Which property of equality justifies the step for solving the equation that is displayed?

The step taken to solve the equation \( 3z/2 - 5 = 10 \) by adding 5 to both sides is justified by the **Addition Property of Equality**. This property states that if you add the same value to both sides of an equation, the equality is maintained.

Which property of equality justifies the step for solving the equation that is displayed?

3z/2 - 5 = 10
3z/2 - 5 + 5 = 10 + 5
3z/2 = 15

Addition Property of Equality

Multiplication Property of Equality

Subtraction Property of Equality

Division Property of Equality

The property of equality that justifies the step taken in the equation is the **Addition Property of Equality**. This property states that you can add the same value to both sides of an equation without changing the equality. In this case, adding 5 to both sides leads to \( 3z/2 = 15 \).

Using the properties of equality, find the solution to the equation 2f/5 - 8/5 + 2f = 8

8/3
12
16/6
4

To solve the equation \( \frac{2f}{5} - \frac{8}{5} + 2f = 8 \), we can follow these steps:

1. Combine like terms. First, express \( 2f \) with a common denominator of 5:
\[
2f = \frac{10f}{5}
\]
Now we can rewrite the equation as:
\[
\frac{2f}{5} - \frac{8}{5} + \frac{10f}{5} = 8
\]
Combining the terms on the left gives:
\[
\frac{(2f + 10f - 8)}{5} = 8
\]
This simplifies to:
\[
\frac{12f - 8}{5} = 8
\]

2. Multiply both sides by 5 to eliminate the fraction:
\[
12f - 8 = 40
\]

3. Add 8 to both sides:
\[
12f = 48
\]

4. Divide both sides by 12:
\[
f = 4
\]

The solution to the equation is \( f = 4 \).

Using the properties of equality, find the solution to the equation 2f/5 - 8/5 + 2f = 8

8/3
12
16/6
4

To solve the equation \( \frac{2f}{5} - \frac{8}{5} + 2f = 8 \), let's follow these steps:

1. **Rewrite \( 2f \)** to have a common denominator:
\[
2f = \frac{10f}{5}
\]

2. **Substitute this back into the equation:**
\[
\frac{2f}{5} - \frac{8}{5} + \frac{10f}{5} = 8
\]

3. **Combine the terms on the left:**
\[
\frac{2f + 10f - 8}{5} = 8
\]
\[
\frac{12f - 8}{5} = 8
\]

4. **Multiply both sides by 5** to eliminate the fraction:
\[
12f - 8 = 40
\]

5. **Add 8 to both sides:**
\[
12f = 48
\]

6. **Divide both sides by 12:**
\[
f = 4
\]

The solution to the equation is \( f = 4 \). Therefore, the correct answer is \( 4 \).

How many solutions will the equation 2/3 (6m + 9) - 6 = 4m have?

two solutions
one solution
no solution
infinite/many solutions

To determine how many solutions the equation

\[
\frac{2}{3}(6m + 9) - 6 = 4m
\]

has, let's simplify and solve it step-by-step.

1. **Distribute \( \frac{2}{3} \)** on the left side:
\[
\frac{2}{3} \cdot 6m + \frac{2}{3} \cdot 9 - 6 = 4m
\]
This simplifies to:
\[
4m + 6 - 6 = 4m
\]

2. The \( +6 \) and \( -6 \) cancel out:
\[
4m = 4m
\]

This means the equation \( 4m = 4m \) is a true statement for all values of \( m \).

Since both sides of the equation are the same, this implies that any value of \( m \) satisfies the equation.

Thus, there are **infinitely many solutions** to the equation.

The answer is **infinite/many solutions**.

How many solutions will the equation 2/3 (6m + 9) - 6 = 4m have?

two solutions
one solution
no solution
infinite/many solutions

To determine how many solutions the equation

\[
\frac{2}{3}(6m + 9) - 6 = 4m
\]

has, let's proceed with solving it.

1. **Distribute \( \frac{2}{3} \)** on the left side:
\[
\frac{2}{3} \cdot 6m + \frac{2}{3} \cdot 9 - 6 = 4m
\]
This simplifies to:
\[
4m + 6 - 6 = 4m
\]

2. The \( +6 \) and \( -6 \) cancel out:
\[
4m = 4m
\]

This tells us that the equation is true for all values of \( m \).

Since both sides of the equation are the same, it implies that any value of \( m \) satisfies the equation.

Thus, there are **infinitely many solutions** to the equation.

The answer is **infinite/many solutions**.

Identify the pair of linear equations that have the same solution set.

4k - 5 = 2k and 8j - 2 = 3j
4 = k/2 + 5 and 4j - 14/3 + 5 = 3
2(3g + 5) - 2g = 2 (4 - g) and - 36h/6 = 2
3(m + 1) = 10 and 6n + 3 = 6 - n

To identify the pair of linear equations that have the same solution set, let's solve each pair of equations one by one.

### 1. First Pair: \(4k - 5 = 2k\) and \(8j - 2 = 3j\)

**Solve the first equation:**
\[
4k - 5 = 2k
\]
Subtract \(2k\) from both sides:
\[
4k - 2k - 5 = 0 \implies 2k - 5 = 0 \implies 2k = 5 \implies k = \frac{5}{2}
\]

**Solve the second equation:**
\[
8j - 2 = 3j
\]
Subtract \(3j\) from both sides:
\[
8j - 3j - 2 = 0 \implies 5j - 2 = 0 \implies 5j = 2 \implies j = \frac{2}{5}
\]

**Conclusion on the first pair**: \( k = \frac{5}{2}, j = \frac{2}{5} \). **Not the same solution set.**

---

### 2. Second Pair: \(4 = \frac{k}{2} + 5\) and \(\frac{4j - 14}{3} + 5 = 3\)

**Solve the first equation:**
\[
4 = \frac{k}{2} + 5
\]
Subtract 5 from both sides:
\[
4 - 5 = \frac{k}{2} \implies -1 = \frac{k}{2} \implies k = -2
\]

**Solve the second equation:**
\[
\frac{4j - 14}{3} + 5 = 3
\]
Subtract 5 from both sides:
\[
\frac{4j - 14}{3} = 3 - 5 \implies \frac{4j - 14}{3} = -2
\]
Multiply both sides by 3:
\[
4j - 14 = -6 \implies 4j = -6 + 14 \implies 4j = 8 \implies j = 2
\]

**Conclusion on the second pair**: \( k = -2, j = 2\). **Not the same solution set.**

---

### 3. Third Pair: \(2(3g + 5) - 2g = 2(4 - g)\) and \(- \frac{36h}{6} = 2\)

**Solve the first equation:**
\[
2(3g + 5) - 2g = 2(4 - g)
\]
Distribute:
\[
6g + 10 - 2g = 8 - 2g
\]
Combine like terms:
\[
4g + 10 = 8 - 2g
\]
Add \(2g\) to both sides:
\[
4g + 2g + 10 = 8 \implies 6g + 10 = 8 \implies 6g = -2 \implies g = -\frac{1}{3}
\]

**Solve the second equation:**
\[
-\frac{36h}{6} = 2 \implies -6h = 2 \implies h = -\frac{1}{3}
\]

**Conclusion on the third pair**: \( g = -\frac{1}{3}, h = -\frac{1}{3}\). **Same solution set.**

---

### 4. Fourth Pair: \(3(m + 1) = 10\) and \(6n + 3 = 6 - n\)

**Solve the first equation:**
\[
3(m + 1) = 10
\]
Distribute:
\[
3m + 3 = 10 \implies 3m = 7 \implies m = \frac{7}{3}
\]

**Solve the second equation:**
\[
6n + 3 = 6 - n
\]
Rearranging gives:
\[
6n + n = 6 - 3 \implies 7n = 3 \implies n = \frac{3}{7}
\]

**Conclusion on the fourth pair**: \( m = \frac{7}{3}, n = \frac{3}{7}\). **Not the same solution set.**

---

### Summary:
The pair of linear equations that have the same solution set is the **third pair:**
- \(2(3g + 5) - 2g = 2(4 - g)\) and \(-\frac{36h}{6} = 2\) both yield the solution of \( g = -\frac{1}{3} \) and \( h = -\frac{1}{3} \).

Tamara, Payton, Krish, and Haruki are analyzing the solution sets to the following linear equations. Tamara says that Equation A and Equation C have the same solution. Payton says Equation B and Equation C have the same solution. Krish says all three have the same solution. Haruki says that none of the equations have the same solution. Identify who is correct about the linear equations.

Equation A: 3 (4 - 5h) - 7 = -12 + 2h
Equation B: 2 (p - 3) - 1 = 6p + 14/2
Equation C: 2/3 * v - 3 + 8/3 = -1
(1 point)
Haruki
Krish
Payton
Tamara

Let's analyze the equations one by one to determine their solution sets.

### Equation A:
\[
3(4 - 5h) - 7 = -12 + 2h
\]

1. Distributing \(3\):
\[
12 - 15h - 7 = -12 + 2h
\]
2. Simplifying the left side:
\[
5 - 15h = -12 + 2h
\]
3. Adding \(15h\) to both sides:
\[
5 = -12 + 17h
\]
4. Adding \(12\) to both sides:
\[
17 = 17h
\]
5. Dividing by \(17\):
\[
h = 1
\]

### Equation B:
\[
2(p - 3) - 1 = \frac{6p + 14}{2}
\]
1. Distributing \(2\):
\[
2p - 6 - 1 = 3p + 7
\]
2. Simplifying the left side:
\[
2p - 7 = 3p + 7
\]
3. Subtracting \(2p\) from both sides:
\[
-7 = p + 7
\]
4. Subtracting \(7\) from both sides:
\[
-14 = p
\]

### Equation C:
\[
\frac{2}{3}v - 3 + \frac{8}{3} = -1
\]
1. Combining the constants on the left:
\[
\frac{2}{3}v + \frac{5}{3} = -1
\]
2. Subtracting \(\frac{5}{3}\) from both sides:
\[
\frac{2}{3}v = -1 - \frac{5}{3}
\]
Convert \(-1\) to \(-\frac{3}{3}\):
\[
\frac{2}{3}v = -\frac{3}{3} - \frac{5}{3} = -\frac{8}{3}
\]
3. Multiplying by \(\frac{3}{2}\):
\[
v = -4
\]

### Summary of Solutions:
- Equation A: \(h = 1\)
- Equation B: \(p = -14\)
- Equation C: \(v = -4\)

### Comparison of Statements:
- **Tamara**: Says Equation A and Equation C have the same solution (False; A has \(h=1\), C has \(v=-4\)).
- **Payton**: Says Equation B and Equation C have the same solution (False; B has \(p=-14\), C has \(v=-4\)).
- **Krish**: Says all three have the same solution (False).
- **Haruki**: Says none of the equations have the same solution (True).

### Conclusion:
The correct answer is **Haruki**.