Question 551297: The value of certain two digit number is four less than six times the sum of it's digits. If the digits of the number are reversed, the resulting number is nine less than the original number. Find the number. Please help me!!!!
Found 2 solutions by TutorDelphia, scott8148:
Answer by TutorDelphia(193)   (Show Source):
You can put this solution on YOUR website! To figure this out logically, for the number to be 9 less when the digits are reversed, they have to be within 1 of each other: for example 23 is 9 less than 32.
Than we can check 98, 87 ...21, 10 to see which works.
Algebraically:
Let's say our two digit number is called XY
And this is not the answer, but if x is 3 and y is 4 the value of xy is 3*10+4 which is 34. So always multiply the digit in the tens place by 10 to find its value.
So with that in mind: The value of certain two digit number is four less than six times the sum of it's digits could be expressed as:
X*10+y=6*(x+y)-4
Distribute
10x+y=6x+6y-4
subtract 6x-6y from both sides
4x-5y=-4
We also know if you reverse the digits the resulting number is nine less. Again, multiply the digit in the tens place by 10
X*10+y-9=y*10+x
9x-9y=9
So now we have the two equations:
4x-5y=-4
9x-9y=9
So we have to multiply the first by 9 and the second by -4
36x-45y=-36
-36x+36y=-36
-9y=-72
y=8
Sub in 8 for y
9*x-9*8=9
9x-72=9
9x=81
x=9
Try it out
98 is 4 less than 6 times 15 (102-4=98)
And 89 is 9 less than 98
Answer by scott8148(6628)  (Show Source):
You can put this solution on YOUR website! t = tens digit , u = units digit
"The value of certain two digit number is four less than six times the sum of it's digits"
___ 10t + u + 4 = 6(t + u)
"If the digits of the number are reversed, the resulting number is nine less than the original number"
___ 10u + t + 9 = 10t + u ___ 9u + 9 = 9t ___ u + 1 = t
substituting ___ 10(u + 1) + u + 4 = 6[(u + 1) + u] ___ 11u + 14 = 12u + 6
solve for u, then substitute back to find t ___ the number is "tu"
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