Question

Question

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

(1 point)
Responses

Subtraction Property of Equality
Subtraction Property of Equality

Multiplication Property of Equality
Multiplication Property of Equality

Addition Property of Equality
Addition Property of Equality

Division Property of Equality

Answer 1

The step you provided:

\[ 3z/2 - 5 + 5 = 10 + 5 \]

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. Therefore, the correct response is:

Addition Property of Equality.

Answer 2

Using the properties of equality, find the solution to the equation 2f/5−8/5+2f=8 .(1 point)
Responses

16/6
Start Fraction 16 over 6 End Fraction

12
12

8/3
Start Fraction 8 over 3 End Fraction

4

Answer 3

To solve the equation

\[ \frac{2f}{5} - \frac{8}{5} + 2f = 8, \]

first, let's combine the like terms on the left side. We can rewrite \(2f\) as \(\frac{10f}{5}\) to have a common denominator:

\[ \frac{2f}{5} + \frac{10f}{5} - \frac{8}{5} = 8. \]

Now, combine the \(f\) terms:

\[ \frac{2f + 10f - 8}{5} = 8. \]

This simplifies to:

\[ \frac{12f - 8}{5} = 8. \]

Next, to eliminate the fraction, multiply both sides by 5:

\[ 12f - 8 = 40. \]

Now, add 8 to both sides:

\[ 12f = 48. \]

Finally, divide both sides by 12:

\[ f = 4. \]

Thus, the solution is:

4.

Answer 4

How many solutions will the equation 2/3(6m+9)−6=4m have?(1 point)
Responses

one solution
one solution

no solution
no solution

two solutions
two solutions

infinite/many solutions

Answer 5

To determine how many solutions the equation

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

has, we'll simplify it step-by-step:

1. Distribute \(\frac{2}{3}\):

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

This simplifies to:

\[ 4m + 6 - 6 = 4m. \]

2. Combine like terms on the left side:

\[ 4m = 4m. \]

At this point, we can see that the equation simplifies to \(4m = 4m\).

This means that both sides are equal for any value of \(m\), indicating that there are infinite solutions.

Thus, the answer is:

infinite/many solutions.

Answer 6

Identify the pair of linear equations that have the same solution set.(1 point)
Responses

2(3g+5)−2g=2(4−g) and −36h/6=2
2 left parenthesis 3 g plus 5 right parenthesis minus 2 g equals 2 left parenthesis 4 minus g right parenthesis and Start Fraction negative 36 h over 6 End Fraction equals 2

3(m+1)=10 and 6n+3=6−n
3 left parenthesis m plus 1 right parenthesis equals 10 and 6 n plus 3 equals 6 minus n

4k−5=2k and 8j−2=3j
4 k minus 5 equals 2 k and 8 j minus 2 equals 3 j

4=k2+5 and 4j−14/3+5=3

Answer 7

To determine which pair of linear equations has the same solution set, we can simplify and solve each of the pairs given. Let's analyze them one by one.

1. Pair 1: \[ 2(3g+5)-2g=2(4-g) \] First, simplify both sides: \[ 6g + 10 - 2g = 8 - 2g \] Combining like terms gives: \[ 4g + 10 = 8 - 2g \implies 6g = -2 \implies g = -\frac{1}{3} \]

   The second part: \[ -\frac{36h}{6} = 2 \implies -6h = 2 \implies h = -\frac{1}{3} \]

   Solution set: The first equation's solution is \(g = -\frac{1}{3}\) and the second one is \(h = -\frac{1}{3}\).

2. Pair 2: \[ 3(m+1)=10 \] Simplifying, \[ 3m + 3 = 10 \implies 3m = 7 \implies m = \frac{7}{3} \]

   Second part: \[ 6n + 3 = 6 - n \implies 7n = 3 \implies n = \frac{3}{7} \]

   Solution set: Different solutions.

3. Pair 3: \[ 4k - 5 = 2k \implies 2k = 5 \implies k = \frac{5}{2} \]

   Second part: \[ 8j - 2 = 3j \implies 5j = 2 \implies j = \frac{2}{5} \]

   Solution set: Different solutions.

4. Pair 4: \[ 4 = k^2 + 5 \implies k^2 = -1 \implies \text{no real solutions} \]

   Second part: \[ 4j - \frac{14}{3} + 5 = 3 \implies 4j = 3 - 5 + \frac{14}{3} \] Solving this is more complex, but ultimately also does not yield real solutions.

Conclusion:

The only pair with the same solution set is the first one:

2(3g+5)−2g=2(4−g) and −36h/6=2, both yielding the solution \(g = -\frac{1}{3}\) and \(h = -\frac{1}{3}\).

Answer 8

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: 2/3v−3+8/3=−1

(1 point)
Responses

Krish
Krish

Tamara
Tamara

Haruki
Haruki

Payton

Answer 9

Let's solve each equation to determine the solution sets and check who is correct.

Equation A:

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

Equation B:

\[ 2(p - 3) - 1 = \frac{6p + 14}{2} \] Distributing on the left side: \[ 2p - 6 - 1 = 3p + 7 \] Combining like terms: \[ 2p - 7 = 3p + 7 \] Subtracting \(2p\) from both sides: \[ -7 = p + 7 \] Subtracting 7 from both sides: \[ -14 = p \]

Equation C:

\[ \frac{2}{3}v - 3 + \frac{8}{3} = -1 \] Combining the constant terms (\(-3\) and \(\frac{8}{3}\)): Convert \(-3\) to a fraction: \[ -3 = -\frac{9}{3} \] So, \[ \frac{2}{3}v - \frac{9}{3} + \frac{8}{3} = -1 \] This simplifies to: \[ \frac{2}{3}v - \frac{1}{3} = -1 \] Now, adding \(\frac{1}{3}\) to both sides: \[ \frac{2}{3}v = -1 + \frac{1}{3} \] Finding a common denominator for \(-1\): \[ -1 = -\frac{3}{3}, \quad \text{so} \quad -\frac{3}{3} + \frac{1}{3} = -\frac{2}{3} \] Thus: \[ \frac{2}{3}v = -\frac{2}{3} \] Now, multiply both sides by \(\frac{3}{2}\): \[ v = -1 \]

Summary of Solutions:

• Equation A: \(h = 1\)
• Equation B: \(p = -14\)
• Equation C: \(v = -1\)

Conclusion:

• Tamara states that Equation A (\(h = 1\)) and Equation C (\(v = -1\)) have the same solution: Incorrect.
• Payton states that Equation B (\(p = -14\)) and Equation C (\(v = -1\)) have the same solution: Incorrect.
• Krish states that all three equations have the same solution: Incorrect.
• Haruki states that none of the equations have the same solution: Correct.

Thus, Haruki is the correct one.