Question

A 2-digit number is one more than 6 times the sum of its digits. If the
digits are reversed, the new number is 9 less than the original number.
Find the original number?​

Answer 1

Answer:

Step-by-step explanation:

Lets say that the digit in the unit's place is y and the digit in the ten's place is x. Then the value of the original number will be 10x+y

Since this number is one more than 6 times the sum of it's digits,

10x+y=1+6(x+y)

upon some simplification,

10x+y=1+6x+6y

4x-5y=1   (we'll call this equation 1)

If the digits are reversed,the value of the new number will be 10y+x

Since this reversed number is 9 less than the original number,

10y+x=10x+y-9

upon some simplification,

-9x+9y= -9

dividing the above equation by 9

-x+y= -1

Now we'll multiply the above equation by 4

-4x +4y= -4 (we'll call this equation 2)

Add equation 1 and equation 2

4x  - 5y = 1

-4x + 4y = -4

-------------------

 0x  - y = -3

therefore,y=3

substituting y=3 in equation 2,

-4x +4*3=-4

-4x+12= -4

-4x= -16

x=4

the solution is x=4,y=3

the original number is 43