NCERT Class 10th Mathematics
7

NCERT Class 10th – Chapter 3- Exercise 3.5 Pair of Two Equations in Two Variables

NCERT full solutions  here you can find full detailed solutions for all the questions  with proper format. class 10 maths ncert solutions chapter 3 exercise 3.5.

You can see the solution for complete chapter here –

EXERCISE:-3.5

QUE :-1. Which of the following pairs of linear equations has unique solution, no solution, or
               infinitely many solutions. In case there is a unique solution, find it by using cross
              multiplication method.
              (i) x – 3y – 3 = 0 ,3x – 9y – 2 = 0 (ii) 2x + y = 5 ,3x + 2y = 8
                (iii) 3x – 5y = 20 ,6x – 10y = 40 (iv) x – 3y – 7 = 0 , 3x – 3y – 15 = 0

   SOL:-

(i) x – 3y – 3 = 0 ,

          3x – 9y – 2 = 0

a1/a2 =1/3 ,b1/b2 =-3/-9=1/3 , c1/c2=-3/-2=3/2

a1/a2 =b1b2 ≠c1/c2

∴ the given set of line will not intersect each other they are parallel to each other the pair of linear equation has            no solution.

(ii) 2x + y = 5 ,

3x + 2y = 8

a1/a2 =2/3 ,b1/b2 =1/2 , c1/c2=-5/-8

a1/a2 ≠ b1/b2

∴ pair of linear equation will intersect each other at an unique point

solution by cross multipication

x/b1c2-b2c1 =y/c1a2-c2a1=1/a1b2-a2b1

x/(-8)-(-10) =y/(-15)-(-16)=1/4-3

x/2 =y/1 =1/1

x/2 =1

x=2

y =1

therefore x=2 and y =1

(iii) 3x – 5y = 20 ,

6x – 10y = 40

a1/a2 =3/6 ,b1/b2 =-5/-10=1/2 , c1/c2=-20/-40=1/2

a1/a2 =b1b2 =c1/c2

therefore the pair line overlaping each other pair of linear equation has  infinitely many solutions.

(iv) x – 3y – 7 = 0 ,

3x – 3y – 15 = 0

a1/a2 =1/3 ,b1/b2 =-3/-3=1/1 , c1/c2=-7/-15=7/15

a1/a2≠b1/b2

∴ pair of linear equation will intersect each other at an unique point

solution by cross multipication

x/b1c2-b2c1 =y/c1a2-c2a1=1/a1b2-a2b1

x/(45)-(21) =y/(-21)-(-15)=1/-3-(-9)

x/24 =y/-6 =1/6

x=4 ,y=-1

QUE:-2. (i) For which values of a and b does the following pair of linear equations have an
              infinite number of solutions?
              2x + 3y = 7
             (a – b) x + (a + b) y = 3a + b – 2
           (ii) For which value of k will the following pair of linear equations have no solution?
              3x + y = 1
            (2k – 1) x + (k – 1) y = 2k + 1

 SOL:-

(i)   2x + 3y = 7
     (a – b) x + (a + b) y = 3a + b – 2

  (a – b) x + (a + b) y -(3a + b – 2) = 0

a1/a2=b1/b2=c1/c2

2/(a – b) =3/(a + b) =7/(3a + b – 2)

2/(a – b) =7/(3a + b – 2)

6a +2b -4 = 7a -7b

a -9b =-4 …………………….(1)

2/(a – b) =3/(a + b)

2a +2b =3a -3b

a -5b =0………………….(2)

substracting equ.(1) and equ.(2)

a -9b =-4

  -a +5b =0

-4b = -4

b = 1

putting the value of b in equ.(1)

a -9 =-4

a = 5

a=5 and b= 1 is the solution for which pair of linear equ. give infinite number of solutions .

(ii)         3x + y -1=
(2k – 1) x + (k – 1) y -2k + 1=  0

a1/a2=b1/b2=c1/c2

3/(2k-1)=1/(k-1) =-1/-2k +1 = 1/2k -1

3/(2k-1) = 1/2k -1

6k -3 =2k -1

k = -4/-2=2

for k = 2 the linear equ.has no solution .

QUE:-3. Solve the following pair of linear equations by the substitution and cross-multiplication
            methods :
                8x + 5y = 9
                3x + 2y = 4

SOL :-

    8x + 5y = 9 …………….(1)
        3x + 2y = 4 ……………..(2)

from the equ. (1) we get

8x + 5y = 9

x= (9 -5y)/8…………….(3)

by substitution value of x in equ.(2)

3(9 -5y)/8 + 2y = 4

27-15y +16y =32

y=32-27

y=5

putting the value of y in equ. (3)

x= (9 -5y)/8 = (9-25)/8=-2

so the value of x =-2 and value of y=5

By cross multypication 

  8x + 5y -9= 0

  3x + 2y -4= 0

x/b1c2-b2c1 =y/c1a2-c2a1=1/a1b2-a2b1

x/-20-(-18)=y/-27 -(-32) =1/16-15

x/-2=y/5=1/1

x=-2

y=5

QUE:-4. Form the pair of linear equations in the following problems and find their solutions (if
          they exist) by any algebraic method :

(i) A part of monthly hostel charges is fixed and the remaining depends on the
   number of days one has taken food in the mess. When a student A takes food for
   20 days she has to pay ` 1000 as hostel charges whereas a student B, who takes
   food for 26 days, pays ` 1180 as hostel charges. Find the fixed charges and the
   cost of food per day.
(ii) A fraction becomes1/3when 1 is subtracted from the numerator and it becomes
    1/4when 8 is added to its denominator. Find the fraction.
(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1
    mark for each wrong answer. Had 4 marks been awarded for each correct answer
    and 2 marks been deducted for each incorrect answer, then Yash would have
    scored 50 marks. How many questions were there in the test?
(iv) Places A and B are 100 km apart on a highway. One car starts from A and another
     from B at the same time. If the cars travel in the same direction at different speeds,
     they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What
     are the speeds of the two cars?
(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by
   5 units and breadth is increased by 3 units. If we increase the length by 3 units and
   the breadth by 2 units, the area increases by 67 square units. Find the dimensions
   of the rectangle.

SOL:-

(i) let the fix charge is of hostel= xrs

and the charge of food in hostel = yrs/day

according to the que.

charge for 20 day =x +20y = 1000……………(1)

charge for 26 days= x +26y =1180……………(2)

subtracting equ.(1) from equ.(2)

x +26y =1180

-x -20y = -1000

6y =180

y=180/6=30

putting the value of y in equ. (1)

x +20y = 1000

x  + 20 ×30 =1000

x = 400

the fix charge is of hostel= 400rs

and the charge of food in hostel = 30rs/day

(ii) let the fraction is x/y

according to the que.

x-1/y =1/3

3x -3 =y

3x -y =3………….(1)

and x/y+8 =1/4

4x =y +8

4x -y =8…………….(2)

substracting equ.(1) from equ.(2)

4x -y =8

-3x +y =-3

x = 5

putting the value of x in equ. (1)

3×5 -y =3

– y = 3-15

y = 12

so the fraction is 5/12

(iii) let number of correct answer is =x

and number of wrong answer is =y

according to the quetion

3x – y =40……………..(1)

and , 4x -2y =50

2x -y =25…………..(2)

substracting equ.(2) from equ. (1)

3x – y =40

-2x +y =-25

x  = 15

putting the value of x in equ.(1)

3×15 – y =40

45 -y = 40

-y = 40-45

-y = -5

y =5

so the total number of quetion in test is x +y =20

(iv) let the speed of first car is =x km/h

and speed of scond car is =y km/h

when cars  move in same direction =(x-y) km/h

when cars  move in opposite direction =(x+y)km/h

according to the que

5(x-y) =100

x-y =20……..(1)

1(x+y) =100

x +y =100………(2)

substracting equ. (1) from equ.(2)

x +y =100

-x+y =-20

2y =80

y =40

putting the value of y in equ. (1)

x-y =20

x =60

the speed of first car is =60 km/h

and speed of scond car is =40km/h

(v) let the length of rectangle is =x

and the breadth of rectangle is= y

area of rectangle =xy

according to the que.

(x-5)(y+3) =xy -9

3x -5y -6 =0 …………..(1)

(x+3)(y+2) =xy +67

2x -3y -61 =0 …………(2)

by cross multyplication

x/b1c2-b2c1 =y/c1a2-c2a1=1/a1b2-a2b1

x/305-(-18)=y/-12-(-183)=1/9-(-10)

x/323=y/171=1/19

x =17 ,y=9

hence the length of rectangle is =17

and the breadth of rectangle is= 9

You can see the solution for complete chapter here –

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