Question 1119269: A number consist of two digit who's sum of 9 if 9 is subtracted from the number the digit interchange there place find the number
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! a number consists of 2 digits whose sum is 9.
if 9 is subtracted from the number, the digits interchange.
find the number.
let a = the first digits and let b equal the second digit.
the value of the number is 10a + b.
the sum of the digits is 9, therefore a + b = 9.
if you substract 9 from the number, then the digits interchange, therefore:
10a + b - 9 = 10b + a
subtract a from both sides of this equation and subtract b from both sides of this equation to get:
9a - 9 = 9b
since a + b = 9, solve for a to get a = 9 - b.
in the equation of 9a - 9 = 9b, replace a with 9 - b to get:
9 * (9 - b) - 9 = 9b
simplify to get:
81 - 9b - 9 = 9b
add 9b to both sides of the equaiton and combine like terms to get:
72 = 18b.
solve for b to get b = 72/18 = 4.
since a + b = 9, then a must be 5.
you have a = 5 and b = 4.
the number is 54.
subtract 9 from it and the number becomes 45.
your solution is that the number is equal to 54.
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