# The sum of the digits of a certain two-digit number is 9. Reversing its digits decreases the number by 9. what is the number?

1
by rezter

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by rezter

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Let x be the tens digit

and y be the unit digit

10x + y = number

The sum of the digits of a number is 9.

x + y = 9 (1)

Reversing the digits decrease the number by 9.

10y + x = 10x + y - 9

10y - y = 10x - x - 9

9y = 9x - 9

9y = 9(x - 1)

Divide above equation by 9, we have

y = x - 1

y-x=-1 ..............(2)

Add (1) and (2)

y - x = -1

x + y = 9

--------------

2y=8 (2)

Divide by 2 both sides

2y/2=8/2

y = 4

Put the value of x in (1)

x+y=9

x+4=9

x=9-4

x=5

Digits are 5 and 4

and number is 54.

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So if you want to be sure if it's right and you want to check it, here it is:

The sum of the digits of a number is 9

5+4=9

9=9

Reversing the digits decrease the number by 9.

45=54-9

45=45

and y be the unit digit

10x + y = number

The sum of the digits of a number is 9.

x + y = 9 (1)

Reversing the digits decrease the number by 9.

10y + x = 10x + y - 9

10y - y = 10x - x - 9

9y = 9x - 9

9y = 9(x - 1)

Divide above equation by 9, we have

y = x - 1

y-x=-1 ..............(2)

Add (1) and (2)

y - x = -1

x + y = 9

--------------

2y=8 (2)

Divide by 2 both sides

2y/2=8/2

y = 4

Put the value of x in (1)

x+y=9

x+4=9

x=9-4

x=5

Digits are 5 and 4

and number is 54.

======

So if you want to be sure if it's right and you want to check it, here it is:

The sum of the digits of a number is 9

5+4=9

9=9

Reversing the digits decrease the number by 9.

45=54-9

45=45