A number consists of two digits whose sum is
5. When the digits are reversed, the number
becomes greater by 9. Find the number​

Answer:

[tex]\bigstar{\bold{The\:number\:is\:23}}[/tex]

Step-by-step explanation:

[tex]\Large{\underline{\underline{\bf{Given:}}}}[/tex]

• The sum of the digits is 5
• When the digits are reversed, the number becomes greater by 9

[tex]\Large{\underline{\underline{\bf{To\:Find:}}}}[/tex]

• The Number

[tex]\Large{\underline{\underline{\bf{Solution:}}}}[/tex]

→ Let the digits of the number be x and y

→ By first case,

  x + y = 5

  x = 5 - y -----equation 1

→ The number is represented by

   Number = 10 x + y

   ∵ x is in tens place and y is in ones place

→ By second case given, reversing the number.

  New number = 10y + x

→ By given,

  10 y + x = 10x + y + 9------equation 2

→ Substitute equation 1 in equation 2

  10y + 5 - y = 10 ( 5 - y) + y + 9

  9y + 5 = 50 - 10y + y + 9

  9y + 5 = 59 - 9y

  18y = 54

     y = 54/18

     y = 3

→ Hence the one's digit is 3

→ Substitute the  value of y i equation 1

  x = 5 - y

  x = 5 - 3

  x = 2

→ Hence x is 2

→ The number is 10x + y = 10×2 + 3 = 23

→ Hence the number is 23

 [tex]\boxed{\bold{The\:number\:is\:23}}[/tex]

[tex]\Large{\underline{\underline{\bf{Verification:}}}}[/tex]

→ The sum of the digits is 5

   x + y = 5

   3 + 2 = 5

          5 = 5

→ When the digists are reserved number becomes greater by 9

   32 = 23 + 9

   32 = 32

→ Hence verified.

[tex]\Large{\underline{\underline{\bf{Notes:}}}}[/tex]

→ A linear equation in 2 variables can be solved by

• Substitution method
• Elimination method
• Cross multiplication method