SOLUTION: THE DIGIT AT TEN,S PLACE OF A TWO DIGIT NUMBER IS TWICE THE DIGIT AT THE ONE,S PLACE . IF THE SUM OF THIS NUMBER AND NUMBER FORMED BY REVERSING THE DIGIT IS 66. FIND THE NUMBERS?

Algebra ->  Numeric Fractions Calculators, Lesson and Practice -> SOLUTION: THE DIGIT AT TEN,S PLACE OF A TWO DIGIT NUMBER IS TWICE THE DIGIT AT THE ONE,S PLACE . IF THE SUM OF THIS NUMBER AND NUMBER FORMED BY REVERSING THE DIGIT IS 66. FIND THE NUMBERS?      Log On


   

Question 319076: THE DIGIT AT TEN,S PLACE OF A TWO DIGIT NUMBER IS TWICE THE DIGIT AT THE ONE,S PLACE . IF THE SUM OF THIS NUMBER AND NUMBER FORMED BY REVERSING THE DIGIT IS 66. FIND THE NUMBERS?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let AB be your digit or numerically 10*A+B.
The reverse digit number is BA or numerically 10*B+A.
1.A=2B
.
.
10A%2BB%2B10B%2BA=66
11A%2B11B=66
2.A%2BB=6
Substitute eq. 1 into eq. 2,
2B%2BB=6
3B=6
highlight%28B=2%29
Then from eq. 1,
highlight%28A=4%29
.
.
.
The original number is 24, the reverse number is 42.