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A two-digit number is such that the sum of its digits is 1/8 of the number. When the digits of the number are reversed and the number is subtracted from the original number, the result obtained is 45. By letting the tens digit of the original number be x and the ones digit of the original number be y, form two simultaneous equations in terms of x and y and solve them to find the original number. ​

Respuesta :

Answer:

The original number = 72

Step-by-step explanation:

Let:

  • the tens digit of the original number = x
  • the ones digit of the original number = y

Then:

  • the value of the original number = 10x + y
  • the sum of the original number's digits = x + y
  • the value of the reversed number = 10y + x

The statement "the sum of its digits is 1/8 of the number" can be written as:

[tex]x+y=\frac{1}{8} (10x+y)[/tex]

[tex]8(x+y)=10x+y[/tex]

[tex]10x-8x=8y-y[/tex]

[tex]\bf 2x=7y\ ...\ [1][/tex]

The statement "When the digits of the number are reversed and the number is subtracted from the original number, the result obtained is 45" (original number - reversed number = 45) can be written as:

[tex](10x+y)-(10y+x)=45[/tex]

[tex]9x-9y=45[/tex]

[tex]\bf x-y=5\ ...\ [2][/tex]

The system of equations will be:

[tex]\displaystyle\left \{ {{2x=7y} \atop {x-y=5}} \right.[/tex]

To solve the system of equations:

2x = 7y     ⇔  2x - 7y = 0

x - y = 5    ⇔  2x - 2y = 10

                    -------------------- (-)

                            -5y = -10

                               y = 2

[2]

x - y = 5

x - 2 = 5

x = 7

Therefore, the original number = 72