If (a-s)│(ab + st), then show that (a-s)│(at + bs)

Solution;

Let’s take,

 (ab + st)- (at + bs)

= ab-at + st-bs

= a(b-t)-s(b-t)

= (b-t)(a-s)

Here,

 (ab + st)- (at + bs)= (b-t)(a-s)

therefore, (ab + st)- (b-t)(a-s) =(at + bs)

According to question it is given that (ab + st) is divisible by (a-s) so we can take (a-s) common from (ab + st)- (b-t)(a-s)

That is,

by considering (ab + st)= (a-s)m, here m is any natural number.

therefore,

 (a-s)[m-(b-t)]= (at + bs)

Since m,b and t are integers then [m-(b-t)] must be an integer.

 Hence (at + bs) is divisible by (a-s) whenever

(a-s)[m-(b-t)]= (at + bs)