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if the average of two numbers is 3Y+6. If one of the numbers is 7 what is the other number.

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Aug 31, 2011
Unknown Number in Known Average
by: Staff

The question:

if the average of two numbers is 3Y+6. If one of the numbers is 7 what is the other number.


The answer:

n₁ = first number = 7
n₂ = second number = unknown
A = average of the two numbers = 3Y + 6

The formula used to compute the average of any two numbers is:

A = (n₁ + n₂)/2

Substitute all known values in the formula

3Y + 6 = (7 + n₂)/2

Solve for unknown number "n₂".

To remove the denominator of 2 from the right side of the equation, multiply each side of the equation by 2

(3Y + 6)*2 = [(7 + n₂)/2]*2

Use the distributive law to eliminate the parentheses on the left side of the equation

3Y*2 + 6*2 = [(7 + n₂)/2]*2

3*2*Y + 6*2 = [(7 + n₂)/2]*2

6Y + 12 = [(7 + n₂)/2]*2

Use the distributive law to eliminate the denominator of 2 on the right side of the equation

6Y + 12 = (7 + n₂)*(2/2)

6Y + 12 = (7 + n₂)*(1)

6Y + 12 = (7 + n₂)

6Y + 12 = 7 + n₂

Subtract 7 from each side of the equation to eliminate the 7 from the right side of the equation

6Y + 12 - 7 = 7 + n₂ - 7

6Y + (12 ? 7) = 7 + n₂ - 7

6Y + 5 = 7 + n₂ - 7

6Y + 5 = 7 ? 7 + n₂

6Y + 5 = (7 ? 7) + n₂

6Y + 5 = (0) + n₂

6Y + 5 = n₂

The final answer is: n₂ = 6Y + 5


Check the solution by substituting the values of n₁ and n₂ into the formula for computing the average of two numbers

A = (n₁ + n₂)/2

3Y + 6 = [7 + (6Y + 5)]/2

3Y + 6 = [7 + 6Y + 5]/2

3Y + 6 = [6Y + 7 + 5]/2

3Y + 6 = [6Y + (7 + 5)]/2

3Y + 6 = [6Y + (12)]/2

3Y + 6 = [6Y + 12]/2

3Y + 6 = [(6Y/1) + (12/1)]*(1/2)

3Y + 6 = (6Y/1)*(1/2) + (12/1)*(1/2)

3Y + 6 = (Y/1)*( 6/2) + (12/2)

3Y + 6 = (Y/1)*( 3) + (6)

3Y + 6 = (Y*3) + (6)

3Y + 6 = (3Y) + (6)

3Y + 6 = 3Y + 6, OK → n₂ = 6Y + 5 is a valid solution




Thanks for writing.

Staff
www.solving-math-problems.com



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