Question 222968: A two digit number is equal to seven times the the sum of the digits. If 18 were subtracted from the number, the position of the digits would be reversed. Find the number.
Answer by brich(7)   (Show Source):
You can put this solution on YOUR website! Let x = the first digit of the two digit number
Let y = the second digit of the two digit number
To represent the two digit number using x and y, take x and multiply it times ten, and add y to it like so:
10x+y
For example, if your number was 56, x = 5, and y = 6.
To get 56:
(10*5)+6
50+6
56
So the two digit number = 10x+y
The next part is easy, it says its equal to seven times the sum of the digits.
or:
7(x+y)
!Dont forget the parentheses around x+y to show that you add x and y before you multiply be 7!
Now set the equations equal:
10x+y = 7(x+y)
Break up this next part of the question into two parts.
If 18 were subtracted from the number...
10x+y - 18
Then the position of the digits would be reversed...
So
Simply switch x and y
10y+x
Note that now y is in the tens place and x is in the ones place.
Set those equal
10x+y - 18 = 10y+x
simply both equations, and solve each for y
Equation1:
7(x+y)=10x+y
7x+7y = 10x+y
7y=3x+y
6y=3x
y=(1/2)x
Equation2:
10x+y - 18 = 10y+x
9x+y-18 = 10y
9x-18 = 9y
y=x-2
Now use substitution to solve:
x-2 = (1/2)x
2x-4=x
x=4
Plug back in for y:
y=x-2
y=4-2
y=2
Finally display your answer:
Two Digit Number = 10x+y
=10(4) + 2
=42
And thats all there is to it.
I hope I was helpful.
Let me know.
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