of the equation are equal when \( x = 0 \), the answer is **yes**, \( 0 \) is a solution.

---

### 4. **Kiran says that a solution to the equation \( x + 4 = 20 \) must also be a solution to the equation \( 5(x + 4) = 100 \).**

To explain this, let’s solve both equations.

First, solve \( x + 4 = 20 \):
\[
x = 20 - 4 = 16
\]

Now, substitute \( x = 16 \) into the second equation:
\[
5(x + 4) = 100
\]
\[
5(16 + 4) = 5(20) = 100
\]

Since \( x = 16 \) satisfies both equations, Kiran is correct. The solution to \( x + 4 = 20 \) also works in \( 5(x + 4) = 100 \). This is because multiplying both sides of the first equation by 5 yields the second equation.

---

### Summary:
- Problem 1: Find the equivalent equation for \( 7x = 1 \).
- Problem 2: The correct answer is **d**.
- Problem 3: \( 0 \) is indeed a solution.
- Problem 4: The solution to \( x + 4 = 20 \) must also satisfy \( 5(x + 4) = 100 \) because both equations are related by multiplication.

---

Do you want more details or have any questions?  
Here are 5 related problems to expand on these:

1. Solve \( 8x = 2 \). What is \( x \)?
2. Solve the equation \( 2(x + 5) = 14 \).
3. Check if \( x = 3 \) is a solution to \( 4x - 5 = 7 \).
4. If \( 3(x - 2) = 12 \), what is \( x \)?
5. Are the equations \( x + 2 = 5 \) and \( 2x + 4 = 10 \) equivalent?

**Tip:** Always check solutions by plugging them back into the original equation.

Practice Problems 1. Match each equation with an equivalent equation. Some of the answer choices are not used. e. 7x = 1 2. Mai says that equations A and B have the same solution. - Equation A: - 3(x + 7) = 24 Equation B: x + 7 = -8 Which statement explains why this is true? a. Adding 3 to both sides of Equation A gives x + 7 = -8. b. Applying the distributive property to Equation A gives x + 7 = -8. c. Subtracting 3 from both sides of Equation A gives x + 7 = -8. d. Dividing both sides of Equation A by -3 gives x + 7 = -8. 3. Is 0 a solution to 2x + 10 = 4x + 10? Explain or show your reasoning. 4. Kiran says that a solution to the equation x + 4 = 20 must also be a solution to the equation 5(x + 4) = 100. Write a convincing explanation as to why this is true.

Explore step-by-step solutions to algebraic problems involving linear equations, distributive property, and equivalence. Learn how to match equations, check solutions, and understand the relation between multiple equations. Perfect for Grades 7-9 students.

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