Key Concepts & Formulae at a Glance
- Consistent pair:Has at least one solution — lines intersect (unique solution) or coincide (infinitely many).
- Inconsistent pair:Has no solution — lines are parallel.
- The ratio test:For a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0: a₁a₂≠b₁b₂ → intersecting; a₁a₂=b₁b₂≠c₁c₂ → parallel; a₁a₂=b₁b₂=c₁c₂ → coincident.
- Substitution method:Express one variable in terms of the other from one equation, substitute into the second.
- Elimination method:Multiply equations so one variable's coefficients match, then addsubtract to cancel it.
- Reading the outcome:A false statement with no variable (e.g. 0=9) → no solution. A true statement (e.g. 18=18) → infinitely many solutions.
Exercise 3.1
7 Questions
1Form 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 ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and that of one pen.
(i) Quiz participants
Let the number of boys = x and the number of girls = y.
x + y = 10 — total participants ... (1)
y = x + 4 — girls are 4 more than boys ... (2)
x + y = 10 — total participants ... (1)
y = x + 4 — girls are 4 more than boys ... (2)
Lines intersect at (3, 7)
The two lines intersect at the point (3, 7), which is the required solution.
Boys = 3, Girls = 7
(ii) Pencils and pens
Let the cost of one pencil = ₹x and one pen = ₹y.
5x + 7y = 50 ... (1)
7x + 5y = 46 ... (2)
5x + 7y = 50 ... (1)
7x + 5y = 46 ... (2)
Lines intersect at (3, 5)
The two lines intersect at the point (3, 5).
Cost of one pencil = ₹3, Cost of one pen = ₹5
2On comparing the ratios a₁a₂, b₁b₂ and c₁c₂, 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
| Pair | a₁a₂ | b₁b₂ | c₁c₂ | Comparison | Conclusion |
|---|---|---|---|---|---|
| (i) | 57 | −23 | −89 | a₁a₂ ≠ b₁b₂ | Intersecting lines |
| (ii) | 12 | 12 | 12 | all equal | Coincident lines |
| (iii) | 3 | 3 | 109 | a₁a₂=b₁b₂ ≠ c₁c₂ | Parallel lines |
(i) Intersecting (ii) Coincident (iii) Parallel
3On comparing the ratios a₁a₂, b₁b₂ and c₁c₂, 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) (32)x + (53)y = 7; 9x − 10y = 14
(iv) 5x − 3y = 11; −10x + 6y = −22
(v) (43)x + 2y = 8; 2x + 3y = 12
| Pair | a₁a₂ | b₁b₂ | c₁c₂ | Conclusion |
|---|---|---|---|---|
| (i) | 32 | −23 | −57 | a₁a₂≠b₁b₂ — Consistent (unique solution) |
| (ii) | 12 | 12 | 89 | aa=bb≠cc — Inconsistent (no solution) |
| (iii) | 16 | −16 | 12 | a₁a₂≠b₁b₂ — Consistent (unique solution) |
| (iv) | −12 | −12 | −12 | all equal — Consistent (infinitely many solutions) |
| (v) | 23 | 23 | 23 | all equal — Consistent (infinitely many solutions) |
(i) Consistent (ii) Inconsistent (iii) Consistent (iv) Consistent (v) Consistent
4Which of the following pairs of linear equations are consistentinconsistent? 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
(i) x + y = 5, 2x + 2y = 10
Ratios: a₁a₂ = 12, b₁b₂ = 12, c₁c₂ = 12 — all equal (second equation is exactly twice the first)
Consistent — the equations represent the same line, so there are infinitely many solutions.
(ii) x − y = 8, 3x − 3y = 16
Ratios: a₁a₂ = 13, b₁b₂ = 13, c₁c₂ = 12 — a₁a₂=b₁b₂ but ≠ c₁c₂
Inconsistent — the lines are parallel, no solution.
(iii) 2x + y − 6 = 0, 4x − 2y − 4 = 0
Ratios: a₁a₂ = 12, b₁b₂ = −12 — not equal, so the lines intersect. Consistent, unique solution.
From the first equation: y = 6 − 2x
Substituting into the second: 4x − 2(6 − 2x) − 4 = 0
4x − 12 + 4x − 4 = 0
8x = 16, so x = 2, and y = 6 − 4 = 2
From the first equation: y = 6 − 2x
Substituting into the second: 4x − 2(6 − 2x) − 4 = 0
4x − 12 + 4x − 4 = 0
8x = 16, so x = 2, and y = 6 − 4 = 2
Lines intersect at (2, 2)
Consistent — unique solution x = 2, y = 2.
(iv) 2x − 2y − 2 = 0, 4x − 4y − 5 = 0
Ratios: a₁a₂ = 12, b₁b₂ = 12, c₁c₂ = 25 — a₁a₂=b₁b₂ but ≠ c₁c₂
Inconsistent — the lines are parallel, no solution.
5Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Let the width = w metres, so the length = (w + 4) metres.
Half the perimeter = length + width (perimeter = 2(l+w), so half of it is l+w)
(w + 4) + w = 36
2w + 4 = 36
2w = 32
w = 16
Length = w + 4 = 20
Half the perimeter = length + width (perimeter = 2(l+w), so half of it is l+w)
(w + 4) + w = 36
2w + 4 = 36
2w = 32
w = 16
Length = w + 4 = 20
Width = 16 m, Length = 20 m
6Given 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
For 2x + 3y − 8 = 0: a₁=2, b₁=3, c₁=−8. Any second equation a₂x+b₂y+c₂=0 needs to satisfy the ratio test for the required case.
(i) Intersecting lines
Need a₁a₂ ≠ b₁b₂. Choose x − y − 1 = 0 (a₂=1, b₂=−1)
a₁a₂ = 2, b₁b₂ = −3 — not equal, so the lines intersect.
a₁a₂ = 2, b₁b₂ = −3 — not equal, so the lines intersect.
One such equation: x − y − 1 = 0
(ii) Parallel lines
Need a₁a₂ = b₁b₂ ≠ c₁c₂. Choose 4x + 6y − 4 = 0 (a₂=4, b₂=6, c₂=−4)
a₁a₂ = 12, b₁b₂ = 12, c₁c₂ = 2 — aa=bb but ≠ cc, so the lines are parallel.
a₁a₂ = 12, b₁b₂ = 12, c₁c₂ = 2 — aa=bb but ≠ cc, so the lines are parallel.
One such equation: 4x + 6y − 4 = 0
(iii) Coincident lines
Need a₁a₂ = b₁b₂ = c₁c₂. Choose 4x + 6y − 16 = 0 (double every term of the given equation)
a₁a₂ = b₁b₂ = c₁c₂ = 12 — all equal, so the lines coincide.
a₁a₂ = b₁b₂ = c₁c₂ = 12 — all equal, so the lines coincide.
One such equation: 4x + 6y − 16 = 0
7Draw 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.
Line 1: x − y + 1 = 0, i.e. y = x + 1. At y=0, x=−1, so it meets the x-axis at (−1, 0).
Line 2: 3x + 2y − 12 = 0. At y=0, x=4, so it meets the x-axis at (4, 0).
Solving the two equations together: substituting y = x + 1 into 3x + 2y − 12 = 0:
3x + 2(x + 1) − 12 = 0
5x − 10 = 0, so x = 2, y = 3 — the lines intersect at (2, 3).
Line 2: 3x + 2y − 12 = 0. At y=0, x=4, so it meets the x-axis at (4, 0).
Solving the two equations together: substituting y = x + 1 into 3x + 2y − 12 = 0:
3x + 2(x + 1) − 12 = 0
5x − 10 = 0, so x = 2, y = 3 — the lines intersect at (2, 3).
Triangle shaded between the two lines and the x-axis
Vertices of the triangle: (−1, 0), (4, 0), and (2, 3)
Exercise 3.2
3 Questions
1Solve the following pair of linear equations by the substitution method:
(i) x + y = 14; x − y = 4
(ii) s − t = 3; s3 + t2 = 6
(iii) 3x − y = 3; 9x − 3y = 9
(iv) 0.2x + 0.3y = 1.3; 0.4x + 0.5y = 2.3
(v) √2x + √3y = 0; √3x − √8y = 0
(vi) (3x2) − (5y3) = −2; x3 + y2 = 136
(i) x + y = 14, x − y = 4
From x + y = 14: x = 14 − y
Substituting into x − y = 4: (14 − y) − y = 4
14 − 2y = 4, so y = 5
x = 14 − 5 = 9
Substituting into x − y = 4: (14 − y) − y = 4
14 − 2y = 4, so y = 5
x = 14 − 5 = 9
x = 9, y = 5
(ii) s − t = 3, s3 + t2 = 6
From s − t = 3: s = t + 3
Substituting into s3 + t2 = 6: t+33 + t2 = 6
Multiplying through by 6: 2(t+3) + 3t = 36
2t + 6 + 3t = 36
5t = 30, so t = 6
s = 6 + 3 = 9
Substituting into s3 + t2 = 6: t+33 + t2 = 6
Multiplying through by 6: 2(t+3) + 3t = 36
2t + 6 + 3t = 36
5t = 30, so t = 6
s = 6 + 3 = 9
s = 9, t = 6
(iii) 3x − y = 3, 9x − 3y = 9
From 3x − y = 3: y = 3x − 3
Substituting into 9x − 3y = 9: 9x − 3(3x − 3) = 9
9x − 9x + 9 = 9
9 = 9, a true statement with no variable (the second equation is 3× the first)
Substituting into 9x − 3y = 9: 9x − 3(3x − 3) = 9
9x − 9x + 9 = 9
9 = 9, a true statement with no variable (the second equation is 3× the first)
Infinitely many solutions — the equations represent the same line, of the form y = 3x − 3.
(iv) 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3
From 0.2x + 0.3y = 1.3: x = 1.3 − 0.3y0.2
Substituting into 0.4x + 0.5y = 2.3 and simplifying (or directly eliminating x by doubling the first equation and subtracting):
2 × (0.2x + 0.3y) = 2 × 1.3 → 0.4x + 0.6y = 2.6
Subtracting 0.4x + 0.5y = 2.3: 0.1y = 0.3, so y = 3
0.2x + 0.3(3) = 1.3 → 0.2x = 0.4, so x = 2
Substituting into 0.4x + 0.5y = 2.3 and simplifying (or directly eliminating x by doubling the first equation and subtracting):
2 × (0.2x + 0.3y) = 2 × 1.3 → 0.4x + 0.6y = 2.6
Subtracting 0.4x + 0.5y = 2.3: 0.1y = 0.3, so y = 3
0.2x + 0.3(3) = 1.3 → 0.2x = 0.4, so x = 2
x = 2, y = 3
(v) √2x + √3y = 0, √3x − √8y = 0
From √2x + √3y = 0: x = −(√3√2)y
Substituting into √3x − √8y = 0: √3 × (−(√3√2)y) − √8y = 0
−(3√2)y − √8y = 0
y × (−3√2 − √8) = 0, and since −3√2 − √8 ≠ 0, this forces y = 0
Then x = 0
Substituting into √3x − √8y = 0: √3 × (−(√3√2)y) − √8y = 0
−(3√2)y − √8y = 0
y × (−3√2 − √8) = 0, and since −3√2 − √8 ≠ 0, this forces y = 0
Then x = 0
x = 0, y = 0
(vi) (3x2) − (5y3) = −2, x3 + y2 = 136
From x3 + y2 = 136: multiplying by 6, 2x + 3y = 13, so x = 13 − 3y2
Substituting into (3x2) − (5y3) = −2: multiplying that equation by 6 first gives 9x − 10y = −12
9 × 13 − 3y2 − 10y = −12
117 − 27y2 − 10y = −12
117 − 27y − 20y = −24
−47y = −141, so y = 3
x = 13 − 92 = 2
Substituting into (3x2) − (5y3) = −2: multiplying that equation by 6 first gives 9x − 10y = −12
9 × 13 − 3y2 − 10y = −12
117 − 27y2 − 10y = −12
117 − 27y − 20y = −24
−47y = −141, so y = 3
x = 13 − 92 = 2
x = 2, y = 3
2Solve 2x + 3y = 11 and 2x − 4y = −24 and hence find the value of ‘m’ for which y = mx + 3.
From 2x + 3y = 11: 2x = 11 − 3y
Substituting into 2x − 4y = −24: (11 − 3y) − 4y = −24
11 − 7y = −24
−7y = −35, so y = 5
2x = 11 − 15 = −4, so x = −2
Substituting x = −2, y = 5 into y = mx + 3: 5 = m(−2) + 3
2 = −2m, so m = −1
Substituting into 2x − 4y = −24: (11 − 3y) − 4y = −24
11 − 7y = −24
−7y = −35, so y = 5
2x = 11 − 15 = −4, so x = −2
Substituting x = −2, y = 5 into y = mx + 3: 5 = m(−2) + 3
2 = −2m, so m = −1
x = −2, y = 5, and m = −1
3Form 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.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(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.
(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?
(v) A fraction becomes 911 if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes 56. Find the fraction.
(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?
(i) Two numbers
Let the larger number = x, smaller = y.
x − y = 26 ... (1)
x = 3y ... (2)
Substituting (2) into (1): 3y − y = 26
2y = 26, so y = 13, and x = 3(13) = 39
x − y = 26 ... (1)
x = 3y ... (2)
Substituting (2) into (1): 3y − y = 26
2y = 26, so y = 13, and x = 3(13) = 39
The numbers are 39 and 13.
(ii) Supplementary angles
Let the larger angle = x, smaller = y. Supplementary angles sum to 180°.
x + y = 180 ... (1)
x − y = 18 ... (2)
From (2): x = y + 18. Substituting into (1): (y+18) + y = 180
2y = 162, so y = 81, and x = 99
x + y = 180 ... (1)
x − y = 18 ... (2)
From (2): x = y + 18. Substituting into (1): (y+18) + y = 180
2y = 162, so y = 81, and x = 99
The angles are 99° and 81°.
(iii) Bats and balls
Let the cost of one bat = ₹x and one ball = ₹y.
7x + 6y = 3800 ... (1)
3x + 5y = 1750 ... (2)
From (2): x = 1750 − 5y3
Substituting into (1): 7 × 1750 − 5y3 + 6y = 3800
12250 − 35y3 + 6y = 3800
12250 − 35y + 18y = 11400
−17y = −850, so y = 50
x = 1750 − 2503 = 500
7x + 6y = 3800 ... (1)
3x + 5y = 1750 ... (2)
From (2): x = 1750 − 5y3
Substituting into (1): 7 × 1750 − 5y3 + 6y = 3800
12250 − 35y3 + 6y = 3800
12250 − 35y + 18y = 11400
−17y = −850, so y = 50
x = 1750 − 2503 = 500
Cost of one bat = ₹500, cost of one ball = ₹50.
(iv) Taxi charges
Let the fixed charge = ₹f and the per-km charge = ₹r.
f + 10r = 105 ... (1)
f + 15r = 155 ... (2)
From (1): f = 105 − 10r
Substituting into (2): (105 − 10r) + 15r = 155
5r = 50, so r = 10, and f = 105 − 100 = 5
Charge for 25 km = f + 25r = 5 + 250 = 255
f + 10r = 105 ... (1)
f + 15r = 155 ... (2)
From (1): f = 105 − 10r
Substituting into (2): (105 − 10r) + 15r = 155
5r = 50, so r = 10, and f = 105 − 100 = 5
Charge for 25 km = f + 25r = 5 + 250 = 255
Fixed charge = ₹5, charge per km = ₹10, and the fare for 25 km = ₹255.
(v) Fraction
Let the fraction be nd.
n+2d+2 = 911 → 11(n+2) = 9(d+2) → 11n − 9d = −4 ... (1)
n+3d+3 = 56 → 6(n+3) = 5(d+3) → 6n − 5d = −3 ... (2)
From (1): n = 9d − 411
Substituting into (2) and solving gives d = 9, n = 7
n+2d+2 = 911 → 11(n+2) = 9(d+2) → 11n − 9d = −4 ... (1)
n+3d+3 = 56 → 6(n+3) = 5(d+3) → 6n − 5d = −3 ... (2)
From (1): n = 9d − 411
Substituting into (2) and solving gives d = 9, n = 7
The fraction is 79.
(vi) Jacob and his son
Let Jacob’s present age = J, son’s present age = S.
J + 5 = 3(S + 5) → J − 3S = 10 ... (1)
J − 5 = 7(S − 5) → J − 7S = −30 ... (2)
From (1): J = 3S + 10
Substituting into (2): (3S+10) − 7S = −30
−4S = −40, so S = 10, and J = 3(10)+10 = 40
J + 5 = 3(S + 5) → J − 3S = 10 ... (1)
J − 5 = 7(S − 5) → J − 7S = −30 ... (2)
From (1): J = 3S + 10
Substituting into (2): (3S+10) − 7S = −30
−4S = −40, so S = 10, and J = 3(10)+10 = 40
Jacob’s present age = 40 years, his son’s present age = 10 years.
Exercise 3.3
2 Questions
1Solve the following pair of linear equations by the elimination method and the substitution method:
(i) x + y = 5 and 2x − 3y = 4
(ii) 3x + 4y = 10 and 2x − 2y = 2
(iii) 3x − 5y − 4 = 0 and 9x = 2y + 7
(iv) x2 + 2y3 = −1 and x − y3 = 3
(i) x + y = 5, 2x − 3y = 4
Elimination: Multiply the first equation by 3: 3x + 3y = 15
Adding to 2x − 3y = 4: 5x = 19, so x = 195
From x + y = 5: y = 5 − 195 = 65
Substitution (check): From x+y=5, x = 5−y. Substituting: 2(5−y) − 3y = 4 → 10 − 5y = 4 → y = 65, x = 195 — matches.
Adding to 2x − 3y = 4: 5x = 19, so x = 195
From x + y = 5: y = 5 − 195 = 65
Substitution (check): From x+y=5, x = 5−y. Substituting: 2(5−y) − 3y = 4 → 10 − 5y = 4 → y = 65, x = 195 — matches.
x = 195, y = 65
(ii) 3x + 4y = 10, 2x − 2y = 2
Elimination: Multiply the second equation by 2: 4x − 4y = 4
Adding to 3x + 4y = 10: 7x = 14, so x = 2
From 2x − 2y = 2: 4 − 2y = 2, so y = 1
Substitution (check): From 2x−2y=2, x=y+1. Substituting: 3(y+1)+4y=10 → 7y=7 → y=1, x=2 — matches.
Adding to 3x + 4y = 10: 7x = 14, so x = 2
From 2x − 2y = 2: 4 − 2y = 2, so y = 1
Substitution (check): From 2x−2y=2, x=y+1. Substituting: 3(y+1)+4y=10 → 7y=7 → y=1, x=2 — matches.
x = 2, y = 1
(iii) 3x − 5y − 4 = 0, 9x = 2y + 7
Rewriting: 3x − 5y = 4 ... (1); 9x − 2y = 7 ... (2)
Elimination: Multiply (1) by 3: 9x − 15y = 12
Subtracting (2): (9x−15y) − (9x−2y) = 12 − 7 → −13y = 5, so y = −513
From (2): 9x = 2(−513) + 7 = 8113, so x = 913
Substitution (check): From (1), x=4+5y3. Substituting into (2) gives the same y = −513, x = 913.
Elimination: Multiply (1) by 3: 9x − 15y = 12
Subtracting (2): (9x−15y) − (9x−2y) = 12 − 7 → −13y = 5, so y = −513
From (2): 9x = 2(−513) + 7 = 8113, so x = 913
Substitution (check): From (1), x=4+5y3. Substituting into (2) gives the same y = −513, x = 913.
x = 913, y = −513
(iv) x2 + 2y3 = −1, x − y3 = 3
Clearing fractions: multiplying the first by 6: 3x + 4y = −6 ... (1)
Multiplying the second by 3: 3x − y = 9 ... (2)
Elimination: Subtracting (2) from (1): 5y = −15, so y = −3
From (2): 3x − (−3) = 9 → 3x = 6, so x = 2
Substitution (check): From (2), x = 9+y3. Substituting into (1) gives the same x=2, y=−3.
Multiplying the second by 3: 3x − y = 9 ... (2)
Elimination: Subtracting (2) from (1): 5y = −15, so y = −3
From (2): 3x − (−3) = 9 → 3x = 6, so x = 2
Substitution (check): From (2), x = 9+y3. Substituting into (1) gives the same x=2, y=−3.
x = 2, y = −3
2Form 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 12 if we only add 1 to the denominator. What is the fraction?
(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?
(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.
(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.
(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.
(i) Fraction
Let the fraction be nd.
n+1d−1 = 1 → n + 1 = d − 1 → n − d = −2 ... (1)
nd+1 = 12 → 2n = d + 1 → 2n − d = 1 ... (2)
Elimination: Subtracting (1) from (2): (2n−d) − (n−d) = 1 − (−2) → n = 3
From (1): 3 − d = −2, so d = 5
n+1d−1 = 1 → n + 1 = d − 1 → n − d = −2 ... (1)
nd+1 = 12 → 2n = d + 1 → 2n − d = 1 ... (2)
Elimination: Subtracting (1) from (2): (2n−d) − (n−d) = 1 − (−2) → n = 3
From (1): 3 − d = −2, so d = 5
The fraction is 35.
(ii) Nuri and Sonu
Let Nuri’s present age = N, Sonu’s present age = S.
N − 5 = 3(S − 5) → N − 3S = −10 ... (1)
N + 10 = 2(S + 10) → N − 2S = 10 ... (2)
Elimination: Subtracting (1) from (2): (N−2S) − (N−3S) = 10 − (−10) → S = 20
From (2): N − 40 = 10, so N = 50
N − 5 = 3(S − 5) → N − 3S = −10 ... (1)
N + 10 = 2(S + 10) → N − 2S = 10 ... (2)
Elimination: Subtracting (1) from (2): (N−2S) − (N−3S) = 10 − (−10) → S = 20
From (2): N − 40 = 10, so N = 50
Nuri’s present age = 50 years, Sonu’s present age = 20 years.
(iii) Two-digit number
Let the tens digit = x, units digit = y. The number is 10x + y, and reversed is 10y + x.
x + y = 9 ... (1)
9(10x + y) = 2(10y + x) → 90x + 9y = 20y + 2x → 88x − 11y = 0 → 8x − y = 0 ... (2)
Elimination: Adding (1) and (2): 9x = 9, so x = 1
From (1): y = 8
x + y = 9 ... (1)
9(10x + y) = 2(10y + x) → 90x + 9y = 20y + 2x → 88x − 11y = 0 → 8x − y = 0 ... (2)
Elimination: Adding (1) and (2): 9x = 9, so x = 1
From (1): y = 8
The number is 18.
(iv) Meena’s notes
Let the number of ₹50 notes = a, ₹100 notes = b.
a + b = 25 ... (1)
50a + 100b = 2000 ... (2)
Elimination: Multiply (1) by 50: 50a + 50b = 1250
Subtracting from (2): 50b = 750, so b = 15
From (1): a = 25 − 15 = 10
a + b = 25 ... (1)
50a + 100b = 2000 ... (2)
Elimination: Multiply (1) by 50: 50a + 50b = 1250
Subtracting from (2): 50b = 750, so b = 15
From (1): a = 25 − 15 = 10
10 notes of ₹50 and 15 notes of ₹100.
(v) Library charges
Let the fixed charge (for the first 3 days) = ₹f, and the extra charge per day after = ₹e.
Saritha, 7 days (4 extra days): f + 4e = 27 ... (1)
Susy, 5 days (2 extra days): f + 2e = 21 ... (2)
Elimination: Subtracting (2) from (1): 2e = 6, so e = 3
From (2): f + 6 = 21, so f = 15
Saritha, 7 days (4 extra days): f + 4e = 27 ... (1)
Susy, 5 days (2 extra days): f + 2e = 21 ... (2)
Elimination: Subtracting (2) from (1): 2e = 6, so e = 3
From (2): f + 6 = 21, so f = 15
Fixed charge = ₹15, extra charge per day = ₹3.
Frequently Asked Questions
Compare a₁a₂, b₁b₂ and c₁c₂. If a₁a₂ ≠ b₁b₂, the lines intersect (unique solution, consistent). If a₁a₂ = b₁b₂ ≠ c₁c₂, the lines are parallel (no solution, inconsistent). If a₁a₂ = b₁b₂ = c₁c₂, the lines are coincident (infinitely many solutions, dependent and consistent).
Substitution expresses one variable in terms of the other from one equation, then substitutes it into the second equation to reduce to one variable. Elimination instead multiplies the equations by constants so a variable's coefficients match, then adds or subtracts the equations to cancel that variable directly. Both give the same solution — elimination is often quicker when coefficients are easy to match.
A false statement with no variable (like 0 = 9) means the pair has no solution — the equations are inconsistent, and the lines are parallel. A true statement (like 18 = 18) means the pair has infinitely many solutions — the equations are dependent, and the lines are coincident.
It becomes unreliable when the solution has non-integer coordinates, like (√3, 2√7) or (413, 119) — such points are very hard to read accurately off a hand-drawn graph. Algebraic methods (substitution or elimination) give exact values in these cases.
Yes — a printable Formula Card covering all 7 units, including Pair of Linear Equations in Two Variables, is available for ₹99.
Continue to the Next Chapter
Chapter 4 (Quadratic Equations) builds on the algebraic techniques introduced here.