SOLUTION: Find two numbers such that one-half their sum is 11 and one third their difference is 2

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Question 947338: Find two numbers such that one-half their sum is 11 and one third their difference is 2
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
Let x= one number, y= the other number
(1/3)(x-y)=2 multiply each side by 3
x-y=6 Add y to each side
x=y+6 Use this equation to substitute for x below
(1/2)(x+y)=11 multiply each side by 2
x+y=22 Substitute for x
(y+6)+y=22
2y+6=22 Subtract 6 from each side
2y=16 Divide each side by 2
y=8 ANSWER 1: one of the numbers is 8
x=y+6=8+6=14 ANSWER 2: The other number is 14
CHECK:
One half the sum is 11:
(1/2)(x+y)=11
(1/2)(14+8)=11
(1/2)(22)=11
11=11
One third difference is 2:
(1/3)(x-y)=2
(1/3)(14-8)=2
(1/3)(6)=2
2=2