NCERT solutions for class 10 Maths chapter 3 Pair of Linear Equations in Two Variables (Exercise 3.2)

NCERT solutions for class 10 Maths


Chapter 3


Pair of Linear Equations in Two Variables


Exercise 3.2


Ex 3.2 Question 1.


1. Form the pair of linear equations in the following problems, and find their solutions graphically.

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

(ii) 5 pencils and 7 pens together cost 50, whereas 7 pencils and 5 pens together cost 46. Find the cost of one pencil and that of one pen.

Solution:

(i) Let the number of girls = x  and the number of boys in the class = y respectively.
x + y = 10
x – y = 4
x + y = 10 Ncert Solutions Cbse Class 10 Mathematics Chapter - Pairs Of Linear Equations In Two Variables x = 10 – y
Three solutions of this equation are

x 5 4 6
y 5 6 4

x – y = 4 Ncert Solutions Cbse Class 10 Mathematics Chapter - Pairs Of Linear Equations In Two Variables x = 4 + y
Three solutions of this equation are

x 5 4 3
y 1 0 -1


Two lines cross each other at the point (7, 3).
∴There are 7 girls and 3 boys in the class. So, x = 7 and y = 3.

(ii) Let the cost of one pencil  = ₹ X and one pen = ₹ y respectively.

5x + 7y = 50
7x + 5y = 46
5x + 7y = 50 or  x = (50-7y)/5
Three solutions of this equation are

x 3 10 -4
y 5 0 10

7x + 5y = 46 or x = (46-5y)/7
Three solutions of this equation are

x 8 3 -2
y -2 5 12


Two lines cross each other at the point (3, 5).
∴The cost of a pencil is 3/- and the cost of a pen is 5/- So, x = 3 and y = 5.


Ex 3.2 Question 2.


On comparing the ratios a1/a2,b1/b1,c1/c2 find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

(i) 5x – 4y + 8 = 0

7x + 6y – 9 = 0

(ii) 9x + 3y + 12 = 0

18x + 6y + 24 = 0

(iii) 6x – 3y + 10 = 0

2x – y + 9 = 0

Solutions:

(i)5x−4y+8 = 0

7x+6y−9 = 0

Comparing these equations with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 We get,

a1 = 5, b1 = -4, c1 = 8

a2 = 7, b2 = 6, c2 = -9

(a1/a2) = 5/7

(b1/b2) = -4/6 = -2/3

(c1/c2) = 8/-9

Since, (a1/a2) ≠ (b1/b2) the given pair of equations intersect at exactly one point.

(ii)9x + 3y + 12 = 0

18x + 6y + 24 = 0

Comparing these equations with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, we ge

a1 = 9, b1 = 3, c1 = 12

a2 = 18, b2 = 6, c2 = 24

(a1/a2) = 9/18 = 1/2

(b1/b2) = 3/6 = 1/2

(c1/c2) = 12/24 = 1/2

Since (a1/a2) = (b1/b2) = (c1/c2) the given pair of equation are coincident.

(iii) 6x – 3y + 10 = 0

2x – y + 9 = 0

Comparing these equations with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, we get

a1 = 6, b1 = -3, c1 = 10

a2 = 2, b2 = -1, c2 = 9

(a1/a2) = 6/2 = 3/1

(b1/b2) = -3/-1 = 3/1

(c1/c2) = 10/9

Since (a1/a2) = (b1/b2) ≠ (c1/c2) the given pair of equation are parallel to each other.


Ex 3.2 Question 3.


On comparing the ratio, a1/a2,b1/b1,c1/c2 find out whether the following pair of linear equations are consistent, or inconsistent.

(i) 3x + 2y = 5 ; 2x – 3y = 7

(ii) 2x – 3y = 8 ; 4x – 6y = 9

(iii)(3/2)x+(5/3)y = 7; 9x – 10y = 14

(iv) 5x – 3y = 11 ; – 10x + 6y = –22

(v)(4/3)x+2y = 8 ; 2x + 3y = 12

Solutions:

(i) 3x + 2y = 5 and 2x – 3y = 7

Comparing these equations with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 We get,

a1 =3, b1 = 2,  c1 =-5

a2 = 2, b2 = -3, c2 = 7

(a1/a2) = 3/2

(b1/b2) =2/-3

(c1/c2) = -5/7

Since, (a1/a2) ≠ (b1/b2) the given pair of equations intersect each other at one point and they have only one possible solution. The equations are consistent.

(ii) 2x – 3y = 8 and 4x – 6y = 9

Comparing these equations with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, we get

a1 = 2, b1 = -3, c1 = 8

a2 = 4, b2 = -6, c2 = -9

(a1/a2) = 2/4 = 1/2

(b1/b2) = -3/-6 = 1/2

(c1/c2) = -8/-9 = 8/9

Since (a1/a2) = (b1/b2) ≠ (c1/c2) the given pair of equations are parallel to each other and they have no possible solution. Hence, the equations are inconsistent.

(iii) (3/2)x + (5/3)y = 7 and 9x – 10y = 14

Comparing these equations with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 We get,

a1 =3/2, b1 = 5/3,  c1 =-7

a2 = 9, b2 = -10, c2 = 14

(a1/a2) = 3/2

(b1/b2) =2/-3

(c1/c2) = -5/7

Since, (a1/a2) ≠ (b1/b2) the given pair of equations intersecting  each other at one point and they have only one possible solution. Hence, the equations are consistent.

(iv) 5x – 3y = 11 and – 10x + 6y = –22

a1= 5, b1 = -3, c1 = -11

a2 = -10, b2 = 6, c2 = 22

(a1/a2) = 5/(-10) = -5/10 = -1/2

(b1/b2) = -3/6 = -1/2

(c1/c2) = -11/22 = -1/2

Since (a1/a2) = (b1/b2) = (c1/c2) the given pair of equations have infinite number of possible solutions. Hence, the equations are consistent.

(v)(4/3)x +2y = 8 and 2x + 3y = 12

a1 = 4/3 , b1= 2 , c1 = -8

a2 = 2, b2 = 3 , c2 = -12

(a1/a2) = 4/(3×2)= 4/6 = 2/3

(b1/b2) = 2/3

(c1/c2) = -8/-12 = 2/3

Since (a1/a2) = (b1/b2) = (c1/c2) the given pair of equations have infinite number of possible solutions. Hence, the equations are consistent.


Ex 3.2 Question 4.


Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

(i) x + y = 5, 2x + 2y = 10

(ii) x – y = 8, 3x – 3y = 16

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

Solutions:

(i)x + y = 5 and 2x + 2y = 10

(a1/a2) = 1/2

(b1/b2) = 1/2

(c1/c2) = 1/2

Since (a1/a2) = (b1/b2) = (c1/c2)the given pair of equations have infinite number of possible solutions. Hence, the equations are consistent.

For, x + y = 5 or x = 5 – y

Ncert solutions class 10 chapter 3-16

For 2x + 2y = 10 or x = (10-2y)/2

Ncert solutions class 10 chapter 3-17

The given pair of equations has infinite possible solutions.

(ii) x – y = 8 and 3x – 3y = 16

(a1/a2) = 1/3

(b1/b2) = -1/-3 = 1/3

(c1/c2) = 8/16 = 1/2

Since, (a1/a2) = (b1/b2) ≠ (c1/c2) the given pair of equations has no solution. Thus, the pair of linear equations is inconsistent.

(iii) 2x + y – 6 = 0 and 4x – 2y – 4 = 0

(a1/a2) = 2/4 = ½

(b1/b2) = 1/-2

(c1/c2) = -6/-4 = 3/2

Since (a1/a2) ≠ (b1/b2) the given pair of equations has only one solution. Hence, the pair of linear equations is consistent.

Now, for 2x + y – 6 = 0 or y = 6 – 2x

Ncert solutions class 10 chapter 3-19

And for 4x – 2y – 4 = 0 or y = (4x-4)/2

Ncert solutions class 10 chapter 3-20

Two lines are intersecting each other at only one point,(2,2). Hence, the solution of the given pair of equations is (2, 2).

(iv) x – 2y – 2 = 0 and 4x – 4y – 5 = 0

(a1/a2) = 2/4 = ½

(b1/b2) = -2/-4 = 1/2

(c1/c2) = 2/5

Since, (a1/a2) = (b1/b2) ≠ (c1/c2) the given pair of equations has no solution. Thus, the pair of linear equations is inconsistent.


Ex 3.2 Question 5.


Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Solution:  Let, the width of the garden is x and length is y.

y – x = 4

y + x = 36

y – x = 4 ⇒ y = x + 4

x 0 8 12
y 4 12 16

For y + x = 36 ⇒y = 36 – x

 

x 0 36 16
y 36 0 20

Two lines intersect each other at the point (16, 20). So, x = 16 and y = 20. Hence, the width of the garden is 16 m and the length is 20 m.


Ex 3.2 Question 6.


Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

(i) Intersecting lines

(ii) Parallel lines

(iii) Coincident lines

Solutions:

(i) Given the linear equation 2x + 3y – 8 = 0.

To find another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines, it should satisfy below condition;

(a1/a2) ≠ (b1/b2)

So, the other linear equation can be 2x – 7y + 9 = 0,

(a1/a2) = 2/2 = 1

(b1/b2) = 3/-7

(ii) Parallel lines

(a1/a2) = (b1/b2) ≠ (c1/c2)

So, the other linear equation can be 6x + 9y + 9 = 0,

(a1/a2) = 2/6 = 1/3

(b1/b2) = 3/9= 1/3

(c1/c2) = -8/9

(iii) Coincident lines

(a1/a2) = (b1/b2) = (c1/c2)

So, the other linear equation can be  4x + 6y – 16 = 0,

(a1/a2) = 2/4 = 1/2 ,(b1/b2) = 3/6 = 1/2, (c1/c2) = -8/-16 = 1/2


Ex 3.2 Question 7.


Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Solution: Given, the equations for graphs are x – y + 1 = 0 and 3x + 2y – 12 = 0.

For, x – y + 1 = 0 or x = 1+y

Ncert solutions class 10 chapter 3-25

For, 3x + 2y – 12 = 0 or x = (12-2y)/3

Ncert solutions class 10 chapter 3-26

The coordinates of the vertices of the triangle so formed are (2, 3), (-1, 0), and (4, 0).


 

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