Welcome Class 10 Students!

Mastering Chapter 3: Pair of Linear Equations in Two Variables is essential for scoring high marks in your Class 10 Mathematics board examinations. Below, you will find complete, step-by-step solutions for all exercises in this chapter, covering Graphical Methods, Substitution Methods, and Elimination Methods in detail.

EXERCISE 3.1

Question 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.

Solution:
Let the number of girls be x and the number of boys be y.
According to the given conditions:
1) x + y = 10 ⇒ y = 10 - x
2) x - y = 4 ⇒ y = x - 4

Table for x + y = 10: (x=5, y=5), (x=7, y=3), (x=3, y=7)
Table for x - y = 4: (x=4, y=0), (x=7, y=3), (x=5, y=1)

Plotting these lines on a graph, both lines intersect at the point (7, 3).
Therefore, the number of girls is 7 and the number of boys is 3.

(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:
Let the cost of 1 pencil = ₹ x and the cost of 1 pen = ₹ y.
Equations formed:
1) 5x + 7y = 50 ⇒ y = (50 - 5x) / 7
2) 7x + 5y = 46 ⇒ y = (46 - 7x) / 5

Table for 5x + 7y = 50: (x=3, y=5), (x=10, y=0), (x=-4, y=10)
Table for 7x + 5y = 46: (x=3, y=5), (x=8, y=-2), (x=-2, y=12)

Plotting both lines, they intersect at point (3, 5).
Hence, the cost of one pencil is ₹ 3 and the cost of one pen is ₹ 5.

Question 2: On comparing ratios a₁/a₂, b₁/b₂, and c₁/c₂, find out whether the lines intersect at a point, are parallel or coincident:

(i) 5x - 4y + 8 = 0 ; 7x + 6y - 9 = 0
Here, a₁/a₂ = 5/7, b₁/b₂ = -4/6 = -2/3.
Since a₁/a₂ ≠ b₁/b₂, the lines intersect at a single point.

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

Question 3: On comparing ratios, find out whether the following pairs of linear equations are consistent or inconsistent:

(i) 3x + 2y = 5 ; 2x - 3y = 7
a₁/a₂ = 3/2, b₁/b₂ = -2/3 ⇒ a₁/a₂ ≠ b₁/b₂. System 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. System is Inconsistent.

(iii) (3/2)x + (5/3)y = 7 ; 9x - 10y = 14
a₁/a₂ = (3/2)/9 = 1/6, b₁/b₂ = (5/3)/(-10) = -1/6 ⇒ a₁/a₂ ≠ b₁/b₂. System is Consistent.

(iv) 5x - 3y = 11 ; -10x + 6y = -22
a₁/a₂ = 5/-10 = -1/2, b₁/b₂ = -3/6 = -1/2, c₁/c₂ = 11/-22 = -1/2 ⇒ Ratios are all equal. System is Consistent.

(v) (4/3)x + 2y = 8 ; 2x + 3y = 12
a₁/a₂ = (4/3)/2 = 2/3, b₁/b₂ = 2/3, c₁/c₂ = 8/12 = 2/3 ⇒ Ratios are equal. System is Consistent.

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
Ratios: 1/2 = 1/2 = 5/10 = 1/2. Coincident lines, infinitely many solutions. System is Consistent.

(ii) x - y = 8, 3x - 3y = 16
Ratios: 1/3 = 1/3 ≠ 8/16 = 1/2. Parallel lines, no solution. System is Inconsistent.

(iii) 2x + y - 6 = 0, 4x - 2y - 4 = 0
Ratios: 2/4 = 1/2, 1/-2 = -1/2. Intersecting lines, unique solution. System is Consistent.
Graph points for 2x + y = 6: (0,6), (2,2), (3,0).
Graph points for 4x - 2y = 4: (0,-2), (2,2), (1,0).
Intersection point is (2, 2). So, x = 2, y = 2.

(iv) 2x - 2y - 2 = 0, 4x - 4y - 5 = 0
Ratios: 2/4 = 1/2, -2/-4 = 1/2, -2/-5 = 2/5 ⇒ 1/2 = 1/2 ≠ 2/5. Parallel lines. System is Inconsistent.

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 width = w metres and length = l = w + 4 metres.
Half perimeter = length + width = 36
⇒ (w + 4) + w = 36
⇒ 2w + 4 = 36 ⇒ 2w = 32 ⇒ w = 16 m.
Length l = 16 + 4 = 20 m.
Answer: Length = 20 m, Width = 16 m.

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: 3x + 2y - 7 = 0
(ii) Parallel lines: 2x + 3y - 12 = 0
(iii) Coincident lines: 4x + 6y - 16 = 0

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:
For x - y + 1 = 0 ⇒ y = x + 1: Points (0, 1), (2, 3), (-1, 0).
For 3x + 2y - 12 = 0 ⇒ y = (12 - 3x)/2: Points (0, 6), (2, 3), (4, 0).
The lines intersect at (2, 3).
The lines cut the x-axis at (-1, 0) and (4, 0).
Vertices of the triangle: (2, 3), (-1, 0), (4, 0).

EXERCISE 3.2

Question 1: Solve the following pair of linear equations by the substitution method:

(i) x + y = 14 ; x - y = 4
From second equation: x = y + 4.
Substitute in first: (y + 4) + y = 14 ⇒ 2y = 10 ⇒ y = 5.
x = 5 + 4 = 9.
Solution: x = 9, y = 5.

(ii) s - t = 3 ; s/3 + t/2 = 6
From first equation: s = t + 3.
Substitute in second equation: (t + 3)/3 + t/2 = 6 ⇒ 2(t + 3) + 3t = 36 ⇒ 5t + 6 = 36 ⇒ 5t = 30 ⇒ t = 6.
s = 6 + 3 = 9.
Solution: s = 9, t = 6.

(iii) 3x - y = 3 ; 9x - 3y = 9
From first equation: y = 3x - 3.
Substitute in second: 9x - 3(3x - 3) = 9 ⇒ 9 = 9, which is true for all values of x.
Solution: Infinitely many solutions where y = 3x - 3.

(iv) 0.2x + 0.3y = 1.3 ; 0.4x + 0.5y = 2.3
Multiply by 10: 2x + 3y = 13 and 4x + 5y = 23.
From first: x = (13 - 3y)/2.
Substitute in second: 4[(13 - 3y)/2] + 5y = 23 ⇒ 26 - 6y + 5y = 23 ⇒ -y = -3 ⇒ y = 3.
x = (13 - 9)/2 = 2.
Solution: x = 2, y = 3.

(v) √2 x + √3 y = 0 ; √3 x - √8 y = 0
From first: x = -(√3 / √2)y.
Substitute in second: √3[-(√3 / √2)y] - √8 y = 0 ⇒ (-3/√2 - 2√2)y = 0 ⇒ y = 0.
Then x = 0.
Solution: x = 0, y = 0.

(vi) (3/2)x - (5/3)y = -2 ; x/3 + y/2 = 13/6
Simplifying: 9x - 10y = -12 and 2x + 3y = 13.
From second: x = (13 - 3y)/2.
Substitute in first: 9[(13 - 3y)/2] - 10y = -12 ⇒ 117 - 27y - 20y = -24 ⇒ -47y = -141 ⇒ y = 3.
x = (13 - 9)/2 = 2.
Solution: x = 2, y = 3.

Question 2: Solve 2x + 3y = 11 and 2x - 4y = -24 and hence find the value of 'm' for which y = mx + 3.

Solution:
2x + 3y = 11 ⇒ 2x = 11 - 3y.
Substitute 2x in second equation: (11 - 3y) - 4y = -24 ⇒ -7y = -35 ⇒ y = 5.
2x = 11 - 15 = -4 ⇒ x = -2.
Now, y = mx + 3 ⇒ 5 = m(-2) + 3 ⇒ 2 = -2m ⇒ m = -1.

Question 3: Form the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them.
Let numbers be x and y (x > y).
x - y = 26 and x = 3y.
Substitute x: 3y - y = 26 ⇒ 2y = 26 ⇒ y = 13.
x = 3(13) = 39.
Numbers are 39 and 13.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Let angles be x and y.
x + y = 180° and x - y = 18° ⇒ x = y + 18°.
(y + 18°) + y = 180° ⇒ 2y = 162° ⇒ y = 81°.
x = 81° + 18° = 99°.
Angles are 99° and 81°.

(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.
Let bat cost = ₹ x, ball cost = ₹ y.
7x + 6y = 3800 and 3x + 5y = 1750 ⇒ x = (1750 - 5y)/3.
7[(1750 - 5y)/3] + 6y = 3800 ⇒ 12250 - 35y + 18y = 11400 ⇒ 17y = 850 ⇒ y = 50.
x = (1750 - 250)/3 = 500.
Cost of 1 bat = ₹ 500, Cost of 1 ball = ₹ 50.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
Let fixed charge = ₹ x, charge/km = ₹ y.
x + 10y = 105 ⇒ x = 105 - 10y.
x + 15y = 155 ⇒ (105 - 10y) + 15y = 155 ⇒ 5y = 50 ⇒ y = 10.
x = 105 - 100 = 5.
Fixed charge = ₹ 5, Charge/km = ₹ 10.
For 25 km: x + 25y = 5 + 25(10) = ₹ 255.

(v) A fraction becomes 9/11, if 2 is added to both numerator and denominator. If 3 is added to both, it becomes 5/6. Find the fraction.
Let fraction = x/y.
(x + 2)/(y + 2) = 9/11 ⇒ 11x - 9y = -4.
(x + 3)/(y + 3) = 5/6 ⇒ 6x - 5y = -3 ⇒ x = (5y - 3)/6.
11[(5y - 3)/6] - 9y = -4 ⇒ 55y - 33 - 54y = -24 ⇒ y = 9.
x = (45 - 3)/6 = 7.
Fraction is 7/9.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son. What are their present ages?
Let Jacob's age = x, Son's age = y.
x + 5 = 3(y + 5) ⇒ x - 3y = 10.
x - 5 = 7(y - 5) ⇒ x - 7y = -30.
From first equation: x = 3y + 10.
Substitute: (3y + 10) - 7y = -30 ⇒ -4y = -40 ⇒ y = 10.
x = 30 + 10 = 40.
Jacob's age = 40 years, Son's age = 10 years.

EXERCISE 3.3

Question 1: Solve the following pair of linear equations by the elimination method and the substitution method:

(i) x + y = 5 and 2x - 3y = 4
By Elimination:
Multiply first equation by 3: 3x + 3y = 15.
Add to second equation: (3x + 3y) + (2x - 3y) = 15 + 4 ⇒ 5x = 19 ⇒ x = 19/5.
Substitute x in first: 19/5 + y = 5 ⇒ y = 5 - 19/5 = 6/5.
Solution: x = 19/5, y = 6/5.

(ii) 3x + 4y = 10 and 2x - 2y = 2
By Elimination:
Multiply second equation by 2: 4x - 4y = 4.
Add to first equation: (3x + 4y) + (4x - 4y) = 10 + 4 ⇒ 7x = 14 ⇒ x = 2.
Substitute in second: 2(2) - 2y = 2 ⇒ 4 - 2y = 2 ⇒ y = 1.
Solution: x = 2, y = 1.

(iii) 3x - 5y - 4 = 0 and 9x = 2y + 7
Rearranging: 3x - 5y = 4 and 9x - 2y = 7.
Multiply first equation by 3: 9x - 15y = 12.
Subtract from second: (9x - 2y) - (9x - 15y) = 7 - 12 ⇒ 13y = -5 ⇒ y = -5/13.
Substitute: 3x - 5(-5/13) = 4 ⇒ 3x = 4 - 25/13 = 27/13 ⇒ x = 9/13.
Solution: x = 9/13, y = -5/13.

(iv) x/2 + 2y/3 = -1 and x - y/3 = 3
Simplifying equations: 3x + 4y = -6 and 3x - y = 9.
Subtract second from first: (3x + 4y) - (3x - y) = -6 - 9 ⇒ 5y = -15 ⇒ y = -3.
Substitute in second: 3x - (-3) = 9 ⇒ 3x = 6 ⇒ x = 2.
Solution: x = 2, y = -3.

Question 2: Form the pair of linear equations in the following problems, and find their solutions by the elimination method:

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?
Let fraction = x/y.
(x + 1)/(y - 1) = 1 ⇒ x - y = -2  --- (1)
x/(y + 1) = 1/2 ⇒ 2x - y = 1  --- (2)
Subtract (1) from (2): (2x - y) - (x - y) = 1 - (-2) ⇒ x = 3.
Substitute x=3 in (1): 3 - y = -2 ⇒ y = 5.
Fraction is 3/5.

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Let Nuri's age = x, Sonu's age = y.
x - 5 = 3(y - 5) ⇒ x - 3y = -10  --- (1)
x + 10 = 2(y + 10) ⇒ x - 2y = 10  --- (2)
Subtract (1) from (2): y = 20.
Substitute in (2): x - 40 = 10 ⇒ x = 50.
Nuri is 50 years old and Sonu is 20 years old.

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Let ten's digit = x, unit's digit = y. Number = 10x + y.
x + y = 9  --- (1)
9(10x + y) = 2(10y + x) ⇒ 90x + 9y = 20y + 2x ⇒ 88x - 11y = 0 ⇒ 8x - y = 0  --- (2)
Add (1) and (2): 9x = 9 ⇒ x = 1.
Substitute in (1): 1 + y = 9 ⇒ y = 8.
Number is 18.

(iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.
Let number of ₹ 50 notes = x, ₹ 100 notes = y.
x + y = 25  --- (1)
50x + 100y = 2000 ⇒ x + 2y = 40  --- (2)
Subtract (1) from (2): y = 15.
Substitute in (1): x + 15 = 25 ⇒ x = 10.
Meena received 10 notes of ₹ 50 and 15 notes of ₹ 100.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Let fixed charge (3 days) = ₹ x, charge/extra day = ₹ y.
Saritha (7 days = 3 + 4 extra): x + 4y = 27  --- (1)
Susy (5 days = 3 + 2 extra): x + 2y = 21  --- (2)
Subtract (2) from (1): 2y = 6 ⇒ y = 3.
Substitute in (2): x + 6 = 21 ⇒ x = 15.
Fixed charge = ₹ 15, Charge per extra day = ₹ 3.