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The sum of a two digit ten digit number and a single digit number is 8. The new number obtained by transposing the ten digit number and a single digit number is 2 times more than the original number. 10. Calculate the original two digit number Please use the linear equation of one variable
Let X be the number on ten, then 8-x be the number on one
（10X+8-X）*2+10=10*（8-X）+X
The solution is that x = 2, so the number on the tens is 2 and the number on the ones is 6
The original double-digit number was 26
For a two digit number, the sum of the ten digit number and the single digit number is 7. If this number is added with 45, it is the number after the transposition of the ten digit number and the single digit number?
Teach me how to solve the problem, fast, urgent
Let this number be 10A + B
Then a + B = 7
10a+b+45=10b+a
So the number is 16
There is a two digit number. The number in the ten digit number is two times more than the number in the single digit number. By swapping the number in the ten digit number with the number in the single digit number, the new number is 45 times smaller than the original number
The equation is used to find this two digit number
Let the two digit number be x, then the ten digit number be 2x + 2
x+10(2x+2)=10x+2x+2+45
x+20x+20=12x+47
21x-12x=47-20
9x=27
x=3
3*2+2=8
A: the double digit is 83
For a two digit number, the sum of the number on the tenth digit and the number on the single digit is 11. If the number on the tenth digit and the number on the single digit are exchanged, the new number will be 63 larger than the original number, and the original two digits will be obtained
Let the original digit be x, then the number on the ten digit is 11-x. according to the meaning of the question, 10 × (11-x) + x = 63 + 10x + (11-x), the solution is: x = 2, the number on the original ten digit is 9, that is, the original two digit 29
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