if d is HCF of 40and 65,find the value of integers X and Y which satisfy d=40x+65y i am not getting the answer !!!

2
by anaisha

2014-09-30T01:49:57+05:30
D= 5 so
x=1/8,y=0
or
x=0,y=1/13
and many more such solutions...
the given nos. are - 40,65
among them- 60>45
applying division lemma ..
60=40*1+25........1
since remainder is not equal to 0 .. we apply lemma to 40 and 25
40=25*1+15 then,..... 2
25=15*1+10.............3
15=10*1+5..............4
10=5*2+0 ..............5
as d=5 ... therefore,
5=40x+65y
we know value for 15,10 and 25 is as follows:
15=40-25*1 and 10=25-15*1
25=60-40*1
5=15-10*1
on putting value
5=40-25*1-(25-15*1)
5=40-25-25+15
5=40-(60-40*1)-(60-40*1)+40-25*1
5=40-60+40-60+40+40-(60-40*1)
5=40-60+40-60+40+40-60+40
5=40*5-60*3
5=40*5+60*(-3)
thus x=5 and y= -3
nice preeti thanks
2014-09-30T07:55:53+05:30
D=5
d=40x+65y
5=40x+65y
so
by this we can assume the values of x,y
there are 2 possibilitiesof x,y.
1] x=0,y=1/13
2] x=1/8,y=0

thsi is how this will be solved
the given nos. are - 40,65
among them- 60>45
applying division lemma ..
60=40*1+25........1
since remainder is not equal to 0 .. we apply lemma to 40 and 25
40=25*1+15 then,..... 2
25=15*1+10.............3
15=10*1+5..............4
10=5*2+0 ..............5
as d=5 ... therefore,
5=40x+65y
we know value for 15,10 and 25 is as follows:
15=40-25*1 and 10=25-15*1
25=60-40*1
5=15-10*1
on putting value
5=40-25*1-(25-15*1)
5=40-25-25+15
5=40-(60-40*1)-(60-40*1)+40-25*1
5=40-60+40-60+40+40-(60-40*1)
5=40-60+40-60+40+40-60+40
5=40*5-60*3
5=40*5+60*(-3)
thus x=5 and y= -3