Question

When the digits of a two-digit number are reversed, the value of the number is increased by 45. The sum of the digits is 11. What is the original number?

Answer 1

Answer:

9x+9y=_45

x_y=_ 5

x+y=11

2x=6

x=3

put x=3 in equation 1

3_y=_5

_y=_2

y=2

the original number is 32

Answer 2

Answer:

The original number is $38$

Step-by-step explanation:

Remember that the decimal number system is a positional number system.  For decimal, that means ones place, tens place, etc.

Let the number be xy [note: x is a digit in tens place and y is another digit in ones place]

Learn to translate the words into formulas:

"When you reverse the digits in a certain two-digit number you increase its value by 45."    means

$10y + x = (10x + y) + 45---------$ [eq1]

"the sum of its digits is 11"means $x + y = 11 -----$           [eq2}

Now, solve (let's use substitution):

            $y = 11-x------- [from eq2]$

Put that into eq1:

$10y + x = (10x + y) + 45 [eq1]\\10(11-x) + x = 10x + (11-x) + 45\\110 - 10x + x = 10x + 11 - x + 45\\ 110 - 9x = 9x + 56\\ -18x = -54\\ x = 3$

Now, put that into either equation to find the value of y:

           $3 + y = 11 ------- [eq2]\\ 3 + y = 11\\ y = 8$

The number is 38.

Check:

   Is    83 = 38 + 45    ?

         83 = 83     ?yes