Linear equation

Question

A number consists of two digits whose sum is 11. Of the digits are reversed, the original number is increased by 26. What is the original number?

Answer 1

Let the number be xy

The total value of the original number is 10x + y

 

x + y = 11     {equation 1}

10y + x = 10x + y + 26   {equation 2}

 

Let's work on equation 2:

10y + x = 10x + y + 26

9y - 9x = 26

-9x + 9y = 26

 

Our equations now are:

x + y = 11     {equation 1}

-9x+9y = 26  {equation 2}

 

multiply equation 1 by 9 and add to solve for y

 

 9x + 9y = 99

-9x + 9y = 26

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

18y = 125

                                           _

This gives a value for y of 6.94  with the 4 repeating...

That doesn't seem to be possible for the problem as

stated.

 

Look at the possibilities:

29 / 92   difference = 63

38 / 83   difference = 45

47 / 74   difference = 27

56 / 65   difference = 9

 

None of the possibilities have the new

number as 26 greater than the original

 

There is a problem with your original statement as posted.

 

 

Answer 2

Hi Cindy,

Let's call the digits of the number x and y
x + y = 11
and the value of the number is 10x + y
When the digits are reversed, the value of the new number is
10y + x
We are given that when the digits are reversed, the original number is increased by 26. Then, we can write
10y + x = 10x + y +26
-9x + 9y = 26
We also know x + y = 11
so we can solve for x and y

-9x + 9y = 26
x + y = 11
I multiplied the second equation by 9 to eliminate x
-9x + 9y = 26
9x + 9y = 99
add the two equations
18y = 125

y = 4 (rounded)

x= 7 

 

however 74-47 is not equal to 26. The values of x and y are not equal to whole numbers.