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Question 129107This question is from textbook Algebra !
: The sum of the digits of a two-digit number is 8. If 16 is added to the orginal number, the result is 3 times the original number with its digits reversed. Find the orginal number. This is the problum and have no idea how to do it because i was gone when we learned this and i have bin working on this for litterly 2 hours please help me
This question is from textbook Algebra !
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! Let x=one of the digits---the 10's digit
And let y=the other digit---the unit digit
Now we are told that x+y=8-------------------------eq1
We know that the original number can be written as:
10x+y now if we add 16 to this, we get (10x+y)+16 -------1
If we reverse the digits of the original number, we get:
10y+x and three time this number is:
3(10y+x)----2
We are told that 1=2, so:
(10x+y)+16=3(10y+x) get rid of parens
10x+y+16=30y+3x subtract 10x and also y from both sides:
10x-10x+y-y+16=30y-y+3x-10x collect like terms
16=29y-7x or
-7x+29y=16-----------------------------------------eq2
substitute x=8-y from eq1 into eq2 and we have:
-7(8-y)+29y=16 get rid of parens
-56+7y+29y=16 add 56 to both sides
-56+56+7y+29y=16+56 collect like terms
36y=72 divide both sides by 36
y=2--------------------------------------the unit digit
substitute y=2 into eq1 and we get:
x+2=8 subtract 2 from both sides
x+2-2=8-2 collect like terms
x=6-----------------the 10's digit
Thus, the original number is 62
CK
6+2=8
8=8
and
62+16=3*26
78=78
Hope this helps---ptaylor
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