149-NCERT-10-3-Pair of Linear Equations in Two Variables - Ex-3.2

NCERT

10th Mathematics

Exercise 3.2

Topic: 3 Pair of Linear Equations in Two Variables

Click here for ⇨ NCERT-10-3-Pair of Linear Equations in Two Variables - Ex-3.1

EXERCISE 3.2

Q1. 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. 

Explanation:

1) Here, let x and y be two variables.

2) Apply the given conditions and frame the equations.

3) We will get two equations from the above two conditions, then solve these

equations to get the values of x and y. 

Solution:

(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.

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

2) As 10 students of class X took part in a mathematics quiz,

x + y = 10

y = 10 - x ------------ equation 1

3) As the number of girls is 4 more than the number of boys,

x = y + 4

        y = x - 4 ------------ equation 2

4) Now, we will represent these equations graphically.

a) We will take 3 points for y = 10 - x.

b) We will take 3 points for y = x - 4.

5) The graphical representation will be as follows.

6) Here the lines intersect at (7,3), so the number of girls is 7 and the number of boys is 3.

(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. 

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

2) As 5 pencils and 7 pens cost Rs 50, we have,

5 x + 7 y = 50

7 y = (50 - 5 x)

   y = (50 - 5 x)/7 ------------ equation 1

3) As 7 pencils and 5 pens cost Rs 46, we have,

7 x + 5 y = 46

5 y = (46 - 7 x)

   y = (46 - 7 x)/5  ------------ equation 2

4) Now, we will represent these equations graphically.

a) We will take 2 points for y = (50 - 5 x)/7. 

Q2. On comparing the ratios a₁/a_2,  b₁/b_{2, and} 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

Explanation:

1) For the equations, a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

a) If a₁/a₂ = b₁/b₂ = c₁/c₂ then the lines are coincident.

b) If a₁/a₂ = b₁/b₂ ≠ c₁/c₂ then the lines are parallel.

c) If a₁/a₂ ≠ b₁/b₂ then the lines are intersecting.

Solution:

(i) 5x – 4y + 8 = 0, 7x + 6y – 9 = 0 

1) Here a₁ = 5, a₂ = 7, b₁ = - 4, b₂ = 6, c₁ = 8, c₂ = - 9.

a₁/a₂ = 5/7 -------------1

b₁/b₂ = - 4/6 -------------2

c₁/c₂ = 8/(- 9) -------------3

2) From 1, 2, and 3, we can say that a₁/a₂ ≠ b₁/b₂, so the lines are intersecting.

(ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0

1) Here a₁ = 9, a₂ = 18, b₁ = 3, b₂ = 6, c₁ = 12, c₂ = 24.

a₁/a₂ = 9/18 = 1/2 -------------1

b₁/b₂ = 3/6 = 1/2 -------------2

c₁/c₂ = 12/24 = 1/2 -------------3

2) From 1, 2, and 3, we can say that a₁/a₂ = b₁/b₂ = c₁/c₂, so the lines are coincident.

(iii) 6x – 3y + 10 = 0, 2x – y + 9 = 0

1) Here a₁ = 6, a₂ = 2, b₁ = - 3, b₂ = - 1, c₁ = 10, c₂ = 9.

a₁/a₂ = 6/2 = 3 -------------1

b₁/b₂ = - 3/(- 1) = 3 -------------2

c₁/c₂ = 10/9 -------------3

2) From 1, 2, and 3, we can say that a₁/a₂ = b₁/b₂ ≠ c₁/c₂, so the lines are parallel.

Q3. On comparing the ratios a₁/a_2,  b₁/b_{2, 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

Explanation:

1) For the equations, a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

a) If a₁/a₂ = b₁/b₂ = c₁/c_2, the lines are coincident with infinitely many solutions

so they are consistent.

b) If a₁/a₂ = b₁/b₂ ≠ c₁/c_2, the lines are parallel with no solutions, so they

are inconsistent.

c) If a₁/a₂ ≠ b₁/b_2, the lines are intersecting with unique solution, so they

are consistent.

Solution:

(i) 3x + 2y = 5 ; 2x – 3y = 7

1) Write our equations as, 3x + 2y - 5 = 0; 2x – 3y - 7 = 0.

2) Here a₁ = 3, a₂ = 2, b₁ = 2, b₂ = - 3, c₁ = - 5, c₂ = - 7.

a₁/a₂ = 3/2 -------------1

b₁/b₂ = 2/(- 3) -------------2 

c₁/c₂ = (- 5)/(- 7) -------------3

3) From 1, 2, 3, we can say that (a₁/a₂ ≠ b₁/b₂), 

so the lines are intersecting with unique solution, so they are consistent.

 

(ii) 2x – 3y = 8 ; 4x – 6y = 9

1) Write our equations as, 2x - 3y - 8 = 0; 4x – 6y - 9 = 0.

2) Here a₁ = 2, a₂ = 4, b₁ = - 3, b₂ = – 6, c₁ = - 8, c₂ = - 9.

a₁/a₂ = 2/4 = 1/2 -------------1

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

c₁/c₂ = (- 8)/(- 9) = 8/9 -------------3

3) From 1, 2, 3, we can say that (a₁/a₂ = b₁/b₂ ≠ c₁/c₂), 

the lines are parallel with no solutions, so they are inconsistent.

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

1) Write our equations as, (3/2)x + (5/3)y - 7 = 0; 9x – 10y - 14 = 0.

2) Here a₁ = (3/2), a₂ = 9, b₁ = (5/3), b₂ = – 10, c₁ = - 7, c₂ = - 14.

a₁/a₂ = (3/2)/(9/1) = 1/6 -------------1

b₁/b₂ = (5/3)/(- 10/1) = (- 1/6) -------------2 

c₁/c₂ = (- 7)/(- 14) = 1/2 -------------3

3) From 1, 2, 3, we can say that (a₁/a₂ ≠ b₁/b₂), 

so the lines are intersecting with unique solution, so they are consistent. 

(iv) 5x – 3y = 11 ; – 10x + 6y = –22

1) Write our equations as, 5x - 3y - 11 = 0; – 10x + 6y + 22 = 0.

2) Here a₁ = 5, a₂ = - 10, b₁ = - 3, b₂ = 6, c₁ = - 11, c₂ = 22.

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

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

c₁/c₂ = (- 11)/(22) = (- 1/2) -------------3

3) From 1, 2, 3, we can say that (a₁/a₂ = b₁/b₂ = c₁/c₂), 

the lines are coincident with infinitly many solutions, so they are consistent.

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

1) Write our equations as, (4/3)x + 2y - 8 = 0; 2x + 3y - 12 = 0.

2) Here a₁ = (4/3), a₂ = 2, b₁ = 2, b₂ = 3, c₁ = - 8, c₂ = - 12.

a₁/a₂ = (4/3)/(2/1) = 2/3 -------------1

b₁/b₂ = 2/3 -------------2 

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

3) From 1, 2, 3, we can say that (a₁/a₂ = b₁/b₂ = c₁/c₂), 

the lines are coincident with infinitly many solutions, so they are consistent.

Q4. 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

Explanation:

1) For the equations, a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

a) If a₁/a₂ = b₁/b₂ = c₁/c_2, the lines are coincident with infinitly many solutions

so they are consistent.

b) If a₁/a₂ = b₁/b₂ ≠ c₁/c_2, the lines are parallel with no solutions, so they

are inconsistent.

c) If a₁/a₂ ≠ b₁/b_2, the lines are intersecting with unique solution, so they

are consistent.

Solution:

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

1) Write our equations as, x + y - 5 = 0; 2x + 2y - 10 = 0.

2) Here a₁ = 1, a₂ = 2, b₁ = 1, b₂ = 2, c₁ = - 5, c₂ = - 10.

a₁/a₂ = 1/2 -------------1

b₁/b₂ = 1/2 -------------2 

c₁/c₂ = (- 5)/(- 10) = 1/2 -------------3

3) From 1, 2, 3, we can say that (a₁/a₂ = b₁/b₂ = c₁/c₂), 

the lines are coincident with infinitly many solutions, so they are consistent.

4) We will simplify our equations x + y = 5 and 2x + 2y = 10,

x + y = 5

so, y = (5 - x) -------------4

2x + 2y = 10

 x + y = 5

so, y = (5 - x) -------------5

5) Now, we will represent these equations graphically.

a) We will take 3 points for y = (5 - x).

b) We will take 3 points for y = (5 - x).

6) The graphical representation will be as follows. 

7) The lines are coincident with infinitely many solutions, so they are consistent.

(ii) x – y = 8, 3x – 3y = 16

1) Write our equations as, x - y - 8 = 0; 3x - 3y - 16 = 0.

2) Here a₁ = 1, a₂ = 3, b₁ = - 1, b₂ = - 3, c₁ = - 8, c₂ = - 16.

a₁/a₂ = 1/3 -------------1

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

c₁/c₂ = (- 8)/(- 16) = 1/2 -------------3

3) From 1, 2, 3, we can say that (a₁/a₂ = b₁/b₂ ≠ c₁/c₂), 

the lines are parallel with no solutions, so they are inconsistent.

4) We will simplify our equations x - y = 8 and 3x - 3y = 16,

x - y = 8

so, y = (x - 8) -------------4

3x - 3y = 16

3(x - y) = 16 

 x - y = 16/3

so, y = x - (16/3) -------------5

5) Now, we will represent these equations graphically.

a) We will take 3 points for y = (x - 8). 

b) We will take 3 points for y = x - (16/3).

6) The graphical representation will be as follows. 

7) The lines are parallel with no solutions, so they are inconsistent.

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0

1) Write our equations as, 2x + y - 6 = 0; 4x - 2y - 4 = 0.

2) Here a₁ = 2, a₂ = 4, b₁ = 1, b₂ = - 2, c₁ = - 6, c₂ = - 4.

a₁/a₂ = 2/4 = 1/2 -------------1

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

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

3) From 1, 2, 3, we can say that (a₁/a₂ ≠ b₁/b₂), 

the lines are intersecting with unique solution, so they are consistent.

4) We will simplify our equations 2x + y - 6 = 0 and 4x - 2y - 4 = 0,

2x + y = 6

so, y = (6 - 2x) -------------4

4x - 2y = 4

2(2x - y) = 4

(2x - y) = 2

 y = 2x - 2 -------------5

5) Now, we will represent these equations graphically.

a) We will take 3 points for y = (6 - 2x). 

b) We will take 3 points for y = (2x - 2).

6) The graphical representation will be as follows.

7) The lines are intersecting with a unique solution, so they are consistent.

8) The solution of these equations is (2, 2)

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

1) Write our equations as, 2x - 2y - 2 = 0; 4x - 4y - 5 = 0.

2) Here a₁ = 2, a₂ = 4, b₁ = - 2, b₂ = - 4, c₁ = - 2, c₂ = - 5.

a₁/a₂ = 2/4 = 1/2 -------------1

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

c₁/c₂ = (- 2)/(- 5) = 2/5 -------------3

3) From 1, 2, 3, we can say that (a₁/a₂ = b₁/b₂ ≠ c₁/c₂), 

the lines are parallel with no solutions, so they are inconsistent.

4) We will simplify our equations 2x - 2y - 2 = 0 and 4x - 4y - 5 = 0,

2x - 2y = 2

so, y = (x - 1) -------------4

4x - 4y = 5

4(x - y) = 5

(x - y) = 5/4

 y = x - (5/4) -------------5

5) Now, we will represent these equations graphically.

a) We will take 3 points for y = (x - 1).

b) We will take 3 points for y = (x - (5/4).

6) The graphical representation will be as follows.

7) The lines are parallel with no solutions, so they are inconsistent.

Q5. 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.

Explanation:

1) Here, let x and y be the length and width of the rectangular garden.

2) Apply the given conditions and frame the equation.

3) We will get two equations from the above two conditions, then solve these equations to get the values of x and y. 

Solution:

1) Let the length be x and the width be y.

2) As the length is 4 m more than its width, we have,

x = y + 4

y = x - 4 ------------ equation 1

3) We know that the perimeter of a rectangle is 2(x + y), 

so half of perimeter is (x + y).

4) Therefore, as half of the perimeter is 36 m, we have,

x + y = 36

y = 36 - x ------------ equation 2

5) Now, we will represent these equations graphically.

a) We will take 3 points for y = (x - 4).

 

b) We will take 3 points for y = (36 - x).

6) The graphical representation will be as follows.

7) The lines are intersecting with a unique solution, so they are consistent.

8) Here, the dimensions of the rectangular garden are as follows.

a) length is 20 m,

b) width is 16 m.

Q6. 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

Explanation:

1) For the equations, a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

a) If a₁/a₂ = b₁/b₂ = c₁/c_2, the lines are coincident with infinitly many solutions

so they are consistent.

b) If a₁/a₂ = b₁/b₂ ≠ c₁/c_2, the lines are parallel with no solutions, so they

are inconsistent.

c) If a₁/a₂ ≠ b₁/b_2, the lines are intersecting with unique solution, so they

are consistent.

Solution:

(i) Intersecting lines

1) For the equations, a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0,

a) If a₁/a₂ ≠ b₁/b_2, the lines will be intersecting with unique solution.

2) Let our equation be a₁x + b₁y + c₁ = 0, and the given equation is 2x + 3y – 8 = 0.

3) So, here a₁ = a₁, a₂ = 2, b₁ = b₁, b₂ = 3, c₁ = c₁, c₂ = - 8.

a₁/a₂ = a₁/2 -------------1

b₁/b₂ = b₁/3-------------2 

c₁/c₂ = c₁/(- 8) -------------3

4) As our lines are intersecting, we have a₁/a₂ ≠ b₁/b_2,

so, we have a₁/2 ≠ b₁/3. Let us take any values for a_1, b₁ and c₁ which satisfies, a₁/2 ≠ b₁/3 

let a₁ = 3, b₁ = (-2), and c₁= 14.

5) So, our equation will be 3x - 2y + 14 = 0, the given equation is 2x + 3y – 8 = 0.

6) We will simplify our equations 2x + 3y – 8 = 0 and 3x - 2y + 14 = 0,

2x + 3y = 8

so, 3y = (8 - 2x) 

 y = (8 - 2x)/3 -------------4

3x - 2y = -14

2y = (3x + 14)

y = (3x + 14)/2 -------------5

7) Now, we will represent these equations graphically.

a) We will take 3 points for y = (8 - 2x)/3.

b) We will take 3 points for y = (3x + 14)/2.

8) The graphical representation will be as follows. 

9) Here the point of intersection is (-2, 4).

(ii) parallel lines

1) For the equations, a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

a) If a₁/a₂ = b₁/b₂ ≠ c₁/c_2, the lines are parallel with no solutions, so they

are inconsistent.

2) Let our equation be a₁x + b₁y + c₁ = 0, and the given equation is 2x + 3y – 8 = 0.

3) So, here a₁ = a₁, a₂ = 2, b₁ = b₁, b₂ = 3, c₁ = c₁, c₂ = - 8.

a₁/a₂ = a₁/2 -------------1

b₁/b₂ = b₁/3-------------2 

c₁/c₂ = c₁/(- 8) -------------3

4) As our lines are parallel, we have a₁/a₂ = b₁/b₂ ≠ c₁/c_2,

so, we have a₁/2 = b₁/3. Let us take any values for a_1, b₁ and c₁ which satisfies, a₁/2 = b₁/3 

let a₁ = 4, b₁ = 6, and c₁= 9.

5) So, our equation will be 4x + 6y + 9 = 0, the given equation is 2x + 3y – 8 = 0.

6) We will simplify our equations 2x + 3y – 8 = 0 and 4x + 6y + 10 = 0,

2x + 3y = 8

so, 3y = (8 - 2x) 

 y = (8 - 2x)/3 -------------4

4x + 6y = 10

6y = (10 - 4x)

y = (10 - 4x)/6 -------------5

7) Now, we will represent these equations graphically.

a) We will take 3 points for y = (8 - 2x)/3.

b) We will take 3 points for y = (10 - 4x)/6.

8) The graphical representation will be as follows. 

9) Here the lines are parallel.

(iii) coincident lines

1) For the equations, a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

a) If a₁/a₂ = b₁/b₂ = c₁/c_2, the lines are coincident with infinitly many solutions.

2) Let our equation be a₁x + b₁y + c₁ = 0, and the given equation is 2x + 3y – 8 = 0.

3) So, here a₁ = a₁, a₂ = 2, b₁ = b₁, b₂ = 3, c₁ = c₁, c₂ = - 8.

a₁/a₂ = a₁/2 -------------1

b₁/b₂ = b₁/3-------------2 

c₁/c₂ = c₁/(- 8) -------------3

4) As our lines are parallel, we have a₁/a₂ = b₁/b₂ = c₁/c_2,

so, we have a₁/2 = b₁/3. Let us take any values for a_1, b₁ and c₁ which satisfies, a₁/2 = b₁/3 

let a₁ = 4, b₁ = 6, and c₁= -16.

5) So, our equation will be 4x + 6y - 16 = 0, the given equation is 2x + 3y – 8 = 0.

6) We will simplify our equations 2x + 3y – 8 = 0 and 4x + 6y - 16 = 0,

2x + 3y = 8

so, 3y = (8 - 2x) 

 y = (8 - 2x)/3 -------------4

4x + 6y = 16

6y = (16 - 4x)

y = (16 - 4x)/6 -------------5

7) Now, we will represent these equations graphically.

a) We will take 3 points for y = (8 - 2x)/3.

 b) We will take 3 points for y = (16 - 4x)/6.

8) The graphical representation will be as follows. 

9) Here the lines are coincident.

Q7. 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.

Explanation:

1) For the equations, a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

a) If a₁/a₂ ≠ b₁/b_2, the lines are intersecting with unique solution, so they

are consistent.

Solution:

1) Write our equations as, x - y + 1 = 0; 3x + 2y - 12 = 0.

2) Here a₁ = 1, a₂ = 3, b₁ = -1, b₂ = 2, c₁ = 1, c₂ = - 12.

a₁/a₂ = 1/3 -------------1

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

c₁/c₂ = 1/(- 12) = -(1/12) -------------3

3) From 1, 2, 3, we can say that (a₁/a₂ ≠ b₁/b₂), 

the lines are intersecting with unique solution, so they are consistent.

4) We will simplify our equations x - y + 1 = 0 and 3x + 2y - 12 = 0,

x - y + 1 = 0

so, y = x + 1 -------------4

3x + 2y = 12

2y = 12 - 3x

y = (12 - 3x)/2 -------------5

5) Now, we will represent these equations graphically.

a) We will take 3 points for y = x + 1.

 

b) We will take 3 points for y = (12 - 3x)/2.

6) The graphical representation will be as follows.

7) The lines are intersecting with unique solutions, so they are consistent.

8) Here, the coordinates of the vertices of the triangle formed by these lines and the

x-axis, and shade the triangular region are A(-1, 0), B (2, 3), and C(4, 0).

Click here for ⇨ NCERT-10-3-Pair of Linear Equations in Two Variables - Ex-3.3

ANIL SATPUTE