Linear Equations

Comprehensive Class 10 Linear Equations Notes for Quick Learning

Class 10 Linear Equations Notes focus on equations in one and two variables, offering clear methods for solving algebraic problems. Students learn graphical representation, substitution, elimination, and cross-multiplication methods through step-by-step explanations and examples. The notes include solved problems and practice exercises to enhance problem-solving speed and conceptual clarity. Understanding linear equations is crucial for real-life applications and advanced mathematics topics. These notes also provide tips and shortcuts to tackle questions quickly in board exams. Go through the NCERT textbook and solve the NCERT questions with the help of the NCERT Solutions for class 10 Maths. With structured notes, students can revise topics efficiently, strengthen foundational skills, and improve performance in algebra sections. Class 10 students benefit from these notes for both regular studies and exam-focused preparation.

Class 10 Linear Equations Notes, solved example & Questions

1.1 Linear Equation in One Variable

The equation of the form ax = b or ax + b = 0, where a and b are two real numbers such that a ≠ 0 and x is a variable is called a linear equation in one variable.

1.2 Linear Equation in Two Variables

The general form of a linear equation in two variables is:

ax + by + c = 0 or ax + by = c

where a, b, c are real numbers and a ≠ 0, b ≠ 0 and x, y are variables.

Key Points:

  • The graph of a linear equation in two variables is a straight line.
  • The graph of a linear equation in one variable is a straight line parallel to x-axis for ay = b and parallel to y-axis for ax = b, where a ≠ 0.
  • A pair of linear equations in two variables is said to form a system of simultaneous linear equations.
  • The value of the variable x and y satisfying each one of the equations in a given system of linear equations in x and y simultaneously is called a solution of the system.

2. METHODS OF SOLVING LINEAR EQUATIONS

Standard Form of Linear Equations

(Standard form refers to all positive coefficient)

a₁x + b₁y + c₁ = 0 ....(i)
a₂x + b₂y + c₂ = 0 ....(ii)

For solving such equations we have three methods:

  1. Elimination by substitution
  2. Elimination by equating the coefficients
  3. Elimination by cross multiplication

Also Read: Arithmetic Progressions Notes

2.1 ELIMINATION BY SUBSTITUTION

Method: Find the value of any one variable in terms of other and then use it to find other variable from the second equation.

  • Step 1: Find the value of one variable, say y in terms of the other variable, i.e., x from either equation, whichever is convenient.
  • Step 2: Substitute this value of y in the other equation, and reduce it to an equation in one variable, i.e., in terms of x, which can be solved.
  • Step 3: Substitute the value of x (or y) obtained in Step 2 in the equation used in Step 1 to obtain the value of the other variable.

Example 1: Solve

x + 4y = 14 ....(i)
7x - 3y = 5 ....(ii)

Solution:

From equation (i): x = 14 - 4y ....(iii)

Substitute the value of x in equation (ii):

⇒ 7(14 - 4y) - 3y = 5

⇒ 98 - 28y - 3y = 5

⇒ 98 - 31y = 5

⇒ 93 = 31y

⇒ y = 93/31 = 3

Now substitute value of y in equation (iii):

⇒ 7x - 3(3) = 5

⇒ 7x = 14

⇒ x = 14/7 = 2

Answer: x = 2 and y = 3

2.2 ELIMINATION BY EQUATING THE COEFFICIENTS

Method: Eliminating one variable by making the coefficient equal to get the value of one variable and then put it in any equation to find other variable.

  • Step 1: First multiply both the equations by some suitable non-zero constants to make the coefficients of one variable (either x or y) numerically equal.
  • Step 2: Then add or subtract one equation from the other so that one variable gets eliminated. If you get an equation in one variable, go to step 3.
  • Step 3: Solve the equation in one variable (x or y) so obtained to get its value.
  • Step 4: Substitute this value of x (or y) in either of the original equations to get the value of the other variable.
 
Must Read: Quadratic Equations Notes

Example 1: Solve

9x - 4y = 8 ...(i)
13x + 7y = 101 ...(ii)

Solution:

Multiply equation (i) by 7 and equation (ii) by 4, we get

63x – 28y = 56
52x + 28y = 404
115x = 460

⇒ x = 460/115 = 4

Substitute x = 4 in equation (i):

9(4) – 4y = 8 ⇒ 36 – 8 = 4y ⇒ 28 = 4y ⇒ y = 28/4 = 7

Answer: x = 4 and y = 7

2.3 COMPARISON METHOD

Method: Find the value of one variable from both the equations and equate them to get the value of other variable.

Let any pair of linear equations in two variables is of the form:

a₁x + b₁y + c₁ = 0 ...(i)
a₂x + b₂y + c₂ = 0 ...(ii)
  • Step 1: Find the value of one variable, say y in terms of other variable, i.e. x from equation (i), to get equation (iii).
  • Step 2: Find the value of the same variable (as in step 1) in terms of other variable from equation (ii) to get equation (iv).
  • Step 3: By equating the variable from equation (iii) and (iv) obtained in above two steps. We get the value of second variable.
  • Step 4: Substituting the value of above said variable in equation (iii), we get the value of another variable.

Example 1: Solve for x and y: 2x + 3y = 8, x - 5y = -9

Solution:

2x + 3y = 8 ...(i)
x – 5y = –9 ...(ii)

From equation (i):

2x = 8 - 3y

or x = (8 - 3y)/2 ...(iii)

From equation (ii):

x = –9 + 5y ...(iv)

From equation (iii) and (iv), we have:

(8 - 3y)/2 = –9 + 5y

8 - 3y = –18 + 10y

8 + 18 = 10y + 3y

26 = 13y

y = 26/13 = 2

Putting value of y in equation (iv), we have:

x = (8 - 3×2)/2 = 1

Answer: x = 1, y = 2

Also Read: Statistics Notes

2.4 CROSS MULTIPLICATION METHOD

Let the equations:

a₁x + b₁y + c₁ = 0 ...(i)
a₂x + b₂y + c₂ = 0 ...(ii)

To obtain the values of x and y, we follow these steps:

Step 1: Multiply Equation (i) by b₂ and (ii) by b₁, we get

b₂a₁x + b₂b₁y + b₂c₁ = 0 ...(iii)

b₁a₂x + b₁b₂y + b₁c₂ = 0 ...(iv)

Step 2: Subtracting Equation (iv) from (iii), we get:

(b₂a₁ - b₁a₂)x + (b₂b₁ - b₁b₂)y + (b₂c₁ - b₁c₂) = 0

i.e. (b₂a₁ - b₁a₂)x = b₁c₂ - b₂c₁

x = (b₁c₂ - b₂c₁)/(a₁b₂ - a₂b₁), if a₁b₂ - a₂b₁ ≠ 0 ...(v)

Step 3: Substituting this value of x in (i) or (ii), we get

y = (c₁a₂ - c₂a₁)/(a₁b₂ - a₂b₁) ...(vi)

We can write the solution given by equations (v) and (vi) in the following form:

x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁) ...(vii)

Diagram to Remember:

x y 1
Row 1 b₁ c₁ a₁
Row 2 b₂ c₂ a₂

Cross multiply diagonally and subtract

For equations of the form a₁x + b₁y = c₁ and a₂x + b₂y = c₂:

x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = -1/(a₁b₂ - a₂b₁)

(Please note position of c₁ and c₂ with equality sign and subsequent change in the third term i.e. –1)

Example 1: Solve the following system of equations in x and y

ax + by – a – b = 0

bx – ay – a – b = 0

Solution:

a₁ = a, b₁ = b, a₂ = b and b₂ = –a

∴ a₁b₂ - a₂b₁ = a(-a) - b(b) = -a² - b² ≠ 0

Hence, the given system of the equation has unique solution.

By cross multiplication, we have:

x/[(-a)(-a+b) - b(-a-b)] = y/[(-a-b)(a) - (-a+b)(b)] = 1/[a(-a) - (b)(b)]

or x/(a² + b²) = y/(a² + b²) = 1/(-a² - b²)

or x/(-a² - b²) = y/(a² + b²) = 1/(-a² - b²)

or x = 1 and y = –1

Example 2: Solve for x and y

x/a + y/b = a + b, x/a² + y/b² = 2

Solution:

Here a₁ = 1/a, b₁ = 1/b, a₂ = 1/a² and b₂ = 1/b²

a₁b₂ - a₂b₁ = (1/a)(1/b²) - (1/a²)(1/b) = 1/(ab²) - 1/(a²b) ≠ 0

Hence, the given system of equation has unique solution.

By cross multiplication:

x/[(2/b) - (a+b)/b²] = y/[(a+b)/a² - 2/a] = 1/[(1/(ab²) - 1/(a²b)]

After simplification:

x = a², y = b²

Example 3: Solve 5x + 3y = 19, 16x – 7y = 11

Solution:

Here a₁ = 5, b₁ = 3 and a₂ = 16, b₂ = –7

∴ a₁b₂ - a₂b₁ = 5(–7) - 16(3) = –35 – 48 = –83 ≠ 0

Hence, the given system of equation has a unique solution. By cross multiplication:

x/[(3)(11) - (-7)(19)] = y/[(19)(16) - (11)(5)] = 1/[(5)(-7) - (16)(3)]

or x/(33 + 133) = y/(304 - 55) = 1/(-35 - 48)

or x/166 = y/249 = 1/(-83)

or x = 166/(-83) = –2 and y = 249/(-83) = –3

or x = 2 and y = 3

Also Read: Probability Notes

3. CONDITIONS FOR SOLVABILITY (OR CONSISTENCY) OF SYSTEM OF EQUATION

3.1 UNIQUE SOLUTION (Consistent System)

Two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, if the denominator a₁b₂ – a₂b₁ ≠ 0 then the given system of equations has unique solution (i.e. only one solution) and the solutions are said to be consistent.

a₁b₂ – a₂b₁ ≠ 0 ⇒ a₁/a₂ ≠ b₁/b₂

Graphical Representation: The two lines intersect at one point

Example: Find the value of 'P' for which the given system of equations has only one solution

Px – y = 2 ....(i)

6x – 2y = 3 ....(ii)

Solution:

a₁ = P, b₁ = –1, c₁ = –2

a₂ = 6, b₂ = –2, c₂ = –3

Conditions for unique solution is a₁/a₂ ≠ b₁/b₂

P/6 ≠ (-1)/(-2)

P/6 ≠ 1/2

P ≠ 3

∴ P can have all real values except 3.

3.2 NO SOLUTION (Inconsistent System)

Two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, if the denominator a₁b₂ – a₂b₁ = 0 then the given system of equations has no solution and solutions are said to be inconsistent.

a₁b₂ – a₂b₁ = 0 ⇒ a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Graphical Representation: The two lines are parallel

Example: Determine the value of k so that the following linear equations has no solution

(3k + 1)x + 3y – 2 = 0

(k² + 1)x + (k – 2)y – 5 = 0

Solution:

Here a₁ = 3k + 1, b₁ = 3 and c₁ = –2

a₂ = k² + 1, b₂ = k – 2 and c₂ = –5

For no solution, condition is a₁/a₂ = b₁/b₂ ≠ c₁/c₂

(3k + 1)/(k² + 1) = 3/(k – 2) ≠ (-2)/(-5)

Now, (3k + 1)/(k² + 1) = 3/(k – 2)

⇒ (3k + 1)(k – 2) = 3(k² + 1)

⇒ 3k² – 5k – 2 = 3k² + 3

⇒ –5k – 2 = 3

⇒ –5k = 5

⇒ k = –1

Clearly, 3/(k – 2) ≠ 2/5 for k = –1.

Hence, the given system of equations will has no solution for k = –1.

3.3 INFINITE SOLUTIONS (Dependent and Consistent System)

Two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, if a₁/a₂ = b₁/b₂ = c₁/c₂ then system of equations has many solution and solutions are said to be consistent.

a₁/a₂ = b₁/b₂ = c₁/c₂

Graphical Representation: The two lines are coincident (overlapping)

Example: Find the value of k for which the system of linear equation has infinite solution

kx + 4y = k – 4

16x + ky = k

Solution:

a₁ = k, b₁ = 4, c₁ = –(k – 4)

a₂ = 16, b₂ = k, c₂ = –k

Here condition is a₁/a₂ = b₁/b₂ = c₁/c₂

⇒ k/16 = 4/k = (k – 4)/k

k/16 = 4/k also 4/k = (k – 4)/k

⇒ k² = 64 ⇒ 4k = k² – 4k

⇒ k = ± 8 ⇒ k(k – 8) = 0

k = 0 or k = 8 but k = 0 is not possible otherwise equation will be one variable.

∴ k = 8 is correct value for infinite solution.

Summary Table

Condition Type of Solution Graphical Representation
a₁/a₂ ≠ b₁/b₂ Unique Solution (Consistent) Lines intersect at one point
a₁/a₂ = b₁/b₂ ≠ c₁/c₂ No Solution (Inconsistent) Lines are parallel
a₁/a₂ = b₁/b₂ = c₁/c₂ Infinite Solutions (Dependent) Lines are coincident

4. GRAPHICAL METHOD

  • Step 1: Read the problem carefully to find the unknowns (variables) which are to be determined.
  • Step 2: Depict the unknowns by x and y etc.
  • Step 3: Use the given conditions in the problem to make equations in unknowns x and y.
  • Step 4: Make the proper tables for both the equations.

4.1 TYPES OF GRAPHS

(i) Graphs of the type ax = b:

Example 1: Draw the graphs of the following equations:

(i) x = 2, (ii) 2x = 1 (iii) x + 4 = 0 (iv) x = 0

Solution:

  • (i) x = 2 - Vertical line passing through (2, 0)
  • (ii) 2x = 1 ⇒ x = 1/2 - Vertical line passing through (1/2, 0)
  • (iii) x + 4 = 0 ⇒ x = –4 - Vertical line passing through (–4, 0)
  • (iv) x = 0 - This is the y-axis itself

All these lines are parallel to y-axis

(ii) Graphs of the type ay = b:

Example 1: Draw the graphs of the following equations:

(i) y = 0, (ii) y - 2 = 0, (iii) 2y + 4 = 0

Solution:

  • (i) y = 0 - This is the x-axis itself
  • (ii) y – 2 = 0 ⇒ y = 2 - Horizontal line passing through (0, 2)
  • (iii) 2y + 4 = 0 ⇒ y = –2 - Horizontal line passing through (0, –2)

All these lines are parallel to x-axis

(iii) Graphs of the type ax + by = 0 (Passing through origin):

Example 1: Draw the graphs of the following:

(i) x = y, (ii) x = –y

Solution:

(i) x = y

x 1 –2 0
y 1 –2 0

(ii) x = –y

x 1 4 –3 0
y –1 –4 3 0

Both lines pass through origin (0, 0)

(iv) Graphs of the Type ax + by + c = 0 (Making Interception with x-axis, y-axis):

Example 1: Solve the following system of linear equations graphically: x - y = 1, 2x + y = 8. Shade the area bounded by these two lines and y-axis. Also determine this area.

Solution:

(i) x – y = 1 ⇒ x – y + 1

x 0 1 2
y –1 0 1

(ii) 2x + y = 8 ⇒ y = 8 – 2x

X 0 1 2
Y 8 6 4

Solution is x = 3 and y = 2

Area of △ABC = 1/2 × BC × AD

= 1/2 × 9 × 3 = 13.5 Sq. unit.

NATURE OF GRAPHICAL SOLUTION:

  1. Let equations of two lines are a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0.
  2. Lines are consistent (unique solution) i.e. they meet at one point condition is a₁/a₂ ≠ b₁/b₂
  3. Lines are inconsistent (no solution) i.e. they do not meet at one point condition is a₁/a₂ = b₁/b₂ ≠ c₁/c₂
  4. Lines are coincident (infinite solution) i.e. overlapping lines (or they are on one another) condition is a₁/a₂ = b₁/b₂ = c₁/c₂

5. EQUATIONS REDUCIBLE TO A PAIR OF LINEAR EQUATIONS

We shall discuss the solution of such pairs of equations which are not linear but can be reduced to linear form by making some suitable substitutions.

Example 1: Solve the following system of equations

2/x + 2/(3y) = 1/6, 3/x + 2/y = 0; x, y ≠ 0

Solution:

Put 1/x = u and 1/y = v

2u + 2v/3 = 1/6 or 6u + 2v = 1/2

or 6u + 2v = 3/6 - 1/2

⇒ 12u + 4v = 1 ....(i)

3u + 2v = 0 ....(ii)

Multiply equation (ii) by 4 to make coefficients of u equal

12u + 4v = 1 ....(iii)

12u + 8v = 0 ....(iv)

Subtracting:

–4v = 1

v = –1/4

Putting value of v in equation (i):

12u + 4(–1/4) = 1

12u = 2

u = 2/12 = 1/6

Hence, 1/x = u or 1/x = 1/6 ⇒ x = 6

1/y = v or 1/y = –1/4 ⇒ y = –4

Solution is x = 6, y = –4

Example 2: Solve the following system of equations

10/(x + y) + 2/(x - y) = 4, 15/(x + y) - 5/(x - y) = -2

Solution:

10/(x + y) + 2/(x - y) = 4 ....(i)

15/(x + y) - 5/(x - y) = -2 ....(ii)

Put 1/(x + y) = u and 1/(x - y) = v

10u + 2v = 4 ....(iii)

15u – 5v = –2 ....(iv)

Multiplying equation (iii) by 5 and equation (iv) by 2 to eliminate v:

50u + 10v = 20

30u - 10v = –4

Adding: 80u = 16

u = 16/80 = 1/5

Putting value of u in equation (iii):

10 × 1/5 + 2v = 4

2 + 2v = 4

2v = 2

v = 1

∴ 1/(x + y) = 1/5; 1/(x - y) = 1

or x + y = 5 ....(v)

x – y = 1 ....(vi)

Adding equation (v) and (vi): 2x = 6 ⇒ x = 3

Putting value of x in equation (v): 3 + y = 5 ⇒ y = 2

Solution is x = 3; y = 2

6. WORD PROBLEMS

For solving daily life problems with the help of simultaneous linear equation in two variables or equations reducible to them proceed as:

  1. Represent the unknown quantities by the same variables x and y, which are to be determined.
  2. Find the conditions given in the problem and translate the verbal conditions into a pair of simultaneous linear equation.
  3. Solve these equations & obtain the required quantities with appropriate units.

Type of Problems:

  1. Determining two numbers when the relation between them is given.
  2. Problems regarding fractions, digits of number ages of persons.
  3. Problems regarding current of a river, regarding time & distance.
  4. Problems regarding mensuration and geometry.
  5. Problems regarding time & work.
  6. Problems regarding mixtures, costs of articles, profit & loss, discount.

Example 1: The sum of digits of a two digits 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 digit at units place = y

Let digit at tens place = x

∴ Given number is 10x + y

According to question: x + y = 9 ....(i)

When the digits are reversed, y becomes tens digit and x becomes units digit.

∴ Reverse number is 10y + x

According to question:

9(10x + y) = 2(10y + x)

90x + 9y = 20y + 2x

90x – 2x = 20y – 9y

88x – 11y = 0

8x – y = 0 ....(ii)

8x = y ....(iii)

Putting value of y in equation (ii):

x + 8x = 9

9x = 9

x = 1

From equation (iii): y = 8

Required number is 18

Example 2: Find two numbers such that the sum of twice the first and thrice the second is 89 and four times the first exceeds five times the second by 13.

Solution:

Let the two numbers be x and y.

Then equations formed are:

2x + 3y = 89 ....(i)

4x – 5y = 13 ...(ii)

On solving eq. (i) & (ii), we get

x = 22 and y = 15

Hence required numbers are 22 & 15

Example 3: The numerator of a fraction is 4 less than the denominator. If the numerator is decreased by 2 and the denominator is increased by 1, then the denominator is eight times the numerator. Find the fraction.

Solution:

Let the numerator and denominator of a fraction be x and y.

Then equations formed are:

y – x = 4 ....(i)

y + 1 = 8(x – 2) ....(ii)

On solving eq. (i) & (ii), we get

x = 3 and y = 7

Hence the fraction is 3/7

Example 4: Ramesh travels 760 km to his home partly by train and partly by car. He takes 8 hours, if he travels 160 km by train and the rest by car. He takes 12 minutes more, if he travels 240 km by train and the rest by car. Find the speed of the train and the car.

Solution:

Let speed of the train be x km/hr & car be y km/hr respectively.

According to the problem:

160/x + 600/y = 8 ....(i)

240/x + 520/y = 41/5 ....(ii)

Solving equation (i) & (ii), we get x = 80 and y = 100.

Hence, speed of the train = 80 km/hr and that of car = 100km/hr

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

Definition: An equation of the form ax + by + c = 0, where a, b, c are real numbers, a or b ≠ 0 is called a linear equation in two variables x and y.

Key Properties:

  • The graph of a linear equation in two variables is always a straight line
  • If the graph of two lines intersects at a point, the system is known as consistent and has only one solution
  • If the graph of two lines is same i.e. coincident, the system is known as dependent and has many solutions

Two simultaneous equations:

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

Type Condition Number of Solutions
Consistent a₁/a₂ ≠ b₁/b₂ Unique solution
Inconsistent a₁/a₂ = b₁/b₂ ≠ c₁/c₂ No Solution
Dependent a₁/a₂ = b₁/b₂ = c₁/c₂ Infinite many solutions

Methods of Solution:

  • Algebraic Methods:
    • Elimination by Substitution
    • Elimination by Equating the Coefficients
    • Comparing Variable
    • Cross Multiplication
  • Graphical Method

Method of Cross-multiplication:

For equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we have by cross multiplication:

x/(b₁c₂ - b₂c₁) = y/(c₁a₂ - c₂a₁) = 1/(a₁b₂ - a₂b₁)

EXERCISE - 1 (Multiple Choice Questions)

  1. The equations 3x - 5y + 2 = 0, and 6x + 4 = 10y have:
    1. No solution
    2. A single solution
    3. Two solutions
    4. An infinite number of solution
  2. If p + q = 1 and the ordered pair (p, q) satisfy 3x + 2y = 1 then it also satisfies:
    1. 3x + 4y = 5
    2. 5x + 4y = 4
    3. 5x + 5y = 4
    4. None of these
  3. If x = y, 3x - y = 4 and x + y + z = 6 then the value of z is:
    1. 1
    2. 2
    3. 3
    4. 4
  4. The system of linear equation ax + by = 0, cx + dy = 0 has no solution if:
    1. ad - bc > 0
    2. ad - bc < 0
    3. ad + bc = 0
    4. ad - bc = 0
  5. The value of k for which the system kx + 3y = 7 and 2x – 5y = 3 has no solution is:
    1. 7 & k ≠ –14/3
    2. 4 & k ≠ –14/3
    3. k ≠ –14/3 & k ≠ –6/5
    4. k ≠ –14/3 & k = –6/5
  6. If 29x + 37y = 103, 37x + 29y = 95 then:
    1. x = 1, y = 2
    2. x = 2, y = 1
    3. x = 2, y = 3
    4. x = 3, y = 2
  7. On solving 40/(x + y) - 25/(x - y) = 1, 15/(x + y) + 2/(x - y) = 5, we get:
    1. x = 8, y = 6
    2. x = 4, y = 6
    3. x = 6, y = 4
    4. None of these
  8. If the system 2x + 3y - 5 = 0, 4x + ky - 10 = 0 has an infinite number of solutions then:
    1. k = –3/2
    2. k = 3/2
    3. k ≠ 6
    4. k = 6
  9. The equation x + 2y = 4 and 2x + y = 5:
    1. Are consistent and have a unique solution
    2. Are consistent and have infinitely many solution
    3. Are inconsistent
    4. Are homogeneous linear equations
  10. If 1/x = 1/y + 1/z then z will be:
    1. y – x
    2. x – y
    3. xy/(y - x)
    4. xy/(y + x)

Questions 11-50 continue with similar format covering various topics including:

  • Graph intersections
  • Sum and reciprocals
  • Word problems (coins, age, distance)
  • Linear equations properties
  • System consistency
  • Parallel and perpendicular lines
  • Cyclic quadrilaterals
  • And more...

ANSWERS TO EXERCISE - 1

Q. No. Answer Q. No. Answer Q. No. Answer Q. No. Answer Q. No. Answer
1 (d) 2 (a) 3 (b) 4 (d) 5 (d)
6 (a) 7 (c) 8 (d) 9 (a) 10 (d)
11 (b) 12 (a) 13 (d) 14 (a) 15 (d)
16 (a) 17 (a) 18 (c) 19 (d) 20 (a)
21 (b) 22 (d) 23 (a) 24 (b) 25 (c)
26 (b) 27 (a) 28 (b) 29 (c) 30 (c)
31 (c) 32 (a) 33 (b) 34 (d) 35 (a)
36 (b) 37 (d) 38 (d) 39 (a) 40 (a)
41 (c) 42 (c) 43 (a) 44 (c) 45 (a)
46 (c) 47 (d) 48 (b) 49 (c) 50 (a)

EXERCISE - 2 (Subjective Questions)

  1. A man purchased 60 stamps for Rs.195. Some of the stamps were of Rs.2 denomination and the rest were of Rs.5 denomination. Represent this situation algebraically and graphically.
  2. Determine graphically the vertices of a trapezium, the equations of whose sides are x = 0, y = 0, y = 4 and 2x + y = 6, also its area.
  3. Solve graphically the following system of linear equations. Also find the coordinates of the points where the lines meet the y-axis: 3x + 2y + 4 = 0 and 3x – 2y + 8 = 0.
  4. Show that the straight lines represented by the equations 3x + y + 5 = 0, 3y – x = 5 and 2x + 5y = 1 are concurrent and find the coordinates of the points where they intersect.
  5. Solve for x and y: (a – b)x + (a + b)y = 2a² – 2b²; (a + b)(x + y) = 4ab
  6. Solve for x and y: ax + by = (a + b)/2; 3x + 5y = 4
  7. Solve for x and y: (7x – 2y)/xy = 5; (8x + 7y)/xy = 15; where x ≠ 0, y ≠ 0
  8. Solve for x and y: 1/(2(x + 2y)) + 5/(3(3x – 2y)) = –3/2; 5/(4(x + 2y)) – 3/(5(3x – 2y)) = 61/60
  9. Solve for x and y: 2/x + 3/y = 9/xy; 4/x + 9/y = 21/xy, where x ≠ 0, y ≠ 0
  10. Solve for x and y: xy/(x + y) = 2/3; xy/(y – x) = 2, where x ≠ 0, y ≠0
  11. Solve for x and y: x/a² + y/b² – 2/ab = 0; x/(2b) – y/(2a) – (a³ – b³)/(2a²b²) = 0
  12. Solve for x and y: 1/(7x) + 1/(6y) = 3; 1/(2x) – 1/(3y) = 5, where x ≠ 0, y ≠ 0
  13. If 2x + y = 35 and 3x + 3y = 65 find the value of x/y.
  14. Solve: 2/x + 2/(3y) = 1/6; 3/x + 2/y = 0 and hence find a for which y = ax – 4.
  15. A person invested some amount at the rate of 12% simple interest and some other amount at the rate of 10% simple interest. He received yearly interest of Rs.130. But if he had interchanged the amounts invested, he would have received Rs.4 more as interest. How much amount did he invest at different rates?
  16. A and B are friends and their ages differ by 2 years. A's father D is twice as old as A and B is twice as old as his sister C. The ages of D and C differ by 40 years. Find the ages of A and B.
  17. After covering a distance of 30 km with a uniform speed there is some defect in a train engine and therefore its speed is reduced to 4/5 of its original speed. Consequently, the train reaches its destination late by 45 minutes. Had it happened after covering 18 kilometer more, the train would have reached 9 minutes earlier. Find the speed of the train and the distance of journey.
  18. Father's age is three times the sum of ages of his two children. After 5 years his age will be twice the sum of ages of two children. Find the age of father.
  19. A man sold a chair and a table together for Rs.1520 thereby making a profit of 25% on the chair and 10% on table. By selling them together for Rs.1535 he would have made a profit of 10% on the chair and 25% on the table. Find the cost price of each.
  20. 8 men and 12 boys can finish a piece of work in 10 days while 6 men and 8 boys can finish it in 14 days. Find the time taken by one man alone and that by one boy alone to finish the work.
  21. If three times the larger of two numbers is divided by the smaller one, we get 4 as the quotient and 3 as the remainder. Also, if seven times the smaller number is divided by the larger one, we get 5 as the quotient and 1 as the remainder. Find the numbers.
  22. The monthly incomes of A and B are in the ratio of 5:4 and their monthly expenditures are in the ratio of 7:5. If each saves Rs.3000 per month, find the monthly income of each.
  23. A chemist has one solution containing 50% acid and a second one containing 25% acid. How much of each should be used to make 10 litres of 40% acid solution?
  24. 3 chocolate bars and 4 ice cream cones cost Rs.135. If the price of a chocolate bar increases by 5% and that of an ice cream cone increases by 10%, the total cost goes up by Rs.9.75. Find the increased price of each chocolate bar and an ice cream cone.
  25. The sum total of the ages of father and son is 55 years. If the father was to live till his son's age equals his present age, the total of their ages would be 93 years. Find their present ages.

ANSWERS TO EXERCISE - 2

  1. x + y = 60, 2x + 5y = 195
    Rs.2 denomination stamps = x, Rs.5 denomination stamps = y
  2. (0, 0), (0, 4), (1, 4) and (3, 6) are vertices of trapezium.
    Area of trapezium = 8 sq. units
  3. x = –2, y = 1; (0, –2) and (0, 4)
  4. (–2, 1)
  5. x = (2ab - a² + b²)/b, y = [(a - b)(a² + b²)]/[b(a + b)]
  6. x = 1/2, y = 1/2
  7. x = 1, y = 1
  8. x = 1/2, y = 5/4
  9. x = 1, y = 3
  10. x = 1, y = 2
  11. x = a/b, y = b/a
  12. x = 1/14, y = 1/6
  13. 8/5
  14. x = 6, y = –4, a = 0
  15. Rs.500 at 12% and Rs.700 at 10%
  16. A = 26 years, B = 24 years
  17. Original speed = 30 km/hr, length of journey = 120 km
  18. 45 years
  19. Cost price of chair = Rs.600, cost price of table = Rs.700
  20. 140 days and 280 days
  21. 25 and 18
  22. Rs.10000, Rs.8000
  23. 6 litres, 4 litres
  24. Increased price of chocolate bar = Rs.26.25
    Increased price of icecream cone = Rs.16.50
  25. Present age of father = 37 years
    and present age of son = 18 years

Frequently Asked Questions

Ans.  A linear equation represents a straight line. Standard forms:

  • One variable: ax + b = 0
  • Two variables: ax + by = c or y = mx + b

Here, m is slope, b is y-intercept.

Ans. You can download Linear Equations Class 10 notes from Home-Tution online Learning platform.

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