da sum of a 2digt # is 11.If da ordr of da dgts s rversed,da rsulting # excds da orig. # by 45.Find da orig. #

Question

da sum of a 2digt # is 11.If da ordr of da dgts s rversed,da rsulting # excds da orig. # by 45.Find da orig. #

guys please answer this question im suffering from headache
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tnx guys...

Answer 1

the digits are x & y so
x+y=11

since 10x+y - (10y+x)=45
9x-9y=45
x-y=5
x=5+y

plug this into top equation

(5+y)+y=11

2y=6 y=3 plug this answer into x=5+y and get

x=8, y=3

this checks out because 8+3 is 11and
83-38=45
since the reversed # exceeds the original by 45 then the
original # is the smaller of 83 & 38 so it is 38

Answer 2

Take
t = the tens
u = the units

the sum of digits:
t+u = 11

The number is:
10t+u

In reverse order:
10u+t

Now, form a system of two equations:
u+t=11
(10u + t) - (10t + u) = 45

t=11-u
10u+t - 10t-u=45

t=11-u
9u-9t==45

9u-9(11-u)=45

9u-99+9u=45

18u-99=45

18u = 144

u=8

t=11-u=3

The number is 38

83-38=45