Word problems w/ substitution/elimination

Question

The sum of the digits of a 2-digit number is 15. If the digits are reversed, the number formed is nine less than the original number. What’s the original number?

Answer 1

By trial and error....

the only possibilities are 6 and 9

7 and 8

87-78 = 9 while 96-69 = 27

so it must be 78 and 87

algebra...

T+U = 15 where T is the tens digit and U is the units digit

the original number is 10T+U

when reversed, 10U + T

which is 9 greater...

so 10U + T = 10T + U + 9

9U = 9T + 9

U = T + 1

Then T+U = 15 becomes

T + T+1 = 15

2T + 1 = 15

2T = 14

T = 7

which forces U=8

so 78 and 87

Answer 2

The sum of the digits of a 2-digit number is 15. If the digits are reversed, the number formed is nine less than the original number. What’s the original number?

Let the number be NM.

e know N+M=15

and MN=NM-9 or 10M+N=10N+M-9

So we have two equations ([1]&[2]) and two unknowns:

N+M=15 [1]

9M-9N=-9 [2]

Simplified:

N+M=15 [1]

M-N=-1 [2]

From [2], M=N-1

[2] into [1]"

N+(N-1)=15

2N=16

N=8

M=N-1=8-1=7

The original number: 87

Check Equation [1]: N+M=8+7=15 OK

Equation [2]: M-N=-1=7-8=-1 OK

Solution: 87