Chapter 3 : Pair of Linear Equations in Two Variables Class 10 Maths NCERT Solutions

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Contents

Class 10 Maths NCERT Solutions - Chapter wise

1

Chapter 1: Real Numbers

2

Chapter 2: Polynomials

3

Chapter 3 : Pair of Linear Equations in Two Variables

4

Chapter 4 : Quadratic Equations

5

Chapter 5 : Arithmetic Progressions

6

Chapter 6 : Triangles

7

Chapter 7 : Coordinate Geometry

8

Chapter 8: Introduction to Trigonometry

9

Chapter 9: Some Applications of Trigonometry

10

Chapter 10 : Circles

11

Chapter 11 : Areas Related to Circles

12

Chapter 12: Surface Areas and Volumes

13

Chapter 13: Statistics

14

Chapter 14 : Probability

Chapter Notes

Chapter 3: Pair of Linear Equations in Two Variables — Exercise 3.1

Ever wondered how to find two unknown values (like the price of a bat and a ball) using just a few clues? This exercise teaches you the “Substitution Method”—a logical way to solve life’s puzzles by replacing one mystery with another!

Q1. Solve the following pair of linear equations by the substitution method:

i) x + y = 14 ; x – y = 4

  • The Logic: Think of this as a “tag team” match. we pick one equation, find the value of one variable (let’s say x), and then “substitute” or put that value into the second equation to kick out the other mystery (y).

  • The Steps:

    1. From the second equation: x – y = 4 => x = 4 + y (Let’s call this eq. 3)

    2. Now, put this value of x into the first equation: (4 + y) + y = 14

    3. 4 + 2y = 14 => 2y = 10 => y = 5

    4. Put y = 5 back into our “x = 4 + y” equation: x = 4 + 5 = 9

  • The Result: x = 9, y = 5

ii) s – t = 3 ; s/3 + t/2 = 6

  • Human Tip: Fractions can look scary! Pro-Tip: Multiply the whole second equation by the LCM of denominators (6) to turn it into a simple line.

  • The Steps:

    1. From eq. 1: s = 3 + t

    2. Simplify eq. 2: 2s + 3t = 36 (Multiplying by 6)

    3. Substitute s: 2(3 + t) + 3t = 36

    4. 6 + 2t + 3t = 36 => 5t = 30 => t = 6

    5. Find s: s = 3 + 6 = 9

  • The Result: s = 9, t = 6

iii) 3x – y = 3 ; 9x – 3y = 9

  • The Logic: Sometimes equations are just clones of each other! If you notice that the second equation is just the first one multiplied by 3, you’ll find they have infinite solutions.

  • The Steps:

    1. From eq. 1: y = 3x – 3

    2. Substitute in eq. 2: 9x – 3(3x – 3) = 9

    3. 9x – 9x + 9 = 9 => 9 = 9.

  • The Result: This statement is true for all values of x. Therefore, there are infinitely many solutions.

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

  • Pro-Tip: Multiply both equations by 10 to get rid of the decimals immediately!

  • The Steps:

    1. New equations: 2x + 3y = 13 and 4x + 5y = 23

    2. From eq. 1: x = (13 – 3y) / 2

    3. Substitute in eq. 2: 4[(13 – 3y) / 2] + 5y = 23

    4. 2(13 – 3y) + 5y = 23 => 26 – 6y + 5y = 23 => -y = -3 => y = 3

    5. Find x: x = (13 – 3(3)) / 2 = 4 / 2 = 2

  • The Result: x = 2, y = 3

v) root 2 x + root 3 y = 0 ; root 3 x – root 8 y = 0

  • The Logic: When the right side of both equations is 0, and there’s no constant term, there’s a very high chance the answer is just zero for both!

  • The Steps:

    1. From eq. 1: x = – (root 3 / root 2) y

    2. Substitute in eq. 2: root 3 [- (root 3 / root 2) y] – root 8 y = 0

      • (3 / root 2) y – root 8 y = 0. Taking ‘y’ common: y [ – (3 / root 2) – root 8 ] = 0

    4. Since the bracket is not zero, y must be 0.

  • The Result: x = 0, y = 0

Q2. Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of ‘m’ for which y = mx + 3.

  • The Logic: First, solve for x and y using substitution. Once you have them, plug them into the final “m” equation.

  • The Steps:

    1. From eq. 1: 2x = 11 – 3y

    2. Substitute ‘2x’ directly into eq. 2: (11 – 3y) – 4y = -24

    3. 11 – 7y = -24 => -7y = -35 => y = 5

    4. Find x: 2x = 11 – 3(5) => 2x = -4 => x = -2

    5. Find m: y = mx + 3 => 5 = m(-2) + 3

    6. 2 = -2m => m = -1

  • The Result: x = -2, y = 5, m = -1

Summary Table 

Step Method Name What to do?
Step 1 Isolate Pick one equation and get one variable alone (e.g., x = …).
Step 2 Substitute Put that “value” into the other equation.
Step 3 Solve Find the value of the first variable.
Step 4 Back-Substitute Use that answer to find the second variable.

Chapter 3: Pair of Linear Equations in Two Variables — Exercise 3.2

This exercise is where the magic happens! We move from drawing lines to using the “Substitution Method”—a powerful algebraic tool that lets you solve for two unknowns by simply swapping one out for the other.

Q1. Solve the following pair of linear equations by the substitution method:

i) x + y = 14 ; x – y = 4

  • The Logic: Think of this like a relay race. We find the value of one variable (like ‘x’) from the first equation and “pass the baton” by plugging it into the second equation.

  • The Steps:

    1. From eq (1): x = 14 – y

    2. Substitute this x in eq (2): (14 – y) – y = 4

    3. 14 – 2y = 4 => -2y = 4 – 14 => -2y = -10

    4. y = 5

    5. Now put y = 5 back into x = 14 – y: x = 14 – 5

    6. x = 9

  • The Result: x = 9, y = 5

ii) s – t = 3 ; s/3 + t/2 = 6

  • Human Tip: Fractions can be scary! Let’s kill the fractions first by multiplying the whole second equation by the LCM of 3 and 2, which is 6.

  • The Steps:

    1. eq (2) becomes: 2s + 3t = 36

    2. From eq (1): s = 3 + t

    3. Substitute in new eq (2): 2(3 + t) + 3t = 36

    4. 6 + 2t + 3t = 36 => 5t = 30

    5. t = 6

    6. Then s = 3 + 6 = 9.

  • The Result: s = 9, t = 6

iii) 3x – y = 3 ; 9x – 3y = 9

  • The Logic: Sometimes equations are just “disguises” of each other. If you multiply the first one by 3, you get the second one!

  • The Steps:

    1. From eq (1): y = 3x – 3

    2. Substitute in eq (2): 9x – 3(3x – 3) = 9

    3. 9x – 9x + 9 = 9 => 9 = 9

  • The Result: This pair has infinitely many solutions because the lines are coincident.

Q2. Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of ‘m’ for which y = mx + 3.

  • The Logic: First, solve for x and y just like before. Once you have those “coordinates,” plug them into the final equation to find the missing ‘m’.

  • The Steps:

    1. From eq (1): 2x = 11 – 3y

    2. Substitute ‘2x’ directly into eq (2): (11 – 3y) – 4y = -24

    3. 11 – 7y = -24 => -7y = -35

    4. y = 5

    5. Put y=5 in 2x = 11 – 3(5) => 2x = 11 – 15 => 2x = -4 => x = -2

    6. Now find ‘m’: 5 = m(-2) + 3

    7. 5 – 3 = -2m => 2 = -2m

  • The Result: x = -2, y = 5, and m = -1

Q3. 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.

  • The Logic: We have two conditions, so we make two equations. Let the numbers be x and y (where x > y).

  • The Steps:

    1. eq (1): x – y = 26

    2. eq (2): x = 3y

    3. Substitute x = 3y into eq (1): 3y – y = 26 => 2y = 26 => y = 13

    4. Then x = 3(13) = 39.

  • The Result: The numbers are 39 and 13.

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

  • Pro-Tip: Remember, supplementary angles always add up to 180 degrees!

  • The Steps:

    1. eq (1): x + y = 180

    2. eq (2): x = y + 18

    3. Substitute: (y + 18) + y = 180 => 2y = 162 => y = 81

    4. Then x = 81 + 18 = 99.

  • The Result: The angles are 99 degrees and 81 degrees.

Summary Table

Step Name What to do? Pro-Tip
Isolate Pick the easiest equation and solve for one variable (x or y). Pick the variable with no coefficient (like just ‘x’).
Substitute Plug that value into the OTHER equation. Don’t plug it back into the same equation!
Solve Find the value of the first variable. Watch your plus/minus signs.
Back-Substitute Plug the answer back to find the second variable. Check if the answers fit both original equations.

Chapter 3: Pair of Linear Equations in Two Variables — Exercise 3.3

Ever wondered how to find two unknown values at once? Exercise 3.3 is your ultimate toolkit for solving word problems that appear in real life, from age calculations to taxi fares!

Q1. 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.

  • The Logic: We have two unknowns, so we need two “clues” (equations). We’ll use the Substitution Method, where we find the value of one variable from one equation and “plug” it into the second one.

  • The Steps:

    1. Let the larger number be ‘x’ and the smaller number be ‘y’.

    2. Clue 1: The difference is 26. So, x – y = 26 (Equation 1).

    3. Clue 2: One is 3 times the other. So, x = 3y (Equation 2).

    4. Substitute the value of x from Eq 2 into Eq 1:

      3y – y = 26

      2y = 26

      y = 13

    5. Now put y = 13 back into Eq 2:

      x = 3(13)

      x = 39

  • The Result: The numbers are 39 and 13.

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

    • Human Tip: Remember the math basics! Supplementary angles always add up to 180 degrees. That is your hidden first equation.

    • The Steps:

      1. Let the larger angle be ‘x’ and smaller be ‘y’.

      2. Eq 1: x + y = 180

      3. Eq 2: x = y + 18 (since it exceeds by 18)

      4. Substitute Eq 2 into Eq 1:

        (y + 18) + y = 180

        2y + 18 = 180

        2y = 162

        y = 81

      5. Find x: x = 81 + 18 = 99.

    • The Result: The angles are 99 degrees and 81 degrees.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

  • The Logic: This is a classic “Market Math” problem. Assign ‘x’ to the bat and ‘y’ to the ball.

  • The Steps:

    1. Eq 1: 7x + 6y = 3800

    2. Eq 2: 3x + 5y = 1750

    3. From Eq 2, let’s find x: 3x = 1750 – 5y => x = (1750 – 5y) / 3

    4. Substitute this x into Eq 1:

      7[(1750 – 5y) / 3] + 6y = 3800

      (12250 – 35y) / 3 + 6y = 3800

      Multiply the whole equation by 3 to remove the fraction:

      12250 – 35y + 18y = 11400

      -17y = 11400 – 12250

      -17y = -850

      y = 50

    5. Find x: x = (1750 – 5*50) / 3 = 1500 / 3 = 500.

  • The Result: Cost of a bat is Rs 500 and a ball is Rs 50.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered…

  • Pro-Tip: In “Taxi Problems,” always separate the Fixed Charge (x) from the Running Charge (y per km).

  • The Steps:

    1. For 10 km: x + 10y = 105

    2. For 15 km: x + 15y = 155

    3. Subtract Eq 1 from Eq 2: (x + 15y) – (x + 10y) = 155 – 105

      5y = 50 => y = 10 (Charge per km)

    4. Put y=10 in Eq 1: x + 10(10) = 105 => x = 5 (Fixed charge)

    5. For 25 km: 5 + 25(10) = 255.

  • The Result: Fixed charge is Rs 5, charge per km is Rs 10, and a 25km journey costs Rs 255.

(v) A fraction becomes 9/11 if 2 is added to both… it becomes 5/6 if 3 is added. Find the fraction.

  • The Steps:

    1. Let fraction be x/y.

    2. Eq 1: (x+2)/(y+2) = 9/11 => 11x + 22 = 9y + 18 => 11x – 9y = -4

    3. Eq 2: (x+3)/(y+3) = 5/6 => 6x + 18 = 5y + 15 => 6x – 5y = -3

    4. Solve via substitution to get: x = 7, y = 9.

  • The Result: 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?

  • Human Tip: “Hence” means future (+) and “Ago” means past (-). Always start from the Present Age.

  • The Steps:

    1. Let Jacob be ‘x’ and Son be ‘y’.

    2. Eq 1 (Future): (x+5) = 3(y+5) => x – 3y = 10

    3. Eq 2 (Past): (x-5) = 7(y-5) => x – 7y = -30

    4. Solve: x = 40, y = 10.

  • The Result: Jacob is 40 years old and his son is 10 years old.

Summary Table

Problem Type Variable Setup Key Tip
Number Based x and y “Is” usually means “=”
Geometry x + y = 180 (Suppl.) x + y = 90 (Compl.)
Fixed Charge x (Fixed) + y (per unit) Common in Taxi/Library problems
Age Problems Jacob (x), Son (y) Apply (+/-) to BOTH ages!

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