Find the numbers whose sum is 15 and the second number is twice the first decreased 6.

Answer 1

x+y=15

y = 2x-6

x + 2x-6 = 15

3x -6 = 15

3x = 21

x=7

7 and 8, where 8 = 2(7)-6 = 14-6

so yes, the official answer is 7 and 8

Answer 2

Hi Taylor T

First number is 7

Second number is 8

Let x = first number

Let y = second number

According the data you are given

x + y = 15

And

y = 2x - 6

Substitute 2x - 6 for y then solve for x

x + y = 15

x + 2x - 6 =15

3x - 6 = 15

Add 6 to both sides of the equation

3x = 15 + 6

3x = 21

Divide both sides by 3

x = 7

y = 2x - 6 = 2(7) - 6 = 14 - 6 = 8

Also since

x + y = 15

y = 15 - x= 15 - 7 = 8

Hope this helps

Answer 3

S=2F-6

S+F = 15

2F-6 + F = 15

3F = 21

F = 7

S =2(7)-6 = 8

Answer 4

Because you are looking for 2 unknown numbers, you can label

x = 1st number

y = 2nd number

Equation 1: x + y = 15

(numbers whose sum is 15)

Equation 2: y = 2x - 6

(second number is twice the first decreased by 6)

You have three options to solve this system of linear equations: substitution, elimination, or graphing.

Because y = 2x - 6 is already isolated for you, substitution is preferred.

Therefore,

Step 1: Setup

x + y = 15

y = 2x - 6

Step 2: Substitution

x + (2x - 6) = 15

Step 3: Solve for x

3x - 6 = 15

3x = 21

x = 7

Step 4: Solve for y

x + y = 15

(7) + y = 15

y = 8

ANSWER: The numbers whose sum is 15 and the second number is twice the first decreased by 6 are 7 and 8.

Hope that helps :)