Pair of Linear Equations Class 10 Maths PPT & Notes (Ch 3)
Section 1 · Pair of linear equations: basics
What is a pair of linear equations?
A pair of linear equations in two variables x and y has the form
a₂x + b₂y + c₂ = 0
where a₁, b₁, c₁, a₂, b₂ and c₂ are real numbers, with a₁² + b₁² ≠ 0 and a₂² + b₂² ≠ 0. Each equation is a straight line, so the pair can have one solution, no solution or infinitely many solutions (coincident lines).
Conditions for consistency and inconsistency
| Case | Condition | Lines | Solutions |
|---|---|---|---|
| Unique solution (consistent) | a₁/a₂ ≠ b₁/b₂ | Intersecting lines | Exactly one |
| No solution (inconsistent) | a₁/a₂ = b₁/b₂ ≠ c₁/c₂ | Parallel lines | None |
| Infinitely many solutions (consistent, dependent) | a₁/a₂ = b₁/b₂ = c₁/c₂ | Coincident lines | Infinitely many |
| Pair | Check | Result |
|---|---|---|
| x + y = 4 and x − y = 2 | 1/1 ≠ 1/(−1) | Unique solution, the point (3, 1) |
| 2x + 3y = 6 and 4x + 6y = 15 | 2/4 = 3/6 ≠ 6/15 (that is 1/2 ≠ 2/5) | Parallel lines, no solution |
| 2x + y = 4 and 4x + 2y = 8 | 2/4 = 1/2 = 4/8 | Coincident lines, infinitely many solutions |
Graphical method: solve and find the triangle
Problem: draw the graphs of 2x + y = 4 and 2x − y = 4. Write the vertices of the triangle formed by these lines and the y-axis, and find its area.
- For 2x + y = 4: the points are (0, 4), (1, 2) and (2, 0)
- For 2x − y = 4: the points are (0, −4), (1, −2) and (2, 0)
- The lines intersect at (2, 0). On the y-axis they cut at (0, 4) and (0, −4)
The vertices of the triangle are A (0, 4), B (2, 0) and C (0, −4). Its base on the y-axis is AC = 4 − (−4) = 8 units and its height (the distance of B from the y-axis) is 2 units, so area = ½ × 8 × 2 = 8 square units.
Section 2 · Algebraic methods
Substitution method
- Express one variable in terms of the other from either equation
- Substitute this value into the other equation
- Solve the single-variable equation
- Back-substitute to find the other variable
Example: solve 3x + 2y = 11 and 2x − y = 2. From the second equation y = 2x − 2. Substituting in the first: 3x + 2(2x − 2) = 11, so 7x − 4 = 11 and x = 15/7. Then y = 2(15/7) − 2 = 16/7.
Check: 3(15/7) + 2(16/7) = 77/7 = 11 and 2(15/7) − 16/7 = 14/7 = 2. Both equations hold.
Elimination method
- Multiply the equations so that the coefficients of one variable become equal (numerically)
- Add or subtract the equations to eliminate that variable
- Solve the resulting single-variable equation
- Substitute back to find the other variable
Example: solve 5x − 3y = 11 and 4x + y = 7. Multiply the second equation by 3: 12x + 3y = 21. Add it to the first: 17x = 32, so x = 32/17. Then y = 7 − 4x = 7 − 128/17 = −9/17.
Check: 5(32/17) − 3(−9/17) = 187/17 = 11 and 4(32/17) − 9/17 = 119/17 = 7.
Finding the value of k
| Problem | Solution |
|---|---|
| Q1. For what value of k do kx + 3y = k − 3 and 12x + ky = k have no solution? | No solution needs k/12 = 3/k ≠ (k − 3)/k. From k/12 = 3/k we get k² = 36, so k = 6 or k = −6. For k = 6 the ratio (k − 3)/k = 3/6 = 1/2 equals the other two ratios (k/12 = 1/2 and 3/k = 1/2), so the lines coincide (infinitely many solutions). For k = −6, k/12 = −1/2 = 3/k but (k − 3)/k = 3/2 ≠ −1/2. So k = −6 |
| Q2. Find k so that 3x − y + 8 = 0 and 6x − ky = −16 represent coincident lines. | Write the second as 6x − ky + 16 = 0. Coincident lines need 3/6 = (−1)/(−k) = 8/16, that is 1/2 = 1/k = 1/2, so k = 2 |
| Q3. For what value of k does 2x + ky = 1, 3x − 5y = 7 have a unique solution? | A unique solution needs a₁/a₂ ≠ b₁/b₂, that is 2/3 ≠ k/(−5), so k ≠ −10/3. The pair has a unique solution for every real k except −10/3 |
Section 3 · Word problems with solutions
Age problems
| Problem | Solution |
|---|---|
| The father’s age is six times his son’s age. Four years hence, the father’s age will be four times the son’s age. Find the present ages. | Let the son be x years and the father y years. Then y = 6x and y + 4 = 4(x + 4). So 6x + 4 = 4x + 16, giving x = 6 and y = 36. Son: 6 years, father: 36 years (check: 36 + 4 = 40 = 4 × 10) |
| Two years ago, Salim was thrice as old as his daughter. Six years later he will be four years older than twice her age. How old are they now? | Let Salim be s and his daughter d. Then s − 2 = 3(d − 2) and s + 6 = 2(d + 6) + 4. So s = 3d − 4 and s = 2d + 10, giving d = 14 and s = 38. Salim: 38 years, daughter: 14 years |
Numbers and digits
| Problem | Solution |
|---|---|
| The sum of a two-digit number and the number obtained by reversing its digits is 66. The digits differ by 2. Find the number. | Number = 10x + y, reversed = 10y + x. The sum is 11(x + y) = 66, so x + y = 6. With x − y = 2, we get x = 4 and y = 2. The number is 42 (if y is the larger digit, it is 24) |
| The sum of the digits of a two-digit number is 9. Nine times this number is twice the number obtained by reversing the digits. Find the number. | x + y = 9 and 9(10x + y) = 2(10y + x), which gives 88x = 11y, so y = 8x. Then 9x = 9, x = 1 and y = 8. The number is 18 (check: 9 × 18 = 162 = 2 × 81) |
| Two numbers are in the ratio 5 : 6. If 8 is subtracted from each, the ratio becomes 4 : 5. Find the numbers. | Let them be 5k and 6k. (5k − 8) ÷ (6k − 8) = 4/5 gives 25k − 40 = 24k − 32, so k = 8. The numbers are 40 and 48 |
Coins and money
| Problem | Solution |
|---|---|
| Aruna has only Re 1 and Rs 2 coins. The total number of coins is 50 and the total amount is Rs 75. Find the number of coins of each type. | Let Re 1 coins = x and Rs 2 coins = y. x + y = 50 and x + 2y = 75. Subtracting, y = 25 and x = 25. 25 coins of Re 1 and 25 coins of Rs 2 |
| Meena withdrew Rs 2000 in Rs 50 and Rs 100 notes only, and got 25 notes in all. How many notes of each type did she receive? | x + y = 25 and 50x + 100y = 2000 (that is x + 2y = 40). Subtracting, y = 15 and x = 10. 10 notes of Rs 50 and 15 notes of Rs 100 |
| The cost of 4 pens and 4 pencil boxes is Rs 100. Three times the cost of a pen is Rs 15 more than the cost of a pencil box. Find the cost of each. | Let a pen cost x and a pencil box y. 4x + 4y = 100 (x + y = 25) and 3x = y + 15. So x + 3x − 15 = 25, giving x = 10 and y = 15. A pen costs Rs 10 and a pencil box costs Rs 15 |
Speed, distance and time
| Problem | Solution |
|---|---|
| A motorboat travels 30 km upstream and 28 km downstream in 7 hours. It travels 21 km upstream and returns in 5 hours. Find the speed of the boat in still water and the speed of the stream. | Let boat speed = x km/h and stream speed = y km/h. Put u = 1/(x − y) and v = 1/(x + y). Then 30u + 28v = 7 and 21u + 21v = 5, so u + v = 5/21. Solving, u = 1/6 and v = 1/14, so x − y = 6 and x + y = 14. Boat: 10 km/h, stream: 4 km/h |
| A person rowing at 5 km/h in still water takes thrice as long to go 40 km upstream as downstream. Find the speed of the stream. | Let the stream be y km/h. 40/(5 − y) = 3 × 40/(5 + y), so 5 + y = 3(5 − y) = 15 − 3y, giving 4y = 10. Speed of the stream = 2.5 km/h |
| Ankita travels 14 km to her home partly by rickshaw and partly by bus. She takes half an hour if she travels 2 km by rickshaw and the rest by bus. If she travels 4 km by rickshaw and the rest by bus, she takes 9 minutes longer. Find the speeds of the rickshaw and the bus. | Let the rickshaw speed be x km/h and the bus speed y km/h, with u = 1/x and v = 1/y. 2u + 12v = 1/2 and 4u + 10v = 1/2 + 9/60 = 13/20. Doubling the first: 4u + 24v = 1, and subtracting gives 14v = 7/20, so v = 1/40 and u = 1/10. Rickshaw: 10 km/h, bus: 40 km/h |
Fractions and angles
| Problem | Solution |
|---|---|
| A fraction becomes 9/11 if 2 is added to both the numerator and denominator. If 3 is added to both, it becomes 5/6. Find the fraction. | Let the fraction be x/y. 11(x + 2) = 9(y + 2) gives 11x − 9y = −4, and 6(x + 3) = 5(y + 3) gives 6x − 5y = −3. Solving, x = 7 and y = 9. The fraction is 7/9 (check: 9/11 and 10/12 = 5/6) |
| The angles of a triangle are x, y and 40°. The difference between x and y is 30°. Find x and y. | x + y + 40 = 180, so x + y = 140, and x − y = 30. Adding, 2x = 170. x = 85° and y = 55° |
| The angles of a cyclic quadrilateral ABCD are ∠A = (6x + 10)°, ∠B = (5x)°, ∠C = (x + y)° and ∠D = (3y − 10)°. Find x and y, and the four angles. | Opposite angles of a cyclic quadrilateral add to 180°: ∠A + ∠C = 180 gives 7x + y = 170, and ∠B + ∠D = 180 gives 5x + 3y = 190. Solving, x = 20 and y = 30. The angles are ∠A = 130°, ∠B = 100°, ∠C = 50°, ∠D = 80° (check: 130 + 50 = 180 and 100 + 80 = 180) |
Profit and pipes
| Problem | Solution |
|---|---|
| Jamila sold a table and a chair for Rs 1050, making 10% profit on the table and 25% on the chair. Had she made 25% profit on the table and 10% on the chair, she would have got Rs 1065. Find the cost price of each. | Let the cost prices be Rs x (table) and Rs y (chair). 1.1x + 1.25y = 1050 and 1.25x + 1.1y = 1065. Adding: 2.35(x + y) = 2115, so x + y = 900. Subtracting: 0.15(x − y) = 15, so x − y = 100. Table: Rs 500, chair: Rs 400 |
| Two pipes together fill a pool in 12 hours. If the larger pipe is used for 4 hours and the smaller for 9 hours, only half the pool is filled. How long would each pipe take alone? | Let the larger pipe take x hours and the smaller y hours. With u = 1/x and v = 1/y: u + v = 1/12 and 4u + 9v = 1/2. Multiplying the first by 4: 4u + 4v = 1/3, and subtracting gives 5v = 1/6, so v = 1/30 and u = 1/20. Larger pipe: 20 hours, smaller pipe: 30 hours |
| A shopkeeper sells a saree at 8% profit and a sweater at 10% discount, getting Rs 1008. At 10% profit on the saree and 8% discount on the sweater she would get Rs 1028. Find the cost price of the saree and the list price of the sweater. | Let the saree cost Rs x and the sweater list price Rs y. 1.08x + 0.90y = 1008 and 1.10x + 0.92y = 1028. Solving, saree cost price = Rs 600, sweater list price = Rs 400 (check: 648 + 360 = 1008 and 660 + 368 = 1028) |
Plans, restaurant and taxi
| Problem | Solution |
|---|---|
| Plan A: fixed Rs 99 per month plus Rs 2 per extra call minute. Plan B: fixed Rs 199 per month plus Rs 1 per extra call minute. At how many minutes will both plans cost the same? | Let the extra minutes be x and the total cost y. Plan A: y = 99 + 2x. Plan B: y = 199 + x. Equal when 99 + 2x = 199 + x, so x = 100 minutes, when both cost Rs 299. Beyond 100 minutes Plan B is cheaper |
| A restaurant sells thalis and cold drinks. Order 1: 3 thalis and 2 cold drinks cost Rs 240. Order 2: 5 thalis and 3 cold drinks cost Rs 390. Find the price of 1 thali and 1 cold drink. | 3x + 2y = 240 and 5x + 3y = 390. Multiplying the first by 3 and the second by 2: 9x + 6y = 720 and 10x + 6y = 780. Subtracting, x = 60 and y = 30. 1 thali = Rs 60, 1 cold drink = Rs 30 |
| A taxi charges a fixed charge of Rs a plus Rs b per km. A 10 km journey costs Rs 105 and a 15 km journey costs Rs 155. Find a and b, and the cost for 25 km. | a + 10b = 105 and a + 15b = 155. Subtracting, 5b = 50, so b = 10 and a = 5. Fixed charge = Rs 5, charge per km = Rs 10. Cost for 25 km = 5 + 25 × 10 = Rs 255 |
Section 4 · Quick revision
| Topic | Key points |
|---|---|
| General form | a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 |
| Unique solution | a₁/a₂ ≠ b₁/b₂ (intersecting lines) |
| No solution | a₁/a₂ = b₁/b₂ ≠ c₁/c₂ (parallel lines) |
| Infinitely many solutions | a₁/a₂ = b₁/b₂ = c₁/c₂ (coincident lines) |
| Substitution | Express x or y, substitute, solve, back-substitute |
| Elimination | Make the coefficients equal, add or subtract, solve, back-substitute |
| Word problems | Age, coins, speed, fractions, profit, digits and pipes |
Section 5 · Practice: multiple choice questions
Tap an option to check it at once. Open Check answer for the reason.
1If a₁/a₂ ≠ b₁/b₂, the pair of linear equations has: [1]
Check answer
The lines intersect at one point.
2Parallel lines represent a pair of equations that has: [1]
Check answer
a₁/a₂ = b₁/b₂ ≠ c₁/c₂.
3Coincident lines represent a pair of equations with: [1]
Check answer
a₁/a₂ = b₁/b₂ = c₁/c₂.
4The solution of x + y = 4 and x − y = 2 is: [1]
Check answer
Adding gives 2x = 6, so x = 3 and y = 1.
5The pair 2x + 3y = 6 and 4x + 6y = 15 has: [1]
Check answer
2/4 = 3/6 ≠ 6/15, so the lines are parallel.
6The pair 2x + y = 4 and 4x + 2y = 8 has: [1]
Check answer
2/4 = 1/2 = 4/8, so the lines coincide.
7The area of the triangle formed by 2x + y = 4, 2x − y = 4 and the y-axis is: [1]
Check answer
The vertices are (0, 4), (2, 0) and (0, −4): ½ × 8 × 2 = 8.
8For 2x + ky = 1 and 3x − 5y = 7 to have a unique solution, k must satisfy: [1]
Check answer
2/3 ≠ k/(−5) gives k ≠ −10/3.
9The value of k for which 3x − y + 8 = 0 and 6x − ky = −16 coincide is: [1]
Check answer
3/6 = 1/k = 8/16 gives k = 2.
10kx + 3y = k − 3 and 12x + ky = k have no solution for k = [1]
Check answer
For k = 6 the lines coincide; for k = −6 they are parallel.
11A father is six times as old as his son, and four years hence he will be four times as old. The son’s present age is: [1]
Check answer
6x + 4 = 4(x + 4) gives x = 6, and the father is 36.
12The two-digit number whose digits add up to 9 and whose 9 times equals twice its reverse is: [1]
Check answer
x + y = 9 and y = 8x give x = 1 and y = 8.
13Two numbers are in the ratio 5 : 6. If 8 is subtracted from each, the ratio becomes 4 : 5. The numbers are: [1]
Check answer
(5k − 8) ÷ (6k − 8) = 4/5 gives k = 8.
14Aruna has 50 coins of Re 1 and Rs 2 worth Rs 75 in all. She has: [1]
Check answer
x + y = 50 and x + 2y = 75 give y = 25 and x = 25.
15Meena gets Rs 2000 in 25 notes of Rs 50 and Rs 100. The number of Rs 100 notes is: [1]
Check answer
x + y = 25 and x + 2y = 40 give y = 15.
16A motorboat goes 30 km upstream and 28 km downstream in 7 hours, and 21 km each way in 5 hours. Its speed in still water is: [1]
Check answer
x − y = 6 and x + y = 14 give x = 10 and y = 4.
17A fraction becomes 9/11 on adding 2 to both parts and 5/6 on adding 3. The fraction is: [1]
Check answer
7/9 gives 9/11 and 10/12 = 5/6.
18Two pipes fill a pool in 12 hours together. If the larger pipe runs for 4 hours and the smaller for 9 hours, half the pool fills. The larger pipe alone takes: [1]
Check answer
1/x = 1/20 and 1/y = 1/30.
19Plan A costs 99 + 2x and Plan B costs 199 + x. Both cost the same at x = [1]
Check answer
99 + 2x = 199 + x gives x = 100.
20A taxi charges Rs 105 for 10 km and Rs 155 for 15 km. The fixed charge is: [1]
Check answer
a + 10b = 105 and a + 15b = 155 give b = 10 and a = 5.
21Assertion (A): The pair x + y = 4 and x − y = 2 has a unique solution.
Reason (R): a₁/a₂ ≠ b₁/b₂, since 1/1 ≠ 1/(−1). [1]
Check answer
Both statements are true, and the condition a₁/a₂ ≠ b₁/b₂ is exactly the test for a unique solution, so R explains A.
22Assertion (A): The pair 2x + 3y = 6 and 4x + 6y = 15 has infinitely many solutions.
Reason (R): The lines are parallel. [1]
Check answer
A is false: 2/4 = 3/6 ≠ 6/15, so the lines are parallel and there is no solution. R is true.
Section 6 · Practice: short answers
Q1State the conditions for a pair of linear equations to be consistent or inconsistent. [3]
Check answer
Q2Solve by the substitution method: 3x + 2y = 11 and 2x − y = 2. [3]
Check answer
Q3Solve by the elimination method: 5x − 3y = 11 and 4x + y = 7. [3]
Check answer
Q4Draw the graphs of 2x + y = 4 and 2x − y = 4. Find the vertices and the area of the triangle formed with the y-axis. [4]
Check answer
Q5Find k for which kx + 3y = k − 3 and 12x + ky = k have no solution. [4]
Check answer
Q6Find the value of k for which 3x − y + 8 = 0 and 6x − ky = −16 represent coincident lines. [3]
Check answer
Q7For what values of k does 2x + ky = 1, 3x − 5y = 7 have a unique solution? [3]
Check answer
Q8The father’s age is six times his son’s age. Four years hence, the father will be four times the son’s age. Find their present ages. [4]
Check answer
Q9Two years ago Salim was thrice as old as his daughter. Six years later he will be four years older than twice her age. Find their ages. [4]
Check answer
Q10The sum of a two-digit number and its reverse is 66, and the digits differ by 2. Find the number. [3]
Check answer
Q11A motorboat travels 30 km upstream and 28 km downstream in 7 hours, and 21 km upstream and back in 5 hours. Find the boat’s speed in still water and the speed of the stream. [5]
Check answer
Q12A fraction becomes 9/11 when 2 is added to both numerator and denominator, and 5/6 when 3 is added. Find the fraction. [4]
Check answer
Q13The angles of a cyclic quadrilateral are (6x + 10)°, (5x)°, (x + y)° and (3y − 10)°. Find x, y and the four angles. [5]
Check answer
Q14Jamila sold a table and a chair for Rs 1050, at 10% profit on the table and 25% on the chair. With 25% and 10% she would have got Rs 1065. Find the cost price of each. [5]
Check answer
Q15Two pipes together fill a pool in 12 hours. If the larger runs for 4 hours and the smaller for 9 hours, half the pool is filled. Find the time each takes alone. [5]
Check answer
Q16A taxi charges a fixed amount plus a rate per km. It costs Rs 105 for 10 km and Rs 155 for 15 km. Find the fixed charge, the rate and the fare for 25 km. [4]
Check answer
Section 7 · Frequently asked questions
What is a pair of linear equations in two variables?
How do I know how many solutions a pair has?
Which is easier, substitution or elimination?
How do I solve word problems on linear equations?
What if the equations involve 1/x and 1/y, as in speed or pipe problems?
Where can I download the Pair of Linear Equations Class 10 PPT?
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