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the sum of the digit of a 2 digit number is 11. if the digit is reversed the resulting number is seven more than twice the original number. what is the original number

Sagot :

Original number   sum of digits  twice the orig#+7   reversed   
29                         11                  65                       ≠   92
38                         11                  83                       =  83
38 is my final answer
Shinalcantaraidn
let 'x' be at the tens place of the two-digit number
    'y' be at the ones place of the two-digit number
then the number would be x(10) + y
why i multiply x by 10? it's in the tens place remember? then the value would be multiplied by 10
since the sum of the two is 11 then you'll have it as:
x + y = 11  ---equation 1
for the second statement you'll have the equivalent mathematical statement as:
y(10) + x = 2(x(10) + y) + 7
10y + x = 20x + 2y + 7
10y - 2y = 20x - x + 7
8y = 19x + 7  ----equation 2
rearranging equation 1 you'll have
x + y =11
y = 11 - x  ---equation 3
substitute equation 3 to equation 2
8y = 19x + 7
8(11-x) = 19x + 7
88 - 8x = 19x +7
19x + 8x = 88 - 7
27x = 81
x = 3
substituting the value of x to equation 3 you'll have
y = 11 - x
y = 11 - 3
y = 8
the original number is defined by 10x + y as stated above then you'll have it as:
10x + y = 10(3) + 8 
            = 38
therefore the original number is 38
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