help, math?

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


i need an equation for this


thanks

Answer 1

Let u = units digit
Let t = tens digit
Then number = 10t+u
So 10t+u = 6(t+u) +1 <-- Eq 1
10u-t = 10t-u - 9 <-- Eq 2
Solve EQ1 and EQ2 simultaniously and you'll have it.

Answer 2

let x be the number in the tenth place
let y be the number in te one place

A 2-digit number is one more than 6 times the sum of its digits, tells you:
10x + y = 6(x + y) + 1
10x + y = 6x + 6y + 1
4x - 5y = 1


digits are reversed, the new number is 9 less than the original number, tells you:
10y + x = 10x + y - 9
9y - 9x = -9
9(y - x) = -9
y - x = -1
y = x - 1

4x - 5(x - 1) = 1
4x - 5x + 5 = 1
-x = -4
x = 4


y = 4 - 1
y = 3

so the number is 43