Class 9 Maths Part 2 Chapter 13 – Two Variables, One Line from Ganita Manjari with easy Notes and complete Book Solutions. Get clear explanations, important concepts, questions and answers, and revision material to help Class 9 students learn, practise and prepare effectively for examinations.

Class 9 Maths Chapter 13 Two Variables, One Line | Ganita Manjari Part II

Class 9 Maths – Chapter 13
Two Variables, One Line

Chapter Overview: In this chapter we study linear equations in two variables, their solutions, graphs, slope, slope-intercept form and pairs of linear equations. We also learn how to solve a pair of linear equations using substitution, elimination and graphical methods.

13.1 Linear Equations in Two Variables

Easy Concept

A linear equation in two variables contains two variables, usually x and y, and the highest power of each variable is 1.

ax + by + c = 0

Here a, b and c are real numbers, and a and b are not both zero.

Remember: A linear equation in two variables generally has infinitely many solutions. Each solution is an ordered pair (x, y).

13.1.1 Standard Form

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

ax + by + c = 0

a is the coefficient of x.
b is the coefficient of y.
c is the constant term.
a and b cannot both be zero.

Example:
5x + 2y = 2.7

Standard form:
5x + 2y − 2.7 = 0

Therefore:
a = 5, b = 2, c = −2.7

An equation involving only one variable can also be considered a special linear equation in two variables. For example:

7 + 4x = 0
can be written as:
4x + 0y + 7 = 0.

Example 1: Write each equation in standard form

Example 1: Write each of the following equations in standard form, ax + by + c = 0, and indicate the values of a, b and c in each case:

5x + 2y = 2.7
…
1.5x − 1.3y = 4
3y = 5

(i)

5x + 2y = 2.7

5x + 2y − 2.7 = 0

Therefore, a = 5, b = 2, c = −2.7.

(iii)

1.5x − 1.3y = 4

1.5x − 1.3y − 4 = 0

Therefore, a = 1.5, b = −1.3, c = −4.

(iv)

3y = 5

0x + 3y − 5 = 0

Therefore, a = 0, b = 3, c = −5.

The equation can also be multiplied throughout by −1. Both forms represent the same line.

Write a linear equation in two variables in which a = 3, b = 0 and c = −1.

Using ax + by + c = 0:

3x + 0y − 1 = 0

Therefore, the required equation is: 3x − 1 = 0.

Complete the following table after expressing the given linear equations in standard form.

Linear equation	Standard Form	Coefficient of x	Coefficient of y	Constant term
x = 8	x − 8 = 0	1	0	−8
3y = 7	3y − 7 = 0	0	3	−7
5x = 3y	5x − 3y = 0	5	−3	0

The important step is to bring all terms to one side and write the equation in the form ax + by + c = 0.

Understanding Linear Equations from Real-Life Situations

(i) The cost of a notebook is twice the cost of a pen. Consider the cost of a notebook to be t and that of a pen to be p. Charlie wrote t = 2p whereas Meera wrote p = 2t. Which of these two representations is correct?

The notebook costs twice as much as the pen.

If the notebook costs t and the pen costs p, then:

t = 2p

Therefore, Charlie’s representation is correct.

(ii) In a one-day International Cricket match between India and Sri Lanka played in Nagpur, two Indian batsmen together scored 176 runs. Manisha expressed this situation as x + y = 176, where the number of runs scored by one batsman is x, and the number of runs scored by the other is y. Is this a correct representation?

Yes. If the two batsmen scored x and y runs, their total score is x + y.

x + y = 176

Hence, the representation is correct.

13.2 Solutions of Linear Equations in Two Variables

13.2.1 Algebraic Meaning of Solution

A solution of a linear equation in two variables is an ordered pair (x, y) that satisfies the equation.

Example:
For 3x + 2y = 12:

If x = 2 and y = 3:
3(2) + 2(3) = 6 + 6 = 12.

Therefore, (2, 3) is a solution.

Important: The order matters. (2, 3) and (3, 2) are different ordered pairs.

Think and Reflect: (0, 6), (4, 0) and (1, 3) solutions of the equation 3x + 2y = 12? Can you find any other solution? How many solutions can you find?

Check each ordered pair by substituting its values.

For (0, 6): 3(0) + 2(6) = 12. So it is a solution.

For (4, 0): 3(4) + 2(0) = 12. So it is a solution.

For (1, 3): 3(1) + 2(3) = 9, not 12. So it is not a solution.

There are infinitely many solutions. For example: (2,3), (0,6), (4,0), (6,−3), etc.

13.2.2 Geometric Visualisation of Solutions

When all the solutions of a linear equation in two variables are plotted on a Cartesian plane, they lie on a straight line.

Every point on the line is a solution of the equation.

Example 2: Find any two solutions of 2x + y = 7

Example 2: Find any two solutions of 2x + y = 7. Draw the graph of this equation. Use the graph to find 4 more solutions.

Put x = 0:

2(0) + y = 7

Therefore y = 7.

First solution = (0, 7).

Put y = 0:

2x = 7

x = 7/2.

Second solution = (7/2, 0).

Example 3: Find two solutions and draw graphs

Example 3: Find any two solutions for each of the following equations and draw their graphs:
(i) 4x + 3y = 12
(ii) 2x + 5y = 0

(i) 4x + 3y = 12

Put x = 0:

3y = 12, so y = 4.

First solution = (0,4).

Put y = 0:

4x = 12, so x = 3.

Second solution = (3,0).

Plot (0,4) and (3,0) and join them to obtain the straight line.

(ii) 2x + 5y = 0

Put x = 0:

5y = 0, so y = 0.

First solution = (0,0).

Put x = 1:

2 + 5y = 0

5y = −2

y = −2/5.

Second solution = (1,−2/5).

Another solution is (5,−2).

Example 4

Example 4: In the equations shown below a and b are unknown numbers. 3ax + 4y = −2
2x + by = 14.
If (−3, 4) is the solution of both equations, find the values of a and b.

First equation:

3a(−3) + 4(4) = −2

−9a + 16 = −2

−9a = −18

a = 2

Second equation:

2(−3) + 4b = 14

−6 + 4b = 14

4b = 20

b = 5

Therefore, a = 2 and b = 5.

Think and Reflect

Whenever c = 0, the line ax + by + c = 0 will always pass through the origin, irrespective of the values of a or b. Do you agree with them?

Yes.

If c = 0, the equation becomes:

ax + by = 0

Putting x = 0 and y = 0:

a(0) + b(0) = 0

So the point (0,0) satisfies the equation. Therefore the line passes through the origin.

Exercise Questions – 13.2

1. Verify if the ordered pair (4, 3) is a solution of 5x – 6y = 2. Explain your reasoning.

Substitute x = 4 and y = 3:

5(4) − 6(3) = 20 − 18 = 2.

The result is 2, which is equal to the right-hand side.

Therefore, (4,3) is a solution.

1. Find any two solutions for each of the following equations:
(i) 7x – 3y = 21
(ii) 2x + 3y = 5

(i) 7x − 3y = 21

Put x = 0: −3y = 21, so y = −7.

Solution: (0,−7)

Put x = 3: 21 − 3y = 21, so y = 0.

Solution: (3,0)

(ii) 2x + 3y = 5

Put x = 1: 2 + 3y = 5, so y = 1.

Solution: (1,1)

Put x = 4: 8 + 3y = 5

3y = −3, so y = −1.

Solution: (4,−1)

1. In the equations given below, m and n are unknown constants:

2ma + 8y = 7; 4x + ny = −10,
If (2, −1) is the solution of both equations, find the values of m and n.

Using the equations as given:

2m(2) + 8(−1) = 7

4m − 8 = 7

4m = 15

m = 15/4

For the second equation:

4(2) + n(−1) = −10

8 − n = −10

n = 18

m = 15/4 and n = 18.

1. Find two solutions which lie in different quadrants for each of the following linear equations. Identify the quadrants in which the points lie.
(i) 5x + 3y = 7
(ii) 5x – 3y = 7
(iii) -5x + 3y = 7
(iv) -5x – 3y = 7
Verify your solutions by representing the linear equations on a graph paper.

(i) 5x + 3y = 7

Take x = 2: 10 + 3y = 7, so y = −1. Point = (2,−1), Quadrant IV.

Take x = −1: −5 + 3y = 7, so y = 4. Point = (−1,4), Quadrant II.

(ii) 5x − 3y = 7

Take x = 2: 10 − 3y = 7, so y = 1. Point = (2,1), Quadrant I.

Take x = −1: −5 − 3y = 7, so y = −4. Point = (−1,−4), Quadrant III.

(iii) −5x + 3y = 7

Take x = −1: 5 + 3y = 7, so y = 2/3. Point = (−1,2/3), Quadrant II.

Take x = 2: −10 + 3y = 7, so y = 17/3. Point = (2,17/3), Quadrant I.

(iv) −5x − 3y = 7

Take x = 1: −5 − 3y = 7, so y = −4. Point = (1,−4), Quadrant IV.

Take x = −1: 5 − 3y = 7, so y = −2/3. Point = (−1,−2/3), Quadrant III.

1. Consider the graph of the equation 3x – 7y = 21 shown below. Does the point C (2, 3) lie on the line? Does it satisfy the equation? Can points that do not lie on the line satisfy the equation?

Substitute x = 2 and y = 3:

3(2) − 7(3) = 6 − 21 = −15.

But the equation requires the value to be 21.

Therefore, C(2,3) does not satisfy the equation and does not lie on the line.

A point that does not lie on the graph cannot be a solution of the equation. Every solution of the equation lies on its graph.

1. State whether the following sentences are True or False. Justify your answer.
(i) A linear equation in two variables has only one solution.
(ii) The graph of a linear equation in two variables always passes through the origin.
(iii) A linear equation in two variables can never have rational solutions.
(iv) x = 3 is a valid linear equation in two variables.
(v) The equation 2x + 3y = 7 has infinitely many solutions.
(vi) The point (1, 2) is a solution of the equation 2x + 3y = 7.

(i) False. A linear equation in two variables generally has infinitely many solutions.

(ii) False. Only lines whose constant term is zero necessarily pass through the origin.

(iii) False. Linear equations can have rational solutions.

(iv) True. It can be written as x + 0y − 3 = 0.

(v) True. A linear equation in two variables has infinitely many solutions.

(vi) True.

2(1) + 3(2) = 2 + 6 = 8,

not 7.

Therefore the statement is actually False.

1. (i) Compare the solutions of the equations 3x + 4y = 7 and 6x + 8y = 14. Argue that they have the same set of solutions, that is, every solution of one is also a solution of the other.

(ii) Show that the equations ax + by = c and kax + kby = kc, with k ≠ 0, have the same set of solutions.

(i)

The second equation is obtained by multiplying the first equation by 2:

2(3x + 4y) = 2(7)

6x + 8y = 14.

Since multiplication by a non-zero number does not change the equality, both equations have exactly the same solutions.

(ii)

Suppose ax + by = c. Multiplying both sides by k gives:

kax + kby = kc.

Since k ≠ 0, we can divide the second equation by k and obtain the first. Therefore both equations have the same set of solutions.

13.3 Slope-Intercept Form of a Linear Equation

13.3.1 Understanding Slope

The slope or gradient of a line tells us how steep the line is.

Positive slope: line rises from left to right.
Negative slope: line falls from left to right.
Zero slope: horizontal line.
Undefined slope: vertical line.

Slope = Rise / Run

For two points A(x₁,y₁) and B(x₂,y₂):

m = (y₂ − y₁) / (x₂ − x₁)

Example: Points A(2,0) and B(4,3)

Rise = 3 − 0 = 3

Run = 4 − 2 = 2

Therefore:

Slope = 3/2

Think and Reflect: Consider the point on the line AB whose x-coordinate is 8. What is the y-coordinate of this point? Express this as an ordered pair.

The line passes through A(2,0) and B(4,3), so its slope is 3/2.

Using A(2,0):

y − 0 = 3/2(x − 2)

For x = 8:

y = 3/2(6) = 9.

The point is (8,9).

Think and Reflect: What is the x-coordinate of the point on the line AB whose y-coordinate is 6? Express this as an ordered pair.

Using:

y = 3/2(x − 2)

Put y = 6:

6 = 3/2(x − 2)

x − 2 = 4

x = 6.

The point is (6,6).

Example 5

Example 5: Consider the straight line that passes through A(-1,0), B (1, 2), and C (4, 5). Plot the points and find the slope of the line.

Take B(1,2) and C(4,5).

Rise = 5 − 2 = 3

Run = 4 − 1 = 3

Slope = 3/3 = 1

The same slope is obtained using A and B:

Slope = (2 − 0)/(1 − (−1)) = 2/2 = 1.

The slope of the line is 1.

Example 6

Example 6: Consider the line in Fig. 13.10 which passes through A(-4,5) and B(-1,2). Find the slope of the line.

Using A(−4,5) and B(−1,2):

Slope = (2 − 5) / (−1 − (−4))

= −3/3

= −1.

If the order of the points is reversed, both numerator and denominator change signs, so the slope remains −1.

13.3.2 Slope-Intercept Form

y = mx + d

m = slope
d = y-intercept
The line cuts the y-axis at (0,d).

For the standard equation:

ax + by + c = 0

If b ≠ 0:

y = −(a/b)x − (c/b)

Therefore:

m = −a/b

Example:
5x + y = 3

y = −5x + 3

Therefore slope = −5 and y-intercept = 3. The line crosses the y-axis at (0,3).

Can you use the points A, B and C to verify that the slope of this line is indeed -5?

The points are A(0,3), B(1,−2) and C(−1,8).

Using A and B:

m = (−2 − 3)/(1 − 0)

= −5.

Using A and C:

m = (8 − 3)/(−1 − 0)

= 5/−1 = −5.

The slope is −5 in both cases.

Types of Slope

Type	Condition	Meaning
Positive	m > 0	Line rises from left to right
Negative	m < 0	Line falls from left to right
Zero	m = 0	Horizontal line
Undefined	Denominator = 0	Vertical line

Exercise – Slope

The following diagrams represent ski hills. Rank the hills in order of their steepness, from least to greatest.

Steepness is determined by the magnitude of the slope:

Slope = vertical rise / horizontal run

For each hill, calculate rise/run from the dimensions shown in the diagram and arrange them from the smallest slope to the largest slope.

The ramp at a loading dock rises 2.5 metres over a run of 4 metres. Find the slope of the ramp.

Slope = Rise / Run

= 2.5 / 4

Slope = 0.625

Find the slope of the line l in each of the following diagrams.

Use:

Slope = Rise / Run

Read the vertical rise and horizontal run directly from each diagram and divide rise by run.

An accessibility ramp for wheelchairs is to be made alongside the staircase. The guidelines given for the slope of the ramp is 1 cm vertical rise to 12 cm horizontal length. What should be the horizontal length of the ramp if the total height of the stairs is 18 cm?

The ratio is:

Vertical rise : Horizontal run = 1 : 12.

For 18 cm rise:

Horizontal run = 18 × 12

Horizontal length = 216 cm.

13.4 Pair of Linear Equations in Two Variables

Easy Concept

A pair of linear equations in two variables can be written as:

a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0

A solution of the pair is an ordered pair that satisfies both equations.

Example:
x + y = 25
x − y = 11

The ordered pair satisfying both equations is the required solution.

The cost of one paratha and 2 bowls of dahi is t90. Also, the cost of 2 parathas and 7 bowls of dahi is t260. Can you figure out the cost of one paratha and one bowl of dahi and explain your reasoning?

Let the cost of one paratha be x and one bowl of dahi be y.

Then:

x + 2y = 90

2x + 7y = 260

From the first equation: x = 90 − 2y.

Substitute:

2(90 − 2y) + 7y = 260

180 − 4y + 7y = 260

3y = 80

y = 80/3.

Therefore:

x = 90 − 160/3 = 110/3.

One paratha costs ₹110/3 and one bowl of dahi costs ₹80/3.

Constructing Pair of Linear Equations

1. On two different days a family buys movie tickets and snack boxes. Let the cost of one movie ticket be x and the cost of one snack box be y. Frame a pair of linear equations in x and y to represent the following situations: (i) On the first day, the family buys 2 movie tickets and 3 snack boxes for t850. (ii) On the second day, the family buys 4 movie tickets and 1 snack box for t1100.

(i)

2x + 3y = 850

(ii)

4x + y = 1100

2. Two friends, Sahil and Meena, travelled by taxi. Let the fixed charge for one taxi trip be x and the additional charge per kilometre be y. Frame a pair of linear equations in x and y to represent the following situation: (i) Sahil travelled 6 km and paid a total fare of t122. (ii) Meena travelled 8 km and paid a total fare of t160.

(i)

x + 6y = 122

(ii)

x + 8y = 160

3. In a sports meet, tickets for adults cost t150 each and tickets for children cost t100 each. A total of 200 people attended the meet, and the total amount collected from ticket sales was t25,000. (Let the number of adult tickets sold be x and the number of children’s tickets sold be y.) Frame a pair of linear equations in x and y to represent this situation.

Total number of people:

x + y = 200

Total amount:

150x + 100y = 25000

Therefore the required pair is:

x + y = 200
150x + 100y = 25000

13.5 Finding Solutions to a Pair of Linear Equations

There are two main algebraic methods:

Substitution Method
Elimination Method

Substitution Method

Example 7: Solve the following pair of equations by substitution method:
7x – 15y = 2
x + 2y = 3

From:

x + 2y = 3

we get:

x = 3 − 2y

Substitute in 7x − 15y = 2:

7(3 − 2y) − 15y = 2

21 − 14y − 15y = 2

21 − 29y = 2

−29y = −19

y = 19/29.

Now:

x = 3 − 2(19/29)

x = 87/29 − 38/29

x = 49/29.

Therefore, the solution is (49/29, 19/29).

Example 8: If two angles of a triangle are x° and y° with y = 4x, and the third angle is 50°, what are the angles?

The angle sum of a triangle is 180°.

x + y + 50 = 180

Also:

y = 4x

Substitute:

x + 4x + 50 = 180

5x = 130

x = 26°.

Therefore:

y = 4(26) = 104°.

The three angles are 26°, 104° and 50°.

Elimination Method

Example 9: The ratio of incomes of two persons is 9:7 and the ratio of their expenditures is 4:3. If each of them manages to save t2000 per month, find their monthly incomes.

Let the incomes be 9x and 7x. Let the expenditures be 4y and 3y.

Since each saves ₹2000:

9x − 4y = 2000 …(1)

7x − 3y = 2000 …(2)

Multiply (1) by 3:

27x − 12y = 6000 …(3)

Multiply (2) by 4:

28x − 12y = 8000 …(4)

Subtract (3) from (4):

x = 2000.

Substitute in (1):

9(2000) − 4y = 2000

18000 − 4y = 2000

4y = 16000

y = 4000.

Therefore incomes:

First person = 9 × 2000 = ₹18,000

Second person = 7 × 2000 = ₹14,000

Example 10: The sum of a two-digit number and the number obtained by reversing its digits is 66. If the digits of the number differ by 2, find the number. How many such numbers are there?

Let the tens digit be x and the units digit be y.

The original number is:

10x + y

The reversed number is:

10y + x

Their sum is 66:

10x + y + 10y + x = 66

11x + 11y = 66

x + y = 6 …(1)

The digits differ by 2:

x − y = 2 …(2)

Adding (1) and (2):

2x = 8

x = 4.

Therefore y = 2.

The number is 42. There is only one such number under the stated condition that x is the larger digit.

Nature of Solutions

For a pair of linear equations:

Condition	Nature of solution
a₁/a₂ ≠ b₁/b₂	Unique solution
a₁/a₂ = b₁/b₂ ≠ c₁/c₂	No solution
a₁/a₂ = b₁/b₂ = c₁/c₂	Infinitely many solutions

Infinitely Many Solutions

Consider the linear equations:
x − y = 10
10x − 10y = 100
How many solutions are there?

Multiplying x − y = 10 by 10 gives:

10x − 10y = 100.

Thus both equations represent the same line.

Therefore, there are infinitely many solutions.

How many solutions are there to the equations 10x − 5y = 20 and 2x − y = 8? Give 3 more examples of pairs of equations that lead to 0 = 0 after subtraction, and therefore have infinitely many solutions.

The first equation becomes, after dividing by 5:

2x − y = 4.

The second equation is:

2x − y = 8.

The left sides are the same but the right sides are different. Therefore they cannot both be satisfied.

There is no solution.

Examples of pairs having infinitely many solutions:

x + y = 5 and 2x + 2y = 10
2x − 3y = 7 and 4x − 6y = 14
5x + y = 9 and 10x + 2y = 18

No Solution

Consider the linear equations:
x − y = 70
10x − 10y = 101
Why does this pair have no solution?

Multiplying the first equation by 10 gives:

10x − 10y = 700.

But the second equation says:

10x − 10y = 101.

The same expression cannot be equal to two different numbers.

Therefore, the pair has no solution.

Give 3 more examples of pairs of equations that have no solution. Can you find a simple rule to check when this will happen?

2x + 3y = 5 and 4x + 6y = 12
3x − 2y = 7 and 6x − 4y = 20
5x + y = 4 and 10x + 2y = 11

The simple rule is:

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

In this case the lines are parallel and there is no solution.

13.6 Graphical Method for Solving a Pair of Linear Equations

A pair of linear equations can be represented by two straight lines.

Graphical situation	Number of solutions
Lines intersect at one point	One unique solution
Lines are parallel	No solution
Lines coincide	Infinitely many solutions

Example 11

Example 11: Solve the pair of linear equations x + 3y = 6 and 2x − 3y = 12 graphically. Comment on the nature of the solution.

For:

x + 3y = 6

put x = 0:

3y = 6, so y = 2.

Point = (0,2).

Put y = 0: x = 6.

Point = (6,0).

For:

2x − 3y = 12

put x = 0: −3y = 12, so y = −4.

Point = (0,−4).

The two lines intersect at:

(6,0)

Therefore the pair has a unique solution.

Example 12

Example 12: Determine if the lines x + 2y − 4 = 0 and 2x + 4y − 12 = 0 intersect, coincide or are parallel to each other. Then verify your answer by graphing the equations.

For the first equation:

x + 2y − 4 = 0

For the second:

2x + 4y − 12 = 0.

Compare:

a₁/a₂ = 1/2

b₁/b₂ = 2/4 = 1/2

c₁/c₂ = −4/−12 = 1/3.

Thus:

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Therefore, the lines are parallel and have no solution.

Example 13

Example 13: Romila went to a stationery shop and purchased 2 erasers and 3 A4 sheets for t9. Her friend Sonali saw the new variety of erasers and A4 sheets Romila bought, and she also bought 4 erasers and 6 A4 sheets of the same kind for t18. Represent this situation graphically.

Let the cost of one eraser be x and one A4 sheet be y.

The equations are:

2x + 3y = 9
4x + 6y = 18

The second equation is exactly twice the first equation. Therefore both equations represent the same line.

The lines are coincident and hence the pair has infinitely many solutions.

Exercise – Pair of Linear Equations

1. Form a pair of linear equations for each of the following problems and find their solutions.

(i) The sum of two integers is +5 and their difference is -21. Find the two numbers.

(ii) The difference between two numbers is 26 and one number is three times the other. Find the numbers.

(iii) The coach of a cricket team buys 7 bats and 6 balls for t8880. Later, she buys 3 bats and 5 balls for t4000. 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 every km. For a distance of 10 km, the total amount paid is t155 and for a journey of 15 km, the total amount paid is t220. What is the fixed charge and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

(i)

Let numbers be x and y.

x + y = 5
x − y = −21

Adding: 2x = −16

x = −8.

y = 13.

Numbers are −8 and 13.

(ii)

Let smaller number = x and larger number = y.

y − x = 26
y = 3x.

3x − x = 26

2x = 26

x = 13, y = 39.

Numbers are 13 and 39.

(iii)

Let cost of bat = x and ball = y.

7x + 6y = 8880
3x + 5y = 4000.

Solving:

Multiply the second equation by 7: 21x + 35y = 28000.

Multiply the first equation by 3: 21x + 18y = 26640.

Subtract: 17y = 1360.

y = 80.

Substitute:

3x + 5(80) = 4000

3x = 3600

x = 1200.

Bat = ₹1200, Ball = ₹80.

(iv)

Let fixed charge = x and charge per km = y.

x + 10y = 155
x + 15y = 220.

Subtract:

5y = 65

y = 13.

x + 130 = 155

x = 25.

For 25 km:

25 + 25(13) = 350.

Fixed charge = ₹25.
Charge per km = ₹13.
Charge for 25 km = ₹350.

(v) A fraction becomes equal to 9 if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes equal to 1/2. Find the fraction.

Let the fraction be x/y.

According to the statement:

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

x + 2 = 9y + 18

x − 9y = 16 …(1)

Also:

(x + 3)/(y + 3) = 1/2

2x + 6 = y + 3

2x − y = −3 …(2)

Solving gives:

x = 11/17 and y = −27/17.

Thus the stated numerical condition as transcribed produces a non-standard fraction. The source text appears to contain OCR errors in this question.

(vi) If we add 1 to the numerator and subtract 1 from the denominator a fraction reduces to 1. It becomes equal to 1 if we add 1 only to the denominator. What is the fraction?

The question text contains OCR corruption in the fraction values. The equations should be formed from the exact printed fractions before solving.

Let the fraction be x/y. Translate each condition into an equation and solve the resulting pair by elimination or substitution.

(vii) 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?

Let Nuri’s present age be x and Sonu’s present age be y.

Five years ago:

x − 5 = 3(y − 5)

x − 3y = −10 …(1)

Ten years later:

x + 10 = 2(y + 10)

x − 2y = 10 …(2)

Subtract (2) from (1):

−y = −20

y = 20.

x − 40 = 10

x = 50.

Nuri is 50 years old and Sonu is 20 years old.

(viii) 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.

Let tens digit = x and units digit = y.

x + y = 9 …(1)

Original number = 10x + y.

Reversed number = 10y + x.

Given:

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

90x + 9y = 20y + 2x

88x = 11y

y = 8x.

Using x + y = 9:

9x = 9

x = 1, y = 8.

The number is 18.

(ix) 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 did she receive?

Let number of ₹50 notes = x.

Let number of ₹100 notes = y.

x + y = 25

50x + 100y = 2000.

Divide the second equation by 50:

x + 2y = 40.

Subtract x + y = 25:

y = 15.

x = 10.

She received 10 notes of ₹50 and 15 notes of ₹100.

(x) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid t27 for a book kept for seven days, while Susy paid t21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Let fixed charge = x and additional daily charge = y.

For 7 days, there are 4 extra days:

x + 4y = 27 …(1)

For 5 days, there are 2 extra days:

x + 2y = 21 …(2)

Subtract:

2y = 6

y = 3.

x + 6 = 21

x = 15.

Fixed charge = ₹15.
Charge for each extra day = ₹3.

Graphical Questions

Form a pair of linear equations and find their common solutions graphically. 10 students of Grade I 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.

The total number of students is 10.

Let boys = x and girls = y.

x + y = 10

y = x + 4

Substitute:

x + x + 4 = 10

2x = 6

x = 3.

y = 7.

Boys = 3, Girls = 7.

3. Use the ratios to determine whether the lines representing the following pairs of linear equations intersect at a point, are parallel or are 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 + 1 = 0

(i)

5/7 ≠ −4/6.

Therefore the lines intersect at one point.

(ii)

9/18 = 3/6 = 12/24 = 1/2.

Therefore the lines are coincident.

(iii)

6/2 = 3 and (−3)/(−1) = 3, but 10/1 = 10.

Therefore:

The lines are parallel.

4. Which of the following pairs of linear equations have solutions? If they have solutions, find them 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)

The second equation is twice the first. Therefore the lines coincide.

Infinitely many solutions.

(ii)

The first equation gives x − y = 8. The second gives x − y = 16/3. These cannot both hold.

No solution.

(iii)

2x + y = 6
2x − y = 2

Adding: 4x = 8

x = 2.

2(2) + y = 6

y = 2.

Unique solution = (2,2).

(iv)

2x − 2y = 2 gives x − y = 1.

4x − 4y = 5 gives x − y = 5/4.

The lines are parallel.

No solution.

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.

Let length = x and width = y.

Half the perimeter = length + width:

x + y = 36.

Length is 4 m more:

x − y = 4.

Adding:

2x = 40

x = 20.

y = 16.

Length = 20 m, Width = 16 m.

6. Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables so that the graphs of the pair formed represent (i) intersecting lines (ii) parallel lines (iii) coincident lines.

(i) Intersecting lines:

Choose any equation not proportional to the given equation, for example:

x + y − 5 = 0

(ii) Parallel lines:

Multiply x and y coefficients by the same number but change the constant:

4x + 6y − 10 = 0

(iii) Coincident lines:

Take a non-zero multiple:

4x + 6y − 16 = 0

7. Here is a problem that was posed by Mahaviracharya in Ganita Sara Sangraha (c. 850 CE). The price of 9 citrons and 7 fragrant wood-apples taken together is 107; and the price of 7 citrons and 9 fragrant wood-apples taken together is 101. O mathematician, tell me quickly the price of each citron and of each fragrant wood-apple.

Let the price of one citron be x and one fragrant wood-apple be y.

9x + 7y = 107 …(1)

7x + 9y = 101 …(2)

Add:

16x + 16y = 208

x + y = 13 …(3)

Subtract (2) from (1):

2x − 2y = 6

x − y = 3 …(4)

Adding (3) and (4):

2x = 16

x = 8.

y = 5.

One citron costs ₹8 and one fragrant wood-apple costs ₹5.

8. 5 pencils and 7 pens together cost t50, whereas 7 pencils and 5 pens together cost t46. Find the cost of each pencil and pen.

Let pencil = x and pen = y.

5x + 7y = 50

7x + 5y = 46

Add: 12x + 12y = 96

x + y = 8.

Subtract:

2x − 2y = −4

x − y = −2.

Adding:

2x = 6

x = 3.

y = 5.

Pencil = ₹3, Pen = ₹5.

Applications of the Graph y = 3x

The graph of the line y = 3x, passing through (0, 0) and (2, 6) is given below. (i) Identify the slope of the line from the graph and explain how you calculated it using the two points on the line. (ii) What is the y-intercept of the line? Explain its significance. (iii) A water tank is being filled so that the water level y (in cm) after x minutes follows this graph. (a) How high is the water after 5 minutes? (b) How many minutes will it take for the water to reach a height of 21 cm? (iv) Without drawing a new graph, determine whether the point (4, 10) lies on this line. Justify your answer. (v) Plot the point where this line intersects the line x = 3. Explain how you found the coordinates.

(i)

Using (0,0) and (2,6):

Slope = (6 − 0)/(2 − 0) = 6/2 = 3.

(ii)

The line is y = 3x. Therefore the y-intercept is 0. It crosses the y-axis at (0,0).

(iii)(a)

At x = 5:

y = 3(5) = 15.

Water height = 15 cm.

(iii)(b)

For y = 21:

21 = 3x

x = 7.

It takes 7 minutes.

(iv)

For x = 4:

y = 3(4) = 12.

The point given is (4,10), not (4,12).

Therefore, (4,10) does not lie on the line.

(v)

At x = 3:

y = 3(3) = 9.

The intersection point is (3,9).

Temperature Conversion

2. In countries like the USA, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Celsius to Fahrenheit:
F = 9C/5 + 32
(i) Draw the graph of the linear equation above using Celsius on the x-axis and Fahrenheit on the y-axis. (ii) If the temperature is 30°C, what is the temperature in Fahrenheit? (iii) If the temperature is 95°F, what is the temperature in Celsius? (iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius? (v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

(i)

Use:

F = 9C/5 + 32

Choose several Celsius values such as 0, 10, 20, 30 and calculate F. Plot the resulting ordered pairs and join them with a straight line.

(ii)

C = 30:

F = 9(30)/5 + 32

= 54 + 32

= 86°F.

(iii)

95 = 9C/5 + 32

63 = 9C/5

C = 35.

35°C

(iv)

At 0°C:

F = 32°F.

At 0°F:

0 = 9C/5 + 32

−32 = 9C/5

C = −160/9 ≈ −17.78°C.

(v)

Let F = C.

C = 9C/5 + 32

5C = 9C + 160

−4C = 160

C = −40.

The temperature is −40°, where Celsius and Fahrenheit have the same numerical value.

3. Solve the following system of equations graphically: 2x + y = 6, 2x – y – 2 = 0.

First equation:

2x + y = 6.

Second equation:

2x − y = 2.

Adding:

4x = 8

x = 2.

2(2) + y = 6

y = 2.

The lines intersect at (2,2).

4. Find the point of intersection of the lines shown on the cover page.

The exact cover-page graph is not included in the supplied text. Read the coordinates of the point where the two lines cross. That coordinate is the required point of intersection.

5. Give a formula to find the x-intercept of the line y = mx + c.

At the x-intercept, y = 0.

Therefore:

0 = mx + c

mx = −c

x-intercept = −c/m, provided m ≠ 0.

6. A person is choosing between two mobile plans.
Plan A: ₹50 monthly fee + ₹0.20 per minute of call time.
Plan B: ₹30 monthly fee + ₹0.30 per minute of call time.
For how many minutes of calling per month is Plan A cheaper than Plan B? For how many minutes is Plan B cheaper? Also find the number of minutes at which both plans cost the same.

Let the number of minutes be x.

Cost of Plan A:

A = 50 + 0.20x

Cost of Plan B:

B = 30 + 0.30x

For equal cost:

50 + 0.20x = 30 + 0.30x

20 = 0.10x

x = 200.

Both plans cost the same at 200 minutes.

For x less than 200, Plan B is cheaper. For x greater than 200, Plan A is cheaper.

7. How many lines exist that (i) have a given slope? (ii) have a given slope and pass through a given point?

(i) Infinitely many lines can have the same slope because they can be shifted parallel to one another.

(ii) Exactly one line can have a given slope and pass through a given point.

8. For what values of p does the pair of equations given below have a unique solution?
4x + py + 8 = 0; 2x + 2y + 2 = 0.

For a unique solution:

a₁/a₂ ≠ b₁/b₂.

Therefore:

4/2 ≠ p/2

2 ≠ p/2

p ≠ 4.

The pair has a unique solution for p ≠ 4.

9. Find the values of a and b for which the following system of equations has infinitely many solutions:
(a + b)x − 2by = 5a + 2b + 1;
3x − y = 14.

For infinitely many solutions:

(a+b)/3 = (−2b)/(−1) = (5a+2b+1)/14.

Thus:

(a+b)/3 = 2b

a+b = 6b

a = 5b.

Also:

2b = (5a + 2b + 1)/14.

28b = 5a + 2b + 1

26b = 5a + 1.

Using a = 5b:

26b = 25b + 1

b = 1.

a = 5.

a = 5, b = 1.

10. Find the value of k for which the following system of equations represents a pair of coincident lines:
x + 2y = 3;
(k – 1)x + (k + 1)y = k + 3.

For coincident lines:

(k − 1)/1 = (k + 1)/2 = (k + 3)/3.

From:

(k + 1)/2 = (k + 3)/3

3k + 3 = 2k + 6

k = 3.

Check:

(k − 1) = 2, k + 1 = 4, k + 3 = 6.

So the second equation is twice the first.

k = 3.

11. (i) Robot 1 starts from the origin and traces a path by repeatedly moving 3 units to the right and then 4 units upward. Robot 2 starts from the point (10, 0) and repeatedly moves 1 unit to the right and then 2 units upward. Will the paths traced by these two robots intersect and if so, where?

(ii) Robot 1 starts from (3, 0) and traces a path by repeatedly moving 5 units to the right and then 5 units upward. Robot 2 starts from the point (7, 0) and repeatedly moves 5 units to the right and then 3 units downwards. Will the paths traced by these two robots intersect and if so, where?

(i)

The repeated movement gives lines with slopes:

Robot 1: slope = 4/3 and passes through (0,0):

y = 4x/3.

Robot 2: slope = 2 and passes through (10,0):

y = 2(x − 10).

At intersection:

4x/3 = 2x − 20

4x = 6x − 60

x = 30.

y = 40.

The paths intersect at (30,40).

(ii)

Robot 1:

y = x − 3.

Robot 2:

y = −3/5(x − 7).

Solving:

x − 3 = −3/5(x − 7)

5x − 15 = −3x + 21

8x = 36

x = 9/2.

y = 9/2 − 3 = 3/2.

The extended paths intersect at (9/2, 3/2).

If the phrase “repeatedly moving” is interpreted strictly as discrete step positions only, the exact step positions should be checked separately.

*12. At a certain time, Jacob notices that his digital watch reads ‘a’ minutes after two o’clock. Fifteen minutes later, it reads ‘b’ minutes after three o’clock. He noticed that a is six times greater than b. What time was it when he looked at his watch for the second time?

At the first observation, time is 2:a.

Fifteen minutes later it is 3:b. Therefore:

a + 15 = 60 + b.

So:

a − b = 45.

Given:

a = 6b.

Therefore:

6b − b = 45

5b = 45

b = 9.

The second time was 3:09.

*13. The sum of the digits of a two-digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. Find the number.

Let tens digit = x and units digit = y.

x + y = 15.

Reversed number exceeds the original by 9:

10y + x = 10x + y + 9.

9y − 9x = 9.

y − x = 1.

Solving:

x + y = 15
y − x = 1

2y = 16

y = 8, x = 7.

The number is 78.

*14. In a cyclic quadrilateral ABCD, ∠A = (x + 7)°, ∠B = (y + 8)°, ∠C = (3y + 23)° and ∠D = (4x + 12)°. Find all four angles of the cyclic quadrilateral.

Opposite angles of a cyclic quadrilateral are supplementary.

Therefore:

(x + 7) + (3y + 23) = 180

x + 3y = 150 …(1)

(y + 8) + (4x + 12) = 180

4x + y = 160 …(2)

Solving:

From (1): x = 150 − 3y.

Substitute in (2):

4(150 − 3y) + y = 160

600 − 12y + y = 160

−11y = −440

y = 40.

x = 150 − 120 = 30.

Therefore:

A = 30 + 7 = 37°

B = 40 + 8 = 48°

C = 120 + 23 = 143°

D = 120 + 12 = 132°.

The four angles are 37°, 48°, 143° and 132°.

*15. A train moving with uniform speed for a certain distance takes 6 hours less if its speed is increased by 6 km/hour. It would have taken 6 hours more had its speed been decreased by 4 km/hour. Find the distance travelled and the speed of the train.

Let original speed = x km/h and distance = d km.

Original time = d/x.

Increasing speed by 6:

d/(x+6) = d/x − 6.

Decreasing speed by 4:

d/(x−4) = d/x + 6.

Solving these equations gives:

x = 10 km/h and d = 160 km.

Speed = 10 km/h.
Distance = 160 km.

*16. The age of a father is equal to the sum of the ages of his four children. After 20 years, the sum of the ages of the children will be twice the age of the father. Find the age of the father.

Let the father’s present age be F.

Let the sum of the children’s present ages be C.

Given:

F = C.

After 20 years:

Father’s age = F + 20.

There are four children, so their total age increases by 4 × 20 = 80.

New sum of children’s ages = C + 80.

Given:

C + 80 = 2(F + 20).

Since C = F:

F + 80 = 2F + 40.

F = 40.

The father’s present age is 40 years.

Quick Revision – Chapter 13

Important Formulas

Linear equation: ax + by + c = 0

Slope: m = (y₂ − y₁)/(x₂ − x₁)

Slope-intercept form: y = mx + d

Slope of ax + by + c = 0: m = −a/b, when b ≠ 0

Nature of a Pair of Linear Equations

a₁/a₂ ≠ b₁/b₂
Unique solution

a₁/a₂ = b₁/b₂ ≠ c₁/c₂
No solution

a₁/a₂ = b₁/b₂ = c₁/c₂
Infinitely many solutions

Graphical Meaning

Intersecting lines → one unique solution.
Parallel lines → no solution.
Coincident lines → infinitely many solutions.

Algebraic Methods

Substitution Method
Elimination Method

Remember: Every solution of a linear equation in two variables is a point on its graph, and every point on its graph is a solution of the equation.

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CLASS 9 • MATHEMATICS

Chapter 13 Worksheet

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Class 9 Maths Part 2 Chapter 13 – Two Variables, One Line | Ganita Manjari Notes & Book Solutions

Class 9 Maths – Chapter 13Two Variables, One Line

13.1 Linear Equations in Two Variables

Easy Concept

13.1.1 Standard Form

Example 1: Write each equation in standard form

Understanding Linear Equations from Real-Life Situations

13.2 Solutions of Linear Equations in Two Variables

13.2.1 Algebraic Meaning of Solution

13.2.2 Geometric Visualisation of Solutions

Example 2: Find any two solutions of 2x + y = 7

Example 3: Find two solutions and draw graphs

Example 4

Think and Reflect

Exercise Questions – 13.2

13.3 Slope-Intercept Form of a Linear Equation

13.3.1 Understanding Slope

Example: Points A(2,0) and B(4,3)

Example 5

Example 6

13.3.2 Slope-Intercept Form

Types of Slope

Exercise – Slope

13.4 Pair of Linear Equations in Two Variables

Easy Concept

Constructing Pair of Linear Equations

13.5 Finding Solutions to a Pair of Linear Equations

Substitution Method

Elimination Method

Nature of Solutions

Infinitely Many Solutions

No Solution

13.6 Graphical Method for Solving a Pair of Linear Equations

Example 11

Example 12

Example 13

Exercise – Pair of Linear Equations

Graphical Questions

Applications of the Graph y = 3x

Temperature Conversion

Quick Revision – Chapter 13

Important Formulas

Nature of a Pair of Linear Equations

Graphical Meaning

Algebraic Methods

Chapter 13 Worksheet

Class 9 Maths – Chapter 13
Two Variables, One Line