# college problem solving

a certain two-digit number is 6 less than the sum of its tens digit and 7 times its units digit. if the digits are reversed the number is increase by 27. fin the original two-digit numbers

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1. let the tens digit be x and the unit digit be y

then the number is 10x + y

10x+y = (x+7y) - 6
9x - 6y = -6
3x - 2y = -2 , #1

the number reversed would be 10y + x

so 10y+x = 10x+y + 27
9x - 9y = -27
3x - 3y = -9 , #2

#1 - #2
y = 7
then from #1, x = 4

the number is 47

check:
reverse it --- 74
is 74 -47 = 27 , so far so good

what is the sum of its tens digit and 7 times its unit digit
4 + 7(7) = 53
is 53 greater than 47 by 6 ? YEAHHHH

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posted by Reiny

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