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Question 113520This question is from textbook Arithmetic and algebra again
: The sum of the digits of a two-digit number is 7. If the digits are reversed, the new number increased by 3 equals 4 times the original number. Find the original number.
here is what I have done:
original number 10x + y
new number 10y + x
x + y = 7
10x + y = 10y + x + 3
10x -x + y -y = 10y -y + x -x + 3
9x = 9y + 3
x = y + 3
x + y = 7
(y+3) + y = 7
2y + 3 = 7
2y +3 -3 = 7 -3
2y = 4
2y/2 = 4/2
y = 2
x + y = 7
2 + y = 7
2 -2 + y = 7 -2
y = 5
x = 2
I get an answer, but it don't check.
Every time I do this problem this my answer or fractional. please help
This question is from textbook Arithmetic and algebra again
Found 3 solutions by kev82, Fombitz, MathLover1:
Answer by kev82(151)   (Show Source):
You can put this solution on YOUR website! It's great to see someone posting their working and, not only that but checking the answer as well, realising they've done something wrong. These are great traits that you should be proud of. Anyway, to the problem, you have only made a very slight mistake:
"If the digits are reversed, the new number increased by 3 equals 4 times the original number"
so 10y+x+3 = 4(10x+y). This tidies up to 6y-39x = -3. Combining this with y=7-x gives 6(7-x)-39x=-3, 42-45x=-3, 45x=45 x=1, y=6.
Answer by Fombitz(32388)  (Show Source):
Answer by MathLover1(20850)  (Show Source):
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