Question #9c419

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Answer 1 · 2016-12-03T08:30:24

58

Let in the two digit number the tens digit be x and units digit be y .
So the number is #" "10x+y#

By the 1st condition

#" "y-x=3.......(1)#

And by the2nd condition

#10x+y-4(x+y)=6#

#=>6x-3y=6#

#=>2x-y=2.....(2)#

Adding (1) and (2) we gwt

#x=5#

Inserting #x=5 # in (1) we get

#y-5=3=>y=8#

Hence the the two digit number is #10x+y=10xx5+8=58#

Answer 2 · 2016-12-03T08:34:51

See explanation.

If we denote the tens digit by #a# and unit digit by #b# then the question leads to the following equations:

Tens digit is 3 less than units digit: #a=b-3#.

The original number is six more than four times than the sum of dygits: #10a+b-6=4*(a+b)#.

If we solve the system we get:

#{(a=b-3),(10a+b-6=4a+4b):}#

#{(a=b-3),(10(b-3)+b-6=4(b-3)+4b):}#

Now from the second equation we calculate #b#:

#10b-30+b-6=4b-12+4b#

#10b+b-8b=36-12#

#3b=24#

#b=8#

So: #a=8-3=5#

Answer: The number is 58.