i need help!

Question

A two-digit number ,the sum of its digits is 11 if the two digits are reversed,then the resulted number is 27 more than the original number what is the original number?
i need a detailed answer because i cant understand when i use X+y only and when i use x+10y in this question

Answer 1

You need two digits, A and B.

One is a tens digit and the other is a ones digit.

 

A + B = 11

When the number is AB, that's:

AB = A*10 + B*1

AB = 10A + B

 

When the number is BA, that's:

BA = 10B + A

 

We know that BA is 27 greater than AB, so:

BA - AB = 27

Substitute for AB and BA:

(10B + A) - (10A + B) = 27

10B + A -10A - B = 27

-9A + 9B = 27

 

Now it's a system of equations with two unknowns.

You can solve by elimination or substitution.

I'll use elimination.

Solve the first equation for A.

A + B = 11

A = 11 - B

Substitute this into the second equation:

-9A + 9B = 27

-9(11-B) +9B = 27

-99 + 9B + 9B = 27

18B = 126

B = 7

 

Then A + B = 11, so:

A + 7 = 11

A = 4

Answer 2

The sum of the digits of a two digit number is 11

 

Let's say the number is "xy" (2-digit number, not multiplication)

Digits are only the numbers used to make up the number

x + y = 11

 

if the two digits are reversed, (yx) then the resulted number is 27 more than the original number what is the original number?

 

Here we want to compare the whole number, xy to another number, yx that is 27 more.

 

This is where you multiply the digit by its placement, (ie,. hundreds, tens, ones etc.)

 

example:

34 = 10(3) + 1(4)

 

10(x)+1(y) + 27 = 10(y) + 1(x)

 

10x + y + 27 = 10y + x

9x + 27 = 9y

divide all by 9

x + 3 = y

 

put this into the first equation,

x + y = 11

x + x + 3 = 11

2x + 3 = 11

2x = 8

x = 4

 

Then put this into either equation to find y

 

x + y = 11

4 + y = 11

y = 7

 

Original number , xy is 47

 

check the solution:

 

4 + 7 = 11

47 + 27 = 74