Chapter 3: Pair of Linear Equations in Two Variables
Exercise & In-Text Questions & Answers

From the table above, you can observe that if the lines represented by the equation a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are:

• (i) intersecting, then a₁/a₂ ≠ b₁/b₂
• (ii) coincident, then a₁/a₂ = b₁/b₂ = c₁/c₂
• (iii) parallel, then a₁/a₂ = b₁/b₂ ≠ c₁/c₂

(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.

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

According to the problem:

x + y = 10 ...(1)

y = x + 4 ⇒ x - y = -4 ...(2)

To solve graphically, we find two solutions for each equation:

For equation (1): x + y = 10

• When x = 0, y = 10
• When x = 10, y = 0

For equation (2): x - y = -4

• When x = 0, y = 4
• When x = 4, y = 8

Plotting these points and drawing the lines, we find they intersect at (3, 7).

Therefore, number of boys = 3 and number of girls = 7.

(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.

Let the cost of one pencil be ₹x and one pen be ₹y.

According to the problem:

5x + 7y = 50 ...(1)

7x + 5y = 46 ...(2)

To solve graphically, we find two solutions for each equation:

For equation (1): 5x + 7y = 50

• When x = 0, y = 50/7 ≈ 7.14
• When y = 0, x = 10

For equation (2): 7x + 5y = 46

• When x = 0, y = 46/5 = 9.2
• When y = 0, x = 46/7 ≈ 6.57

Plotting these points and drawing the lines, we find they intersect at (3, 5).

Therefore, cost of one pencil = ₹3 and cost of one pen = ₹5.

(i) 5x – 4y + 8 = 0; 7x + 6y – 9 = 0

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, the lines intersect at a point.

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

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 lines are coincident.

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

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, the lines are parallel.

(i) 3x + 2y = 5; 2x – 3y = 7

a₁ = 3, b₁ = 2, c₁ = -5

a₂ = 2, b₂ = -3, c₂ = -7

a₁/a₂ = 3/2, b₁/b₂ = 2/-3 = -2/3

Since a₁/a₂ ≠ b₁/b₂, the equations are consistent.

(ii) 2x – 3y = 8; 4x – 6y = 9

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 equations are inconsistent.

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

Multiplying first equation by 6: 9x + 10y = 42

a₁ = 9, b₁ = 10, c₁ = -42

a₂ = 9, b₂ = -10, c₂ = -14

a₁/a₂ = 9/9 = 1, b₁/b₂ = 10/-10 = -1

Since a₁/a₂ ≠ b₁/b₂, the equations are consistent.

(iv) 5x – 3y = 11; –10x + 6y = –22

a₁ = 5, b₁ = -3, c₁ = -11

a₂ = -10, b₂ = 6, c₂ = 22

a₁/a₂ = 5/-10 = -1/2, b₁/b₂ = -3/6 = -1/2, c₁/c₂ = -11/22 = -1/2

Since a₁/a₂ = b₁/b₂ = c₁/c₂, the equations are consistent (dependent).

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

Multiplying first equation by 3: 4x + 6y = 24

a₁ = 4, b₁ = 6, c₁ = -24

a₂ = 2, b₂ = 3, c₂ = -12

a₁/a₂ = 4/2 = 2, b₁/b₂ = 6/3 = 2, c₁/c₂ = -24/-12 = 2

Since a₁/a₂ = b₁/b₂ = c₁/c₂, the equations are consistent (dependent).

Example 1: Check graphically whether the pair of equations x + 3y = 6 and 2x – 3y = 12 is consistent. If so, solve them graphically.

The two lines intersect at point B(6, 0). So, the solution is x = 6 and y = 0.

(i) x + y = 14; x – y = 4

From equation (2): x = y + 4

Substituting in equation (1): (y + 4) + y = 14 ⇒ 2y = 10 ⇒ y = 5

Then x = 5 + 4 = 9

Solution: x = 9, y = 5

(ii) s – t = 3; s/3 + t/2 = 6

From equation (1): s = t + 3

Substituting in equation (2): (t + 3)/3 + t/2 = 6

Multiplying by 6: 2(t + 3) + 3t = 36 ⇒ 2t + 6 + 3t = 36 ⇒ 5t = 30 ⇒ t = 6

Then s = 6 + 3 = 9

Solution: s = 9, t = 6

(iii) 3x – y = 3; 9x – 3y = 9

From equation (1): y = 3x - 3

Substituting in equation (2): 9x - 3(3x - 3) = 9 ⇒ 9x - 9x + 9 = 9 ⇒ 9 = 9

This is a true statement for all values of x. Therefore, the equations have infinitely many solutions.

(iv) 0.2x + 0.3y = 1.3; 0.4x + 0.5y = 2.3

Multiplying both equations by 10: 2x + 3y = 13; 4x + 5y = 23

From first equation: x = (13 - 3y)/2

Substituting in second equation: 4(13 - 3y)/2 + 5y = 23 ⇒ 2(13 - 3y) + 5y = 23 ⇒ 26 - 6y + 5y = 23 ⇒ -y = -3 ⇒ y = 3

Then x = (13 - 9)/2 = 2

Solution: x = 2, y = 3

(v) √2x + √3y = 0; √3x - √8y = 0

From first equation: x = -(√3/√2)y

Substituting in second equation: √3(-(√3/√2)y) - √8y = 0 ⇒ -(3/√2)y - 2√2y = 0 ⇒ y(-3/√2 - 2√2) = 0

Since (-3/√2 - 2√2) ≠ 0, we have y = 0

Then x = 0

Solution: x = 0, y = 0

(vi) (3x/2) - (5y/3) = -2; x/3 + y/2 = 13/6

Multiplying first equation by 6: 9x - 10y = -12

Multiplying second equation by 6: 2x + 3y = 13

From second equation: x = (13 - 3y)/2

Substituting in first equation: 9(13 - 3y)/2 - 10y = -12 ⇒ (117 - 27y)/2 - 10y = -12

Multiplying by 2: 117 - 27y - 20y = -24 ⇒ -47y = -141 ⇒ y = 3

Then x = (13 - 9)/2 = 2

Solution: x = 2, y = 3

(i) x + y = 5 and 2x – 3y = 4

Elimination Method:

Multiplying first equation by 3: 3x + 3y = 15

Adding to second equation: (3x + 3y) + (2x - 3y) = 15 + 4 ⇒ 5x = 19 ⇒ x = 19/5

Substituting in first equation: 19/5 + y = 5 ⇒ y = 5 - 19/5 = 6/5

Solution: x = 19/5, y = 6/5

Substitution Method:

From first equation: y = 5 - x

Substituting in second equation: 2x - 3(5 - x) = 4 ⇒ 2x - 15 + 3x = 4 ⇒ 5x = 19 ⇒ x = 19/5

Then y = 5 - 19/5 = 6/5

Solution: x = 19/5, y = 6/5

(ii) 3x + 4y = 10 and 2x – 2y = 2

Elimination Method:

Multiplying second equation by 2: 4x - 4y = 4

Adding to first equation: (3x + 4y) + (4x - 4y) = 10 + 4 ⇒ 7x = 14 ⇒ x = 2

Substituting in second equation: 2(2) - 2y = 2 ⇒ 4 - 2y = 2 ⇒ y = 1

Solution: x = 2, y = 1

Substitution Method:

From second equation: x = y + 1

Substituting in first equation: 3(y + 1) + 4y = 10 ⇒ 3y + 3 + 4y = 10 ⇒ 7y = 7 ⇒ y = 1

Then x = 1 + 1 = 2

Solution: x = 2, y = 1