NCERT Class 10th Mathematics
7

NCERT Class 10th – Chapter 3- Exercise 3.4 Pair of Two Equations in Two Variables

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EXERCISE -3.4

  QUE:-1. Solve the following pair of linear equations by the elimination method and the substitution
              method :
             (i) x + y = 5 and 2x – 3y = 4                      (ii) 3x + 4y = 10 and 2x – 2y = 2
              (iii) 3x – 5y – 4 = 0 and 9x = 2y + 7          (iv) x/2 +2y/3 =-1 and x-y/3 =3

  SOL :-

   (i) by substitution

             x + y = 5 ………….(1)

             2x – 3y = 4 …………….(2)   

from equ. (1) we get

x =5 – y ……………..(3)

putting the value of x in equ (2)

 2(5 -y)  – 3y = 4

10 -2y -3y = 4

-5y =-6

y =6/5

putting the value of y in equ (3)

x =5 – 6/5 =19/6

so the value of x is =19/6

and value of y is = 6/5

By elimination

multiplying by 2 in equ. (1)

2x +2y = 10 ……………..(4)

subtracting the equ.(2) from (3)

2x +2y = 10

–    (2x – 3y = 4)

5y =6

y= 6/5

putting the value of y in equ. (4)

2x +2(6/5) = 10

10x +12 = 50

x=38/10=19/5

so value of x =19/5 and value of y = 6/5

(ii) By subsititution

       3x + 4y = 10 ……………….(1)

       2x – 2y = 2……………………(2)

from the equ.(2)

x=(1+y )……….(3)

putting the value of x in equ.(1)

   3(1 +y) + 4y = 10

7y =7

y = 1

putting the value of y in equ.(3)

x=(1+y )

x= 1+1=2

by elimination

muliplying by -2 equ.(2)

-2×( 2x – 2y = 2) 

-4x+4y =-2……………(4)

substracting equ (4) by equ.(1)

-4x+4y =-4

   – (3x + 4y = 10)

-7 x      = -14

x =  14 /7=2

putting the value of x in equ. (1)

3x+ 4y = 10

6 + 4y=10

y =(10-6)/4=1

so value of x is 2 and value of y is 1

(iii) 3x – 5y – 4 = 0………………….(1)

        9x = 2y + 7………………………..(2) 

By subsititution

from equ. (1) we get

x =( 4+5y)/3………………(3)

putting the value of x in equ.(2)

    9x = 2y + 7

9 (( 4+5y)/3) =2y +7

36 +45y = 6y +21

45y -6y = 21 -36

39y =-15

y=-15 /39=-5/13

putting the value of y in equ. (3)

x =( 4+5y)/3

x =( 4+5(-5/13))/3 =(52 -25)/39 =27/39=9/13

so the value of x is 9/13 and value of y is -5/13

by elimination

multyplying equ.(1) by 3

(3x – 5y – 4 = 0) ×3

9x -15y -12 =0

9x =12 +15y ………….(4)

subtracting equ.(4) by equ.(2)

9x =12 +15y

-9x = -2y – 7 

0 = 5 +13y

y = -5/13

putting the value of y in equ. (1)

3x – 5y – 4 = 0

3x +25/13 -4 =0

3x -27/13 =0

39 x= 27

x= 27/39= 9/13

so value of x is 9/13 and value y is -5/13

  (iv)  x/2 +2y/3 =-1 …………….(1)

x-y/3 =3………………..(2)

By subsititution

from the equ.(2) we get

x=(9+y)/3 …………(3)

putting the value of x in equ. (1)

x/2 +2y/3 =-1

(9+y)/3/2 +2y/3 =-1

27 +3y +12y =-18

15y = -45

y = -45/15 =-3

putting the value of y in equ. (3)

x=(9+y)/3 = (9 -3)/3 =2

so the value of x is 2 and value y is -3

by elimination

multiplying by 2 in equ. (1)

x/2 +2y/3 =-1

x +4y/3 =-2…………….(4)

subtracting equ.(4) by equ.(2)

x +4y/3 =-2

    -( x-y/3 =3 )    

5y /3 =-5

y=-15/5 = -3

putting the value of  y in equ. (2)

x-y/3 =3

x  +3 /3 =3

3x +3 =9

x = (9 -3)/3 =2

so the value of x=2 and y=-3

QUE :-2. Form the pair of linear equations in the following problems, and find their solutions
             (if they exist) by the elimination method :
        (i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces
            to 1. It becomes1/2if we only add 1 to the denominator. What is the fraction?
      (ii) 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?
      (iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is
          twice the number obtained by reversing the order of the digits. Find the number.
     (iv) Meena went to a bank to withdraw ` 2000. She asked the cashier to give her
         ` 50 and ` 100 notes only. Meena got 25 notes in all. Find how many notes of
        ` 50 and ` 100 she received.
      (v) A lending library has a fixed charge for the first three days and an additional charge
          for each day thereafter. Saritha paid ` 27 for a book kept for seven days, while Susy
           paid ` 21 for the book she kept for five days. Find the fixed charge and the charge
          for each extra day.

  SOL:-

(i) let the fraction is x/y

according to the que

(x+1)/(y-1)=1

x+1 =y-1

x – y =-2…………………(1)

x /y+1 =1/2

2x  = y+ 1

2x -y =1 ……………………(2)

subtracting equ.(2) by equ.(1)

2x -y =1

  – ( x – y =-2)

x   =3

putting the value of x =3 in equ.(1)

x – y =-2

3 -y =-2

-y = -5

y=5

so the fraction is 3/5

(ii) let the age of nuri =x

and age of sonu is =y

according to the que.

five year ago

x-5 =3( y-5)

x-5 = 3y -15

x-3y = -10 ………………..(1)

10 year leter

x+10 =2(y +10)

x +10 =2y +20

x-2y =10 ………….(2)

subtracting equ.(1) by equ.(2)

x-3y = -10

– x+2y =-10  

-y  = -20

y= 20

putting the value of y in equ. (1)

x-3×20 = -10

x -60 =-10

x =-10 +60

x =50

so the present age of nuri 50 and age of sonu is 20 year

(iii) let the digits of the number is x and y

the number is 10y +x

according to the que.

x +y =9………….(1)

also

9 (10y +x) = 2 ( 10x +y)

90y +9x = 20x +2y

90y +9x -20x -2y=0

88y -11x =0 ……………..(2)

multiplying by 11 in equ.(1)

11x +11y =99 ………………..(3)

adding equ.(1) by equ.(2)

88y -11x =0

+ 11x +11y =9

99y  = 99

y =1

putting the value of y in equ (1)

x +1=9

x=8

so the number is 18

(iv) let the number 50rs note is x

and the number of 100 rs note is y

according to the que.

x + y =25…………………….(1)

50x +100y =2000…………(2)

multypling by 100 in equ.(1)

100x +100y =2500………..(3)

substracting equ.(3) by equ.(2)

100x +100y =2500

-50x -100y = -2000

50x  = 500

x =10

putting the value of x in equ. (1)

10 + y =25

y  = 25 -10 =15

so the number of 50 rs note 10and the number of 100 rs note is 15

(v) let the fix charge of first three day =x rs

additional charge of each day after =y rs

according to the que.

x +4y =27…………..(1)

x + 2y =21 ………….(2)

substracting equ. (1) by equ.(2)

x +4y =27

-x – 2y =-21

2y =6

y=3

putting the value of y in equ. (1)

x +4×3 =27

x +12 =27

x  = 15

so the fix charge of first three day =15 rs

and additional charge of each day after =3 rs /day

You can see the solution for complete chapter here –

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