PAIR OF LINEAR EQUATIONS IN TWO VARIABLES - Q&A

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 problem:
1) Total students are 10:
x + y = 10    ...(i)
2) Number of girls is 4 more than boys:
x = y + 4    OR    x − y = 4    ...(ii)

To solve graphically, we find coordinates for each line:
For x + y = 10:
If x = 5, y = 5 (Point 5, 5)
If x = 7, y = 3 (Point 7, 3)

For x − y = 4:
If x = 4, y = 0 (Point 4, 0)
If x = 5, y = 1 (Point 5, 1)

Plotting these lines on a graph, they intersect at the point (7, 3).
x = 7 (Number of girls)
y = 3 (Number of boys)

Answer: Number of girls = 7, Number of boys = 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 one pencil be ₹ x and the cost of one pen be ₹ y.
Algebraic representation:
1) 5x + 7y = 50    ...(i)
2) 7x + 5y = 46    ...(ii)

Coordinates for graphing:
For 5x + 7y = 50:
If x = 3, 15 + 7y = 50 → 7y = 35 → y = 5. Point (3, 5)
If x = 10, 50 + 7y = 50 → y = 0. Point (10, 0)

For 7x + 5y = 46:
If x = 3, 21 + 5y = 46 → 5y = 25 → y = 5. Point (3, 5)
If x = 8, 56 + 5y = 46 → 5y = -10 → y = -2. Point (8, -2)

Observing the common point, the lines intersect at (3, 5).
Answer: Cost of one pencil is ₹ 3 and cost of one pen is ₹ 5.

2. On comparing the ratios a₁/a₂, b₁/b₂ 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

Solution:
Here, 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.
Answer: The lines intersect at a point.

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

Solution:
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.
Answer: The lines are coincident.

(iii) 6x − 3y + 10 = 0
2x − y + 9 = 0

Solution:
a₁/a₂ = 6/2 = 3/1
b₁/b₂ = -3/-1 = 3/1
c₁/c₂ = 10/9

Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂.
Answer: The lines are parallel.

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

Solution:
Rearranging to standard form: 3x + 2y − 5 = 0 and 2x − 3y − 7 = 0.
a₁/a₂ = 3/2
b₁/b₂ = 2/-3
Since a₁/a₂ ≠ b₁/b₂, they intersect at a unique point.
Answer: Consistent.

(ii) 2x − 3y = 8 ; 4x − 6y = 9

Solution:
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 lines are parallel.
Answer: Inconsistent.

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

Solution:
a₁/a₂ = (3/2) / 9 = 3/18 = 1/6
b₁/b₂ = (5/3) / -10 = 5/-30 = -1/6
Since 1/6 ≠ -1/6, a₁/a₂ ≠ b₁/b₂.
Answer: Consistent.

(iv) 5x − 3y = 11 ; −10x + 6y = −22

Solution:
a₁/a₂ = 5/-10 = -1/2
b₁/b₂ = -3/6 = -1/2
c₁/c₂ = 11/-22 = -1/2
Since all ratios are equal, lines are coincident.
Answer: Consistent (Dependent).

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

Solution:
a₁/a₂ = (4/3) / 2 = 4/6 = 2/3
b₁/b₂ = 2/3
c₁/c₂ = -8/-12 = 2/3
Since all ratios are equal.
Answer: Consistent (Dependent).

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

Solution:
a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = -5/-10 = 1/2.
Since ratios are equal, lines are coincident.
Answer: Consistent. Infinite solutions. (Graph will show two overlapping lines).

(ii) x − y = 8 ; 3x − 3y = 16

Solution:
a₁/a₂ = 1/3, b₁/b₂ = -1/-3 = 1/3, c₁/c₂ = -8/-16 = 1/2.
a₁/a₂ = b₁/b₂ ≠ c₁/c₂. Lines are parallel.
Answer: Inconsistent. No solution.

(iii) 2x + y − 6 = 0 ; 4x − 2y − 4 = 0

Solution:
a₁/a₂ = 2/4 = 1/2, b₁/b₂ = 1/-2.
Since ratios are unequal, it is Consistent.
Graphical Solution:
Line 1 (2x + y = 6): Points (0, 6), (3, 0).
Line 2 (4x − 2y = 4 → 2x − y = 2): Points (0, -2), (1, 0).
Intersection point:
Adding equations: (2x + y) + (2x − y) = 6 + 2 → 4x = 8 → x = 2.
Put x=2 in eq 1: 2(2) + y = 6 → 4 + y = 6 → y = 2.
Answer: x = 2, y = 2.

(iv) 2x − 2y − 2 = 0 ; 4x − 4y − 5 = 0

Solution:
a₁/a₂ = 2/4 = 1/2, b₁/b₂ = -2/-4 = 1/2, c₁/c₂ = -2/-5 = 2/5.
Ratios of coefficients are equal but not constant term.
Answer: Inconsistent (Parallel lines).

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 Length = x and Width = y.
1) Length is 4 more than width: x = y + 4 → x − y = 4.
2) Half perimeter is 36: 1/2 * 2(x + y) = 36 → x + y = 36.

Solving the pair:
x = y + 4
Substitute in second eq: (y + 4) + y = 36
2y = 32 → y = 16 m.
x = 16 + 4 = 20 m.
Answer: Length = 20 m, Width = 16 m.

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

Answer: Any equation where a₁/a₂ ≠ b₁/b₂.
Example: 3x + 2y − 7 = 0 (Since 2/3 ≠ 3/2).

(ii) parallel lines

Answer: Any equation where a₁/a₂ = b₁/b₂ ≠ c₁/c₂.
Example: 4x + 6y − 12 = 0 (Since 2/4 = 3/6 ≠ -8/-12).

(iii) coincident lines

Answer: Any equation where all ratios are equal.
Example: 6x + 9y − 24 = 0 (Multiplied original by 3).

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:
Line 1: x − y = -1
Points: (0, 1), (-1, 0), (2, 3)

Line 2: 3x + 2y = 12
Points: (0, 6), (4, 0), (2, 3)

Intersection of lines: Point (2, 3).
Intersection with x-axis (y=0):
Line 1: x - 0 = -1 → x = -1. Point (-1, 0).
Line 2: 3x = 12 → x = 4. Point (4, 0).

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

1. Solve the following pair of linear equations by the substitution method.

(i) x + y = 14 ; x − y = 4

Solution:
From 2nd eq: x = y + 4.
Substitute in 1st eq: (y + 4) + y = 14
2y = 10 → y = 5.
x = 5 + 4 = 9.
Answer: x = 9, y = 5.

(ii) s − t = 3 ; s/3 + t/2 = 6

Solution:
From 1st eq: s = t + 3.
Substitute in 2nd eq:
(t + 3)/3 + t/2 = 6
Multiply by 6 to clear denominator: 2(t + 3) + 3t = 36
2t + 6 + 3t = 36
5t = 30 → t = 6.
s = 6 + 3 = 9.
Answer: s = 9, t = 6.

(iii) 3x − y = 3 ; 9x − 3y = 9

Solution:
From 1st eq: y = 3x − 3.
Substitute in 2nd eq: 9x − 3(3x − 3) = 9
9x − 9x + 9 = 9
9 = 9 (True statement).
Answer: Infinitely many solutions (The equations are identical).

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

Solution:
Multiply both equations by 10 to remove decimals:
2x + 3y = 13   ...(i)
4x + 5y = 23   ...(ii)
From (i): x = (13 − 3y) / 2.
Substitute in (ii): 4[(13 − 3y)/2] + 5y = 23
2(13 − 3y) + 5y = 23
26 − 6y + 5y = 23
-y = -3 → y = 3.
x = (13 − 3(3)) / 2 = (13 − 9)/2 = 4/2 = 2.
Answer: x = 2, y = 3.

(v) √2x + √3y = 0 ; √3x − √8y = 0

Solution:
From 1st eq: x = -(√3 / √2)y.
Substitute in 2nd eq: √3 [ -(√3 / √2)y ] − √8y = 0
-(3/√2)y − 2√2y = 0
y [ -3/√2 - 2√2 ] = 0
Since the constant part is not zero, y must be 0.
If y = 0, x = 0.
Answer: x = 0, y = 0.

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

Solution:
Eq 1: Multiply by 6 → 9x − 10y = -12.
Eq 2: Multiply by 6 → 2x + 3y = 13.
From Eq 2: 2x = 13 − 3y → x = (13 − 3y)/2.
Substitute in Eq 1: 9[(13 − 3y)/2] − 10y = -12
Multiply by 2: 9(13 − 3y) − 20y = -24
117 − 27y − 20y = -24
-47y = -141 → y = 3.
x = (13 − 3(3))/2 = 4/2 = 2.
Answer: x = 2, y = 3.

2. Solve 2x + 3y = 11 and 2x − 4y = −24 and hence find the value of ‘m’ for which y = mx + 3.

Solution:
Equations:
1) 2x + 3y = 11
2) 2x − 4y = -24
Subtract (2) from (1): (2x - 2x) + (3y - (-4y)) = 11 - (-24)
7y = 35 → y = 5.
Substitute y=5 in (1): 2x + 3(5) = 11 → 2x = 11 - 15 = -4 → x = -2.
Solution is x = -2, y = 5.

Find m in y = mx + 3:
5 = m(-2) + 3
5 - 3 = -2m
2 = -2m → m = -1.
Answer: x = -2, y = 5, m = -1.

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.

Solution:
Let numbers be x and y (where x > y).
x − y = 26
x = 3y
Substitute x=3y into first eq: 3y − y = 26 → 2y = 26 → y = 13.
x = 3(13) = 39.
Answer: The numbers are 39 and 13.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Solution:
Let angles be x and y.
Supplementary means sum is 180: x + y = 180.
Difference is 18: x = y + 18.
Substitute: (y + 18) + y = 180 → 2y = 162 → y = 81.
x = 81 + 18 = 99.
Answer: The 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.

Solution:
Let bat cost = x, ball cost = y.
1) 7x + 6y = 3800
2) 3x + 5y = 1750 → 3x = 1750 − 5y → x = (1750 − 5y)/3.
Substitute in 1: 7[(1750 − 5y)/3] + 6y = 3800
Multiply by 3: 7(1750 − 5y) + 18y = 11400
12250 − 35y + 18y = 11400
-17y = 11400 - 12250 = -850
y = 850 / 17 = 50.
x = (1750 − 5(50)) / 3 = 1500 / 3 = 500.
Answer: Bat = ₹ 500, 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?

Solution:
Let fixed charge = x and charge per km = y.
x + 10y = 105
x + 15y = 155
From eq 1: x = 105 − 10y.
Substitute in eq 2: (105 − 10y) + 15y = 155
105 + 5y = 155 → 5y = 50 → y = 10.
x = 105 − 10(10) = 5.
Fixed charge = ₹ 5, Per km = ₹ 10.
For 25km: Cost = x + 25y = 5 + 25(10) = 255.
Answer: Fixed charge = ₹ 5, Charge per km = ₹ 10. Cost for 25 km = ₹ 255.

(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.

Solution:
Let fraction be x/y.
1) (x + 2)/(y + 2) = 9/11 → 11x + 22 = 9y + 18 → 11x − 9y = -4.
2) (x + 3)/(y + 3) = 5/6 → 6x + 18 = 5y + 15 → 6x − 5y = -3.
From 2: 6x = 5y - 3 → x = (5y-3)/6.
Substitute in 1: 11[(5y-3)/6] - 9y = -4
Multiply by 6: 11(5y-3) - 54y = -24
55y - 33 - 54y = -24
y = -24 + 33 = 9.
x = (5(9)-3)/6 = 42/6 = 7.
Answer: The 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?

Solution:
Let Jacob's age = x, Son's age = y.
1) After 5 years: (x + 5) = 3(y + 5) → x - 3y = 10.
2) 5 years ago: (x − 5) = 7(y − 5) → x - 7y = -30.
From 1: x = 10 + 3y.
Substitute in 2: (10 + 3y) − 7y = -30
-4y = -40 → y = 10.
x = 10 + 3(10) = 40.
Answer: Jacob is 40 years old, Son is 10 years old.

1. Solve the following pair of linear equations by the elimination method:

(i) x + y = 5 and 2x − 3y = 4

Solution:
1) x + y = 5
2) 2x − 3y = 4
Multiply eq (1) by 3: 3x + 3y = 15.
Add to eq (2): (2x − 3y) + (3x + 3y) = 4 + 15
5x = 19 → x = 19/5.
Substitute x in eq (1): 19/5 + y = 5 → y = 5 − 19/5 = 6/5.
Answer: x = 19/5, y = 6/5.

(ii) 3x + 4y = 10 and 2x − 2y = 2

Solution:
Divide eq 2 by 2: x − y = 1 → x = y + 1.
Substitute into eq 1: 3(y + 1) + 4y = 10
3y + 3 + 4y = 10
7y = 7 → y = 1.
x = 1 + 1 = 2.
Answer: x = 2, y = 1.

(iii) 3x − 5y − 4 = 0 and 9x = 2y + 7

Solution:
Rearrange:
1) 3x − 5y = 4
2) 9x − 2y = 7
Multiply eq (1) by 3: 9x − 15y = 12.
Subtract eq (2) from this: (9x − 15y) − (9x − 2y) = 12 − 7
-13y = 5 → y = -5/13.
From eq (1): 3x − 5(-5/13) = 4
3x + 25/13 = 4 → 3x = 4 − 25/13 = (52-25)/13 = 27/13.
x = 9/13.
Answer: x = 9/13, y = -5/13.

(iv) x/2 + 2y/3 = −1 and x − y/3 = 3

Solution:
Multiply eq 1 by 6: 3x + 4y = -6.
Multiply eq 2 by 3: 3x − y = 9.
Subtract eq 2 from 1: (3x + 4y) − (3x − y) = -6 − 9
5y = -15 → y = -3.
From modified eq 2: 3x − (-3) = 9 → 3x + 3 = 9 → 3x = 6 → x = 2.
Answer: x = 2, y = -3.

2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) 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?

Solution:
Let fraction = x/y.
1) (x+1)/(y-1) = 1 → x+1 = y-1 → x - y = -2.
2) x/(y+1) = 1/2 → 2x = y+1 → 2x - y = 1.
Subtract eq 1 from 2: (2x - y) - (x - y) = 1 - (-2)
x = 3.
Put x=3 in eq 1: 3 - y = -2 → y = 5.
Answer: The 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?

Solution:
Nuri = x, Sonu = y.
1) (x - 5) = 3(y - 5) → x - 3y = -10.
2) (x + 10) = 2(y + 10) → x - 2y = 10.
Subtract eq 1 from 2: (x - 2y) - (x - 3y) = 10 - (-10)
y = 20.
Put y=20 in eq 2: x - 40 = 10 → x = 50.
Answer: Nuri is 50, Sonu is 20.

(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.

Solution:
Let unit digit = y, tens digit = x. Number = 10x + y.
1) x + y = 9.
2) 9(10x + y) = 2(10y + x) → 90x + 9y = 20y + 2x.
88x - 11y = 0 → divide by 11 → 8x - y = 0 → y = 8x.
Substitute into eq 1: x + 8x = 9 → 9x = 9 → x = 1.
y = 8(1) = 8.
Answer: The 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.

Solution:
Let ₹ 50 notes = x, ₹ 100 notes = y.
1) Total notes: x + y = 25.
2) Total value: 50x + 100y = 2000 → divide by 50 → x + 2y = 40.
Subtract eq 1 from 2: (x + 2y) - (x + y) = 40 - 25.
y = 15.
x + 15 = 25 → x = 10.
Answer: 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.

Solution:
Let fixed charge (for 3 days) = x, extra daily charge = y.
Saritha (7 days = 3 fixed + 4 extra): x + 4y = 27.
Susy (5 days = 3 fixed + 2 extra): x + 2y = 21.
Subtract Susy from Saritha: (x + 4y) - (x + 2y) = 27 - 21.
2y = 6 → y = 3.
Substitute y=3 in Susy's eq: x + 2(3) = 21 → x = 21 - 6 = 15.
Answer: Fixed charge = ₹ 15, Extra charge per day = ₹ 3.