SOLUTION: I don't know how to set this one up properly.
The sum of the digits of a two-digit number is nine. If the original number is doubled, the result is eighteen less than twice the
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Question 2245: I don't know how to set this one up properly.
The sum of the digits of a two-digit number is nine. If the original number is doubled, the result is eighteen less than twice the original number reversed. What is the original number?
Answer by longjonsilver(2297) (Show Source): You can put this solution on YOUR website!
Define...
first digit = x
second digit = y
x+y = 9 --eqn1
now for the tricky one... original number is (10x+y), so
2(10x+y) = (2(10y+x)-18 --eqn2
so expand eqn2 to give 20x+2y = 20y+2x-18
18x-18y=-18 --eqn3
sub in 3 for x (from eqn1)
18(9-y)-18y=-18
162-18y-18y=-18
-36y=180
y=5
so, x=4
CHECK!
45 doubled is 90
54 doubled is 108-18 is 90, so correct!
jon
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