Question

Find the 2 digit number such that if either I is added to 8 times the sum of the digits or if 2 is added to 13 times difference of the digit is the number itself is obtained​

Answer 1

Answer:

Let’s take the tens place to be x and the units to be y. The number is then x|y, next to each other.

Making an equation from the first part:

10x + y = 8(x+y) + 1

10x + y = 8x + 8y + 1

2x - 7y = 1

From the second:

10x + y = 13(x-y) + 2

10x + y = 13x - 13y + 2

-3x + 14y = 2

Now we have two equations, we can solve this simple set.

-3x + 14y = 2

4x - 14y = 2 (doubling to eliminate y)

Adding:

x = 4

Substituting into an equation:

2*4 - 7y = 1

8 - 7y = 1

-7y = -7

y = 1

Therefore, our number is 41. Double-checking:

1 + 8*(4+1)

1 + 40

41

2 + 13(4–1)

2 + 39

41

HOPE IT HELPS YOU...