Consider the equation of the form ax + pax+ q = r where all a, p,q,r a

1.	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 1Take 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 axQ.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» 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..

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 1Take 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 axQ.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»

Answer:

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

5x+4y=11p−3qand5x+4y=

7q+11r

At LEAST one SOLUTION MEANS we can have infinite solution Applying the conditions of infinite

solutions we get 11p−3q=

7q+11r

⇒5p−2q−r=0

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