A two digit number is seven times the sum of it's digits. The number formed by reversing the digits is 6 more than half of the original number. Find the difference of the digits of the given number .
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2 answers

a = first number
b = second number
Your number = 10 a + b
A two digit number is seven times the sum of it's digits meaans:
10 a + b = 7 ( a + b )
The number formed by reversing the digits is 10 b + a
The number formed by reversing the digits is 6 more than half of the original number means:
10 b + a = ( 10 a + b ) / 2 + 6
Now you must solve system of two equations:
10 a + b = 7 ( a + b )
10 b + a = ( 10 a + b ) / 2 + 6
____________________________
First equation:
10 a + b = 7 a + 7 b
Subtract 7 a to both sides
3 a + b = 7 b
Subtract b to both sides
3 a = 6 b
Divide both sides by 3
a = 2 b
Replace a with 2 b in equation:
10 b + a = ( 10 a + b ) / 2 + 6
10 b + 2 b = ( 10 • 2 b + b ) / 2 + 6
12 b = ( 20 b + b ) / 2 + 6
12 b = 21 b / 2 + 6
Multiply both sides by 2
24 b = 21 b + 12
Subtract 21 b to both sides
3 b = 12
Divide both sides by 3
b = 4
a = 2 b
a = 2 • 4
a = 8
a  b = 8  4 = 4
Check result:
A two digit number is seven times the sum of it's digits.
Your number = 10 a + b = 10 • 8 + 4 = 84
Sum of digits = 8 + 4 = 12
84 = 7 •12
The number formed by reversing the digits is 6 more than half of the original number.
The number formed by reversing the digits is 10 b + a = 10 • 4 + 8 = 48
Half of the original number =
84 / 2 = 42
48 = 42 + 6 👍
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answered by Bosnian 
a = first digit
b = second digit 👍
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answered by Bosnian
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