if d is the hcf of 30,72. find the value of x and y satisfying d =30x+72y
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72=30*2+12-------(1)
30=12*2+6-------(2)
12=6*2+0
Hcf is 6
6=30x+72y
From ---2
6=30-12*2
From-----1
12=72-30*2
Then 6=30-(72-30*2)*2
30=12*2+6-------(2)
12=6*2+0
Hcf is 6
6=30x+72y
From ---2
6=30-12*2
From-----1
12=72-30*2
Then 6=30-(72-30*2)*2
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