Which property of equality justifies the step for solving the equation that is displayed? 3z2−5=10 3z2−5+5=10+5 3z2=15 (1 point) Responses Addition Property of Equality Addition Property of Equality Division Property of Equality Division Property of Equality Subtraction Property of Equality Subtraction Property of Equality Multiplication Property of Equality

Answer 1

Addition Property of Equality

Answer 2

Using the properties of equality, find the solution to the equation 2f5−85+2f=8 .(1 point) Responses 166 Start Fraction 16 over 6 End Fraction 4 4 12 12 83

Answer 3

To solve the equation 2f5−85+2f=8, we need to simplify and isolate the variable f.

First, we combine like terms on the left side of the equation:
2f + 2f = 4f
-85 + 5 = -80

So the equation becomes:
4f - 80 = 8

Next, we isolate the variable f by adding 80 to both sides of the equation:
4f - 80 + 80 = 8 + 80
4f = 88

Finally, we solve for f by dividing both sides of the equation by 4:
4f/4 = 88/4
f = 22

Therefore, the solution to the equation 2f5−85+2f=8 is f = 22.

Answer 4

How many solutions will the equation 23(6m+9)−6=4m have?(1 point) Responses two solutions two solutions no solution no solution infinite/many solutions infinite/many solutions one solution

Answer 5

The equation is 23(6m+9)−6=4m. To determine the number of solutions, we need to simplify and solve for m.

First, we distribute the 23 to the terms inside the parentheses:
138m + 207 - 6 = 4m

Next, we combine like terms:
138m + 201 = 4m

To isolate the variable m, we subtract 4m from both sides of the equation:
138m - 4m + 201 = 4m - 4m
134m + 201 = 0

At this point, it is evident that there is no value of m that satisfies the equation.

Therefore, the equation 23(6m+9)−6=4m has no solution.

Answer 6

Identify the pair of linear equations that have the same solution set.(1 point) Responses 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 4=k2+5 and 4j−143+5=3 4 equals Start Fraction k over 2 End Fraction plus 5 and Start Fraction 4 j minus 14 over 3 End Fraction plus 5 equals 3 4k−5=2k and 8j−2=3j 4 k minus 5 equals 2 k and 8 j minus 2 equals 3 j 2(3g+5)−2g=2(4−g) and −36h6=2

Answer 7

The pair of linear equations that have the same solution set is:

3(m+1)=10 and 6n+3=6−n

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+142 Equation C: 23v−3+83=−1 (1 point) Responses Payton Payton Tamara Tamara Haruki Haruki Krish

Answer 9

To determine who is correct about the linear equations, let's analyze each equation:

Equation A: 3(4−5h)−7=−12+2h
Equation B: 2(p−3)−1=6p+142
Equation C: 23v−3+83=−1

We can simplify each equation to find their solution sets:

Equation A:
12 - 15h - 7 = -12 + 2h
5h + 5 = 2h - 5
3h = -10
h = -10/3

Equation B:
2p - 6 - 1 = 6p + 142
2p - 7 = 6p + 142
-4p = 149
p = -149/4

Equation C:
23v - 3 + 83 = -1
23v + 80 = -1
23v = -81
v = -81/23

From the solutions obtained, we can see that Equation A and Equation B have different solutions.

Tamara is incorrect.

Similarly, Equation B and Equation C also have different solutions.

Payton is incorrect.

Krish's statement cannot be confirmed as the solutions obtained for the equations are different.

Haruki's statement is correct. None of the equations have the same solution.

Therefore, Haruki is correct about the linear equations.