Question

there is a two digit number, consider tge digit in unit's place as x and digit in ten's place as y.
1] write the two digit number and the number obtained by interchanging the digits.
2) this two digit number is 7 times the sum of its digit. write the first equation. the number obtained by interchanging the digit is 18 less than the original number. write the second equation.
3) using equation 1 and 2, find the original number.

Answer 1

units place = x and tens place = y

1. number = 10y + x

interchanged = 10x + y

2. 10y + x = 7(x + y)

10x + y = 10y + x - 18

3. 10y + x = 7x + 7y and 10x + y = 10y + x - 18

= 3y - 6x = 0 and 9x - 9y = -18

= y - 2x = 0 and y - x = 2

solving these equations, we get

x = 2 and y = 4

original number = 42

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Answer 2

Answer:

Step-by-step explanation:

1) the two digit number is 10y+x and the number by interchanging their digits id 10x+y.

2)the first equation will be 10y+x = 7(x+y). And the second equation will be (10y+x) - 18 = 10x+y

After solving both equations, we get x = 2 and y = 4. So the two digit number should be 42