Linear equations in two variables – Exercise 3.2 – Class 10

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

Solution:

(i) Let the number of girls be x and the number of boys be y.

According to the question,

the algebraic representation is

x + y = 10

x − y = 4

For x + y = 10, x = 10 – y

x546
y564

For x − y = 4,

x = 4 + y

x543
y10-1

the graphic representation is,

Linear equations in two variables - Exercise 3.2 - Class 10

From the graph, it can be observed that these lines intersect each other at point (7, 3).

Therefore, the number of girls and boys in the class are 7 and 3 respectively.

(ii)Let the cost of 1 pencil be Rs x and the cost of 1 pen be Rs y.

According to the question, the algebraic representation is 5x + 7y = 50

7x + 5y = 46

For 5x + 7y = 50,

x = (50-7y)/5

x310-4
y5010

7x + 5y = 46

x = (46-5y)/7

x83-2
y-2512

Linear equations in two variables - Exercise 3.2 - Class 10

From the graph, it can be observed that these lines intersect each other at point (3, 5). Therefore, the cost of a pencil and a pen are Rs 3 and Rs 5 respectively.


2. On comparing the ratios, a1/a2 ; b1/b2 and c1/c2 find out whether the lines representing the following pairs of linear equations at a point, are parallel or coincident:

(1) 5x – 4y + 8 = 0

7x + 6y – 9 = 0

(2) 9x + 3y + 12 = 0

18x + 6y + 24 = 0

(3) 6x – 3y + 10 = 0

2x – y + 9 = 0

Solution:

(1) 5x – 4y + 8 = 0

7x + 6y – 9 = 0

Compare these equation 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

Since a1/a2 ≠ b1/b2

Thus, the lines representing the given pair of equations have a unique solution and the pair of lines intersects at exactly one point.

 

(2) 9x + 3y + 12 = 0

18x + 6y + 24 = 0

Compare these equation with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, we get,

a1 = 9, b1 = 3;  c1 = 12

a2 = 18, b2 = 6;  c2 = 24

a1/a2 = 9/18 = 1/2

b1/b2 = 3/6  = 1/2

Since a1/a2 = b1/b2

Thus, the lines representing the given pair of equations are coincident and there are infinite possible solutions for the given pair of equations.

(3) 6x – 3y + 10 = 0

2x – y + 9 = 0

Compare these equation 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

b1/b2 = -3/-1  = 3

c1/c2 = 10/9

Since a1/a2 = b1/b2 ≠ c1/c2

Thus, the lines representing the given pair of equations are parallel to each other and hence, these lines will never intersect each other at any point or there is no possible solution for the given pair of equations.


3. On comparing the ratios a1/a2, b1/b2 and 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

Solution:

(i) 3x + 2y = 5

2x – 3y = 7

a1/a2 = 3/2

b1/b2 = -2/3

c1/c2 = 5/7

a1/a2 ≠ b1/b2

These linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

(ii)2x − 3y = 8

4x − 6y = 9

a1/a2 = 2/4 = 1/2

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

c1/c2 = 8/9

a1/a2 = b1/b2 ≠ c1/c2

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.

(iii) 3/2 x + 5/3 y = 7

9x – 10y = 14

a1/a2 = (3/2)/9 = 3/18  = 1/6

b1/b2 = (5/3)x(-10) = -1/6

c1/c2 = 7/14 = 1/2

a1/a2 ≠ b1/b2

These linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

 

(iv) 5x – 3y = 11

-10x + 6y = -22

a1/a2 = 5/-10 = -12

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

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

a1/a2 = b1/b2 = c1/c2

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

 

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

2x + 3y = 12

a1/a2 = (4/3)/2 = 23

b1/b2 = 2/3

c1/c2 = 8/12 = 2/3

a1/a2 = b1/b2 = c1/c2

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.


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

(1) x + y = 5

2x + 2y = 10

(2) x – y = 8

3x – 3y = 16

(3) 2x + y – 6 = 0

4x – 2y – 4 = 0

(4) 2x – 2y – 2 = 0

4x – 4y – 5  = 0

Solution:

(1) x + y = 5

2x + 2y = 10

a1/a2  = 1/2

b1/b2  = 1/2

c1/c2  = 5/10 = 1/2

So, we have, a1/a2  = b1/b2  = c1/c2

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

x + y = 5,

x = 5 – y

x432
y123

and 2x + 2y = 10

x = (10-2y)/2

x432
y123

the graphic representation is,

Linear equations in two variables - Exercise 3.2 - Class 10

From the graph, we can observe that these lines are overlapping each other. Therefore, infinite solutions are possible for the given pair of equations.

(2) x – y = 8

3x – 3y = 16

a1/a2  = 1/3

b1/b2  = -1/-3 = 1/3

c1/c2  = 8/16 = 1/2

So, we have, a1/a2  = b1/b2  ≠ c1/c2

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent

(3) 2x + y – 6 = 0

4x – 2y – 4 = 0

a1/a2  = 2/4 = 1/2

b1/b2  = -1/2

c1/c2  = -6/-4 = 3/2

So, we have, a1/a2  ≠ b1/b2   

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

2x + y − 6 = 0,

y = 6 − 2x

x012
y642

and 4x − 2y − 4 = 0 y = (4x-4)/2

x123
y024

the graphic representation is,

Linear equations in two variables - Exercise 3.2 - Class 10

From the graph we can observe that these lines are intersecting each other at the only point i.e., (2, 2) and it is the solution for the given pair of equations.

(iv) 2x – 2y – 2 = 0

4x – 4y – 5  = 0

a1/a = 2/4 = 1/2

b1/b = -2/-4 = 1/2

c1/c = 2/5 = 2/5

So, we have, a1/a = b1/b ≠ c1/c

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent


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 be x and length be y.

By data given in the problem

y − x = 4 ………………. (1)

y + x = 36 ……………..(2)

y − x = 4

x0812
y41216

y + x = 36

x03616
y36020

the graphic representation is,

Linear equations in two variables - Exercise 3.2 - Class 10

From the graph we can observe that these lines are intersecting each other at only point i.e., (16, 20). Therefore, the length and width of the given garden is 20 m and 16 m respectively.


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

(i) Intersecting lines

(ii) Parallel lines

(iii) Coincident lines 

Solution:

(i) Intersecting lines

For this,

a1/a2 ≠ b1/b2

The second line such that it is intersecting the given line is

2x + 4y – 6 = 0 as a1/a2 = 2/2  = 1 and  b1/b2 = 3/4  ; a1/a2 ≠ b1/b2

(ii) Parallel lines:

For this condition,

a1/a2 = b1/b2≠ c1/c2

Hence the second line can be

4x + 6y – 8 = 0

as a1/a2 = 2/4  = 1/2 ;  b1/b2 = 3/6 =1/2 ; c1/c2 = -8/-8 = 1

Clearly, a1/a2 = b1/b2 ≠ c1/c2

(iii) coincident lines

For this,

a1/a2 = b1/b2 = c1/c2

Hence 6x + 9y – 24 = 0 be the second line

a1/a2 = 2/6  = 1/3;  b1/b2 = 3/9 = 1/3 ; c1/c2 = -8/-24 = 1/3

Thus a1/a2 = b1/b2 = c1/c2


  1. 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:

x − y + 1 = 0

x = y – 1

x012
y123

3x + 2y – 12 = 0

x = (12-2y)/3

x420
y036

the graphic representation is

Linear equations in two variables - Exercise 3.2 - Class 10

From the figure, it can be observed that these lines are intersecting each other at point (2, 3) and x-axis at (−1, 0) and (4, 0).

Therefore, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).

 

2 thoughts on “Linear equations in two variables – Exercise 3.2 – Class 10

  1. It’s shocking how I didn’t get very far past the first few problems when I was already lost. I mean, I was an English major but still. Struggling through to the end makes me feel a bit better than I was yesterday though so thank you.

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