Question

Consider the equation of the form ax + pax+ q = r where all a, p,q,r are natural numbers greater than 1 . This equation can be solved for x by using the following procedure Step 1



Take ax common from LHS then we get ax(1 + paq) =r Step 2 ax = r/ (1+ paq)



Step 3 guess x from the value of ax



Q.1 solve 2x + 3(2x+4 ) = 196 for the value of x using the procedure given above . find x and show your calculation .







Q.2 solve 2(3x+2) + 3x = 513 for the value of x using the procedure given above . find x and show your calculation .



​

Answer 1

Answer:

Eliminating z form the 1st and 2nd equation and from the 2nd and the 3rd equation we get

5x+4y=11p−3qand5x+4y=

5

7q+11r

At least one solution means we can have infinite solution Applying the conditions of infinite

solutions we get 11p−3q=

5

7q+11r

⇒5p−2q−r=0

mark it as branlist

..follow me pls..

Answer 2

hey please massage me