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A number consist of two digits whose sum is 15. If 9 is added to the number, then the digits change their places. Then the equations formed are x + y = 15 ; x – y = 9x – y = 15 ; x + y = 1x + y = 15 ; 9x – 9y = –9x + y = 15 ; x – y = –9

Question

A number consist of two digits whose sum is 15. If 9 is added to the number, then the digits change their places. Then the equations formed are x + y = 15 ; x – y = 9x – y = 15 ; x + y = 1x + y = 15 ; 9x – 9y = –9x + y = 15 ; x – y = –9

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Solution

The question seems to be a bit jumbled, but I believe you're asking for help solving a system of equations based on a two-digit number. The conditions given are:

  1. The sum of the digits is 15.
  2. If 9 is added to the number, the digits switch places.

From these conditions, we can form two equations:

  1. x + y = 15 (where x and y are the tens and units digits of the number, respectively)
  2. 10x + y + 9 = 10y + x (which represents the condition that if 9 is added to the number, the digits switch places)

Let's solve this system of equations step by step:

Step 1: Simplify the second equation: 9x - 9y = -9

Step 2: Divide the entire equation by 9 to simplify: x - y = -1

Now we have a system of two equations:

  1. x + y = 15
  2. x - y = -1

Step 3: Add these two equations together: 2x = 14

Step 4: Solve for x: x = 7

Step 5: Substitute x = 7 into the first equation: 7 + y = 15

Step 6: Solve for y: y = 8

So, the two-digit number is 78.

This problem has been solved

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