SOLVING LINEAR EQUATIONS WORD PROBLEMS USING ELIMINATION METHOD

Problem 1 :

If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1, it becomes 1/2 if we only add 1 to the denominator. What is the fraction?

Solution :

Let x/y be the required fraction.

(x + 1)/(y – 1)  =  1

 (x + 1)  =  (y - 1)

 x – y  =  -1 - 1

x – y  =  -2 -----(1)

x/(y  + 1)  =  1/2

 2 x  =  (y + 1)

 2x – y  =  1 -----(2)

(1) - (2) :

x  =  3

Substitute 3 for x in (1).

(1)-----> 3 - y  =  -2

-y  =  -5

y  =  5

Then,

x/y  =  3/5

So, the required fraction is 3/5.

Problem 2 :

Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu ?

Solution :

Let 'x' be the present age of Nuri.

Let 'y' be the present age of Sonu.

Five years ago, Nuri was thrice as old as sonu. 

(x – 5)  =  3(y – 5)

x – 5  =  3y – 15

 x – 3y  =  -10 -----(1)

Ten years later, Nuri will be twice as old as Sonu.

(x + 10)  =  2(y + 10)

x + 10  =  2y + 20

x – 2y  =  10 -----(2)

(1) - (2) :

y  =  20

Substitute y  =  20 in (2). 

(2)-----> x - 2(20)  =  10

x - 40  =  10

x  =  50

So, the present ages of Nuri and Sonu 50 years and 20 years respectively.  

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