find the whole numbers a and b, if 29/6=a+1/(1+1/b)

Question

home work year 7

Answer 1

(a+1) / [1+(1/b)] =29/6 . . . . so [1+(1/b)] / (a+1)=6/29 . . . or [(b+1)/b] / (a+1)=6/29 . . . (b+1)/b=(6/29)*(a+1) . . . 1 anser is a=b=-1....0=0 . . .

Answer 2

29/6=a+1/(1+1/b)

29/6=a+b/(1+b)

29/6=a+(c-1)/c  where c = b+1 

29c = 6ac + 6c - 6

6ac - 23c = 6   and now let d = ac

6d - 23c = 6 -------------------- (1)

Now we have a linear Diophantine equation, so let's manipulate the coefficients of d and c.

23 = 3*6 + 5

6 = 1*5 + 1

rearranging the above two equations,

1 = 6 - 1*5 = 6 - 1*(23 - 3*6) = -1*23 + 4*6

1 = 6(4) - 23(1)

6 = 6(24) - 23(6)  --------------- (2)

Comparing (1) with (2),

d = 24

c = 6

Since d = ac = 6a then a = 4, and c = 6 gives b = 5

Answer: a = 4, b = 5