The tens digit of a two-digit number is 3 greater than the unit digit. Eight times the sum of the digits is 1 less than the number. Find the number

Question

This is supposed to find the ordered pair

Answer 1

xy is the number

x=y+3

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

substitute

8(y+3+y)=(10[y+3]+y)-1

8(2y+3)=(10y+30+y)-1

16y+24=11y+29

16y-11y=29-24

5y=5

y=1

x=1+3

x=4

41 is the number

Answer 2

We have x, the unit digit, and x+3, the tens digit. So our two-digit number, x+3 x, could be 30, 41, 52, 63, 74, 85, or 96.

 

8*(x+x+3)+1=the two-digit number, x+3 x

16x+24+1="                                              "

16x+25="                                                  "

 

One way we could do this is subtract 25 from each possibility and see which one is divisible by 16 after the subtraction.

(So we would find 16x="the two-digit number, x+3 x"-25.)

 

The possible remainders are 5, 16, 27, 38, 49, or 60. Only one of these is divisible by 16. We got 16 by subtracting 25 from 41. Thus our answer is 41.

 

Check to make sure.

8*[(1)+(1+3)]+1=8*5+1=40+1=41.

 

So yep.