Question

The sum of the certain two digit number is 13 and the number is2 more than 7 times the unit digit ,find the number​

Answer 1

Let the ten's digit and the one's digit be x and y respectively.

Given

The sum of the certain two digit number is 13

$\implies x+y=13--(1)$

The number is 2 more than 7 times the unit digit.

$\implies 10x+y=7y+2\\\implies10x+y-7y=2\\\implies 10x-6y=2\\\implies 5x-3y=1--(2)$

Solving (1) and (2), we get,

$5x+5y=65\\\underline{5x-3y=1}\\\underline{\underline{8y=64}}\\\implies y = \dfrac{64}{8}$

$\implies y= 8$

Substituting y in the first equation, we get,

$\implies x+8=13\\\implies x = 13-8\\\implies x = 5$

$\boxed{\boxed{\bold{Therefore, \ the \ number \ is \ 58}}}}}$

                                                         

Answer 2

Answer:

Let the tens digit =x

then ones digit = 13-x

Thus,

number = 10x+13-x = 9x+13

The number is 7( x -13) + 2

= 93 – 7x

9x+13=93–7x

16x=80

x=5

thus tens digit is 5

and unit digit=13–5=8

Hence the number is 58 .