NCERT Solutions Class 9 Maths Chapter 13 Two Variables One Line

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 9 · M AT H S

NCERT Solutions + ×
Chapter 13: Two
Variables, One Line

NCERT Textbook — Ganita Manjari

BOOK PAGES SECTIONS QUESTIONS MEDIUM

Part II, 85 – 122 21 72 English

Every question Step by step Exam ready
Activities, in-text questions The working shown, not Written to match how
and all exercises. just the final answer. marks are awarded.

Solutions, notes, sample papers & more at 77 pages

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

CLASS 9 · MATHS · GANITA MANJARI

NCERT Solutions — Chapter 13: Two Variables, One Line
Chapter 13 takes linear equations to two variables. You will write equations like 60x + 50y = 280 in the
standard form ax + by + c = 0, see that such an equation has infinitely many solutions which together form a
straight line, and measure how steep that line is with its slope (rise ÷ run). Then you will solve a pair of linear
equations by substitution, by elimination and by graphs, and use the ratios a1/a2, b1/b2, c1/c2 to tell in
advance whether a pair has one solution, none or infinitely many.

TEXTBOOK BOOK PAGES
Ganita Manjari (Class 9) Part II, 85 – 122

SECTIONS QUESTIONS
21 72

MEDIUM
English

Exercise Set 13.1 — Page 87
Section 13.1 Linear Equations in Two Variables

Q1 Write a linear equation in two variables in which a = 3, b = 0 and c = −1/5.

Put the values into the standard form ax + by + c = 0:

3x + 0·y + (−1/5) = 0
i.e. 3x + 0·y − 1/5 = 0

So the equation is 3x + 0·y − 1/5 = 0 (which is the same as 3x − 1/5 = 0). It is still an equation in
two variables, because the y-term is present with coefficient 0.

Why: the book treats an equation such as ax + c = 0 as a special case of a
linear equation in two variables, written as ax + 0×y + c = 0. Only a and b must
not both be zero; here a = 3 ≠ 0, so the equation is allowed.

Page 1 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Tip: multiplying by 5 removes the fraction: 15x + 0·y − 1 = 0. This has exactly
the same solutions, but its coefficients are now 15, 0 and −1, so to show a = 3, b
= 0, c = −1/5 keep the first form.

Check it yourself: every point with x = 1/15 satisfies the equation, whatever y is
— for example (1/15, 0) and (1/15, 7). Its graph is a vertical line.

Q2 Complete the following table after expressing the given linear equations in standard
form.

LINEAR STANDARD Coefficient Coefficient CONSTANT
EQUATION IN FORM of x of y TERM
TWO VARIABLES

y − 15 = √2x

3y − 2x = 0

5x = 3y

x=8

3y = 1

Bring every term to the left-hand side so that the right-hand side is 0, then read off a, b and c.

LINEAR STANDARD Coefficient of Coefficient of CONSTANT
EQUATION FORM x y TERM

y − 15 = √2x √2x − y + 15 = 0 √2 −1 15

3y − 2x = 0 −2x + 3y + 0 = 0 −2 3 0

5x = 3y 5x − 3y + 0 = 0 5 −3 0

x=8 x + 0·y − 8 = 0 1 0 −8

3y = 1 0·x + 3y − 1 = 0 0 3 −1

A missing variable has coefficient 0, and a missing number term means the constant is 0.

Page 2 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Why: in ax + by + c = 0, a and b are the numbers multiplying x and y, and c is
the number standing alone. For y − 15 = √2x, subtract y and add 15 on both
sides: 0 = √2x − y + 15.

Tip: the standard form is not unique. Multiplying the whole equation by −1 is
also correct, e.g. 2x − 3y = 0 for the second row (then a = 2, b = −3). Just make
sure the coefficients you write match the form you chose.

Common mistake: writing the constant of x = 8 as +8. Moving 8 to the left gives
x − 8 = 0, so the constant is −8.

Q3 (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?
(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?

(i) Charlie’s equation t = 2p is correct. The notebook is the costlier item, so its cost t equals
two times the pen’s cost p. Meera’s p = 2t says the pen costs twice the notebook, which is the
opposite.
(ii) Yes. The runs of the two batsmen add up to the total, so x + y = 176 is correct. In standard
form it is x + y − 176 = 0.

Check it yourself: test (i) with numbers. If a pen costs ₹10, a notebook costs
₹20. Charlie: 20 = 2 × 10 ✔. Meera: 10 = 2 × 20 ✘.

Page 3 of 77

as e
Class 9 Maths Chapter 13 Two Variables, One Line
ag l AglaSem · NCERT Solutions

Why: in (ii), only some solutions make sense: runs are whole numbers and
. com
m
cannot be negative, so x and y must be whole numbers from 0 to 176,
a s emsuch as
. agl but not the
coor (88, 88). A solution like (180, −4) satisfies the equation
m
(100, 76)
ase
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situation.

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Think and Reflect — Page 88agla
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Section 13.2.1 Algebraic Meaning of Solution

co m
m.
THINK AND REFLECT

m l ase
Q1 em
.co ag 3x + 2y = 12?
las
Are (1, 9/2), (0, 6), (4, 1) and (1/2, 21/4) solutions of the equation

ag
Substitute each pair in the left-hand side 3xco+m a s
. agl
2y and compare with 12.

a s em
l
(1, 9/2): 3(1) + 2(9/2) = 3 + 9a=g12 ✔

(0, 6): 3(0) + 2(6) = 12 ✔
com
m .
(4, 1): 3(4) + 2(1) = 12 + 2 = 14 ≠ 12 ✘
m 3(1/2) + 2(21/4) = 3/2 + 21/2 = 24/2 = 12 ✔ as e
. co21/4): agl
em
(1/2,
a s
agl (1, 9/2), (0, 6) and (1/2, 21/4) are solutions; (4, 1) is not.
se m
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Why: a pair (u, v) is a solution only if putting x = u and y = v makes both sides
equal. For (4, 1) the left side is 14, not 12. m co
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g them up is the most
.c Tip: the first number of an ordered pair is always x.aMixing
a s em
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common error: (6, 0) is not a solution, since 3(6) + 2(0) = 18.
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m a
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m as e
.co

ag l Page 4 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Q2 Can you find any other solution? How many solutions can you find?

Yes, many more. Choose any value of x, put it in the equation and solve for y:

x = 2: 6 + 2y = 12 ⇒ y = 3, so (2, 3)

x = 4: 12 + 2y = 12 ⇒ y = 0, so (4, 0)
x = −2: −6 + 2y = 12 ⇒ y = 9, so (−2, 9)
x = 6: 18 + 2y = 12 ⇒ y = −3, so (6, −3)

Every value of x gives one value of y, and there are infinitely many values of x. So the equation
has infinitely many solutions.

Why: after substituting x = u, the equation becomes 2y = 12 − 3u, a linear
equation in one variable, which always has exactly one answer y = (12 − 3u)/2.
Since u can be any real number, the list of solutions never ends.

Did you know? all these solutions lie on one straight line, the line y = −(3/2)x +
6 drawn in Fig. 13.1. That is the idea behind the chapter title, Two Variables, One
Line.

Think and Reflect — Page 92

Page 5 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Section 13.2.2 Geometric Visualisation of Solutions
THINK AND REFLECT

Q1 Four friends — Ranju, Meena, Farhan, and Toshi — are solving problems.
Ranju: I noticed something! If c = 0 in the standard form of a line ax + by + c = 0, then
the line must pass through the origin. Look, if I substitute x = 0 and y = 0, in equation
ax + by = 0, the equation is satisfied. So, the origin lies on the line! We can also say
that the line passes through the origin.
Farhan: Let us try for the equation 2x + 3y = 0. Here a = 2, b = 3 but c = 0. If we
substitute x = 0, then we get 3y = 0 or y = 0. This means (0, 0) lies on the line. So yes,
this line passes through the origin.
Toshi: Suppose our equation has b = c = 0, say, 5x = 0. This becomes x = 0 which is
the equation of the y-axis. And the y-axis passes through the origin.
Meena: And if we take a = c = 0, say, 7y = 0, that means y = 0, which is the equation of
the x-axis that also passes through the origin.
So, they conclude: 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, I agree. If c = 0 the equation is ax + by = 0. Putting x = 0, y = 0:

a(0) + b(0) = 0 + 0 = 0 = RHS

This is true for every choice of a and b, so (0, 0) is always a solution and the line always passes
through the origin. Toshi’s and Meena’s cases (the y-axis and the x-axis) are just the special
cases b = 0 and a = 0.

Why: a point lies on the graph exactly when its coordinates satisfy the equation.
Ranju’s substitution works for any a and b, so the conclusion holds in general,
not only for the examples tried.

Did you know? the converse is also true: if a line ax + by + c = 0 passes
through the origin, then a(0) + b(0) + c = 0, so c must be 0. So “c = 0” and “the
line passes through (0, 0)” mean the same thing.

Page 6 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Tip: this gives a quick check: 4x − 9y = 0 passes through the origin, while 4x −
9y = 1 does not.

Exercise Set 13.2 — Pages 93–94
Section 13.2 Solution of Linear Equation in Two Variables

Q1 Verify if the ordered pair (4, 3) is a solution of 5x − 6y = 2. Explain your reasoning.

Put x = 4 and y = 3 in the left-hand side:

5x − 6y = 5(4) − 6(3) = 20 − 18 = 2 = RHS

The two sides are equal, so (4, 3) is a solution of 5x − 6y = 2. Equivalently, the point (4, 3) lies
on the graph of this line.

Why: an ordered pair is a solution exactly when substituting its x-value and y-
value makes LHS = RHS. Here both sides are 2.

Common mistake: swapping the order. (3, 4) gives 15 − 24 = −9, which is not 2,
so (3, 4) is not a solution. The first number always goes in place of x.

Q2 Find any two solutions for each of the following equations:
(i) 7x − 3y = 21 (ii) 2x + 3y = 5

(i) 7x − 3y = 21

Page 7 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

x = 0: −3y = 21 ⇒ y = −7 → (0, −7)

y = 0: 7x = 21 ⇒ x = 3 → (3, 0)

(ii) 2x + 3y = 5

x = 1: 2 + 3y = 5 ⇒ y = 1 → (1, 1)

x = 4: 8 + 3y = 5 ⇒ y = −1 → (4, −1)

Two solutions: (i) (0, −7) and (3, 0); (ii) (1, 1) and (4, −1).

Check it yourself: (i) 7(3) − 3(0) = 21 ✔ and 7(0) − 3(−7) = 21 ✔. (ii) 2 + 3 = 5 ✔
and 8 − 3 = 5 ✔.

Tip: for (ii), putting x = 0 gives y = 5/3 and putting y = 0 gives x = 5/2. These,
(0, 5/3) and (5/2, 0), are correct too, but choosing an x that makes 5 − 2x
divisible by 3 (such as x = 1, 4, −2) gives whole-number answers.

Why: every equation of this kind has infinitely many solutions, so any two
correct pairs are acceptable. Other answers for (i): (6, 7), (−3, −14).

Q3 In the equations given below, m and n are unknown constants: 2mx + 3y = 7; 4x + ny =
−10. If (2, −1) is the solution of both equations, find the values of m and n.

Substitute x = 2, y = −1 in each equation.

Page 8 of 77

as e
Class 9 Maths Chapter 13 Two Variables, One Line
ag l AglaSem · NCERT Solutions

co m
e m.
First: 2m(2) + 3(−1) = 7
m l as
.co
4m − 3 = 7 ⇒ 4m = 10 ⇒ m = 5/2
m ag
la se
agSecond: 4(2) + n(−1) = −10

com
ag
8 − n = −10 ⇒ n = 18
m .
la se
m = 5/2 (= 2.5) and n = 18. ag

co m
m.

om l ase
m .c it yourself: with m = 5/2 the first equation is 5x + 3yag= 7: 10 − 3 = 7 ✔.
Check
e
ag lasWith n = 18 the second is 4x + 18y = −10: 8 − 18 = −10 ✔.

omsatisfy each equation. Each equation a s
c agl
Why: since (2, −1) is a solution, it must
then has only one unknown lefte(m .
m or n), which we solve just as in Example 4.
glas
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Find two solutions which lie in different quadrants for each of sthe
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se (i) 5x + 3y = 7 (ii) 5x − 3y = 7 (iii) −5x + 3y = 7 (iv) −5x − 3y = 7
la
ag Verify your solutions by representing the linear equations on a graph paper.
se m
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m . a
l a se y values, then check signs: I (+, +), II (−, +), III (−, −),
Choose x values that give whole-number
IV (+, −). ag

co m QUADRANT
.
EQUATION SOLUTION 1 QUADRANT SOLUTION 2
se m
m la
.co ag
5x + 3y = 7 (2, −1) IV (−1, 4) II

a s em 5x − 3y = 7
agl
(2, 1) I (−1, −4) III

.c
−5x + 3y = 7 (1, 4) I (−2, −1) III
s e m
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−5x − 3y = 7 (−2, 1) (1, −4) IV

la
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Each pair satisfies its equation;
both points of each pair lie on that line.
co m
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m as e
.co

ag l Page 9 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

(i) 5x + 3y = 7 (ii) 5x − 3y = 7
y y
4 4
(−1, 4)

(2, 1)
x x
−2 0 2 −2 0 2
(2, −1)

−4 (−1, −4) −4

(iii) −5x + 3y = 7 (iv) −5x − 3y = 7
y y
4 4
(1, 4)

(−2, 1)
x x
−2 0 2 −2 0 2
(−2, −1)

−4 −4 (1, −4)

Page 10 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

The four lines, each with the two solutions found. Grid squares are 1 unit.

Check it yourself: (ii) 10 − 3 = 7 and −5 + 12 = 7; (iii) −5 + 12 = 7 and 10 − 3 = 7;
(iv) 10 − 3 = 7 and −5 + 12 = 7.

Why: a straight line that does not pass through the origin crosses exactly three
quadrants. For example, line (i) has intercepts (7/5, 0) and (0, 7/3), both
positive, and slope −5/3, so it runs through II, I and IV and never enters III. A
third choice such as (1, 2/3) for (i) lies in quadrant I.

Tip: the four equations differ only in signs, and their graphs are mirror images
of each other in the axes: (ii) is (i) reflected in the x-axis, (iii) is (i) reflected in
the y-axis.

Page 11 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Q5 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?

y-axis
4
C (2, 3)
3

2

1
A (7, 0)
−2 −1 0 1 2 3 4 5 6 7 8 x-axis
−1

−2

−3
B (0, −3)

Graph for Q5, page 93 — the line 3x − 7y = 21 through A (7, 0) and B (0, −3), with the point C
(2, 3).

No, C (2, 3) does not lie on the line; the graph shows it well above the line. It does not satisfy
the equation either:

3(2) − 7(3) = 6 − 21 = −15 ≠ 21

No point off the line can satisfy the equation. The graph of an equation is made of exactly its
solutions: every solution lies on the line, and every point on the line is a solution.

Why: the line is drawn through all the points (x, y) with 3x − 7y = 21. If a point
satisfied the equation, it would be one of these points and so would be on the
line. C fails the test, which is why it sits off the line.

Page 12 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Check it yourself: A and B do satisfy it: 3(7) − 7(0) = 21 and 3(0) − 7(−3) = 21.
The point of the line with x = 2 is (2, −15/7), about (2, −2.1), not (2, 3).

Q6 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. It has infinitely many solutions: each value of x gives a value of y.
(ii) False. Only when c = 0. For x + y = 1: 0 + 0 ≠ 1, so (0, 0) is not on it.
(iii) False. 3x + 2y = 12 has the rational solutions (2, 3) and (1/2, 21/4).
(iv) True. It can be written x + 0·y − 3 = 0, with a = 1, b = 0 (not both zero). Its graph is a
vertical line.
(v) True. For every x, y = (7 − 2x)/3, e.g. (2, 1), (5, −1), (0, 7/3), …
(vi) False. 2(1) + 3(2) = 8 ≠ 7.

Why: for (iii): whenever a, b, c are rational and b ≠ 0, any rational x gives a
rational y = −(ax + c)/b. So such an equation has infinitely many rational
solutions, not none.

Tip: for (vi), a point that does satisfy 2x + 3y = 7 is (2, 1).

Page 13 of 77

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Class 9 Maths Chapter 13 Two Variables, One Line
ag l AglaSem · NCERT Solutions

.
(i) Compare the solutions of the equations 3x + 4y = 7 and 6x + 8y = 14. Argue comthat
ema solution of
Q7
m a s
they have the same set of solutions, that is, every solution of one is also
.co agl
em
the other.
la s
(ii) Show that the equations ax + by = c and kax + kby = kc, with k ≠ 0, have the same
ag set of solutions.

co m
. ag
m
(i) Let (p, q) be any solution of 3x + 4y =s7,e so 3p + 4q = 7. Then
a
agl
m
6p + 8q = 2(3p + 4q) = 2 × 7 = 14
co
m.
e 2 gives 3p + 4q =
sby
o m l a
g4), (5, −2).
.cequations have exactly the same solutions, e.g. (1, 1), (−3,
so (p, q) solves the second equation. Conversely, if 6p + 8q = 14, dividing
m a
seap + bq = c, multiplying by k gives kap + kbq = kc. If kap + kbq = kc, then, because k ≠ 0,
7. So both
l a
agwe can divide by k to get ap + bq = c. Each solution of one is a solution of the other, so the
(ii) If

m a s
.co agl
solution sets are the same: the two equations draw the same line.

se m
la
ag

Why: we must show the statement in both directions. Multiplying always
c o m works,
se m.
but undoing it needs division by k. That is exactly why k ≠ 0 is required.
om gl a
m.c a
a se Check it yourself: with k = 0 the second equation becomes 0x + 0y = 0, which
agl every pair satisfies, for example (100, 100), even though 3(100) + 4(100) ≠ 7. So
the result really fails when k = 0.
se m
com g l a
m . a
e
Did you know? this factlisasused throughout the chapter: we can multiply an
ag number (to clear fractions, or to match coefficients in
equation by any non-zero
the elimination method) without changing its solutions.
co m
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m la se
.co ag
a s em
agl Think and Reflect — Page 97 c
m .
m a s e
. co agl
a sem
agl

co m
m .
m as e
.co

ag l Page 14 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Section 13.3.1 Understanding Slope
THINK AND REFLECT

Q1 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.

y-axis D
9
8
7
C
6
5
4
B (4, 3)
3
2
1
A (2, 0)
−2 −1 0 1 2 3 4 5 6 7 8 9 x-axis
−1

Fig. 13.8, page 96 — the line AB extended beyond B to the points C and D.

The slope of AB is rise/run = 3/2: every 2 steps right, the line rises 3. From B (4, 3) to x = 8 is a
run of 4 = 2 × 2, so the rise is 2 × 3 = 6.

y=3+6=9

The point is (8, 9). This is the point D in Fig. 13.8.

Page 15 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Why: the slope is the same between any two points of a line. So whatever the
run, rise = (3/2) × run. A run of 4 gives a rise of (3/2) × 4 = 6.

Check it yourself: the line through A (2, 0) with slope 3/2 is y = (3/2)(x − 2), i.e.
3x − 2y = 6. At x = 8: y = (3/2)(6) = 9 ✔, and 3(8) − 2(9) = 24 − 18 = 6 ✔.

Q2 What is the x-coordinate of the point on the line AB whose y-coordinate is 6? Express
this as an ordered pair.

From B (4, 3) the line must rise 6 − 3 = 3. A rise of 3 needs a run of 2 (slope 3/2).

x=4+2=6

The point is (6, 6), which is the point C in Fig. 13.8.

Check it yourself: using the equation 3x − 2y = 6 with y = 6: 3x = 18, so x = 6
✔.

Tip: the slopes of the steps in the figure also agree: B to C is a run of 2 and a
rise of 3, and C to D is again a run of 2 and a rise of 3. Each gives 3/2.

Think and Reflect — Page 99

Page 16 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Section 13.3.2 Slope-Intercept Form
THINK AND REFLECT

Q1 The line 5x + y = 3 is rewritten as y = −5x + 3. Its table of values is:

x 0 1 −1

y 3 −2 8

giving the points A (0, 3), B (1, −2) and C (−1, 8). Can you use the points A, B and C to
verify that the slope of this line is indeed −5?

Use slope = (y2 − y1)/(x2 − x1) for each pair of points:

A to B: (−2 − 3)/(1 − 0) = −5/1 = −5
A to C: (8 − 3)/(−1 − 0) = 5/(−1) = −5
C to B: (−2 − 8)/(1 − (−1)) = −10/2 = −5

Every pair gives slope = −5, the same as m in y = −5x + 3. So the slope is verified.

Why: the slope of a line does not depend on which two of its points we use. All
three pairs give −5, which also confirms that A, B and C really lie on one straight
line.

Tip: a negative slope means the line falls from left to right: here, for every 1 step
to the right, y drops by 5.

Exercise Set 13.3 — Pages 102–103

Page 17 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Section 13.3 Slope-intercept Form of a Linear Equation in Two Variables

Q1 The following diagrams represent ski hills. Rank the hills in order of their steepness,
from least to greatest.

A B C

80 m
60 m 60 m

70 m 110 m 100 m
Ski hills A, B and C, drawn to scale (page 102). In each, the slope rises from left to right; the
horizontal and vertical lengths are marked.

Steepness = slope = rise ÷ run.

A: 60/70 = 6/7 ≈ 0.86

B: 60/110 = 6/11 ≈ 0.55
C: 80/100 = 4/5 = 0.80

From least to greatest steepness: B, C, A.

Why: a hill is steeper when it rises more for each metre you move across. B and
A rise the same 60 m, but B spreads it over 110 m instead of 70 m, so B is
gentler. C rises the most (80 m), but over 100 m, so its slope 0.80 is still a little
less than A’s 0.86.

Common mistake: ranking by height alone (C highest) or by base alone.
Steepness needs the ratio of the two.

Page 18 of 77

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Class 9 Maths Chapter 13 Two Variables, One Line
ag l AglaSem · NCERT Solutions

. com=
Check it yourself: compare A and C without decimals: 6/7 = 30/35 and 4/5
sem a
com
28/35, so A > C.
m . agl
la se
ag
Q2
c om over a run of 4 metres. Find the slope of
The ramp at a loading dock rises 2.5 metres
. ag
the ramp. m se
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The loading-dock ramp as a right triangle, to scale (page 102). The book shows a photograph

a se
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of a ramp leading up to a loading bay, with a parcel on it and a truck behind.
. a g
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co m
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slope = rise/run = 2.5/4 = 25/40 = 5/8 = 0.625
m la
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a s emThe slope of the ramp is 5/8 (0.625): it rises 0.625 m for every metre of horizontal distance.
agl c
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Why: rise and run areaboth
number. agl

co m
m .
m as e
.co

ag l Page 19 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Tip: multiplying top and bottom by 10 first (25/40) avoids dividing a decimal.

Q3 Find the slope of the line l in each of the following diagrams.

(i) (ii)
y y
l

x x

l

(iii) (iv)
y l y
l

x x

The four diagrams of Q3 (pages 102–103). Each grid square has the same width and height.

Count squares between two points on l (the marked dots where given): slope = rise/run, with a
fall counted as negative.
(i) 0. l is horizontal: the rise between any two points is 0, so slope = 0/run = 0.

Page 20 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

(ii) −1. From the upper dot to the lower dot: run = 4 squares right, rise = −4 (down 4). Slope =
−4/4 = −1.
(iii) 2/3. From the lower dot to the upper dot: run = 3 squares, rise = 2 squares. Slope = 2/3.
(iv) Undefined. l is vertical: the run is 0, and division by 0 is not defined.

Why: the diagrams have no numbers on the axes, but slope is a ratio. As long
as a square has equal width and height, counting squares gives the correct
slope whatever one square stands for.

Check it yourself: taking one square as 1 unit, the dots in (ii) are (−2, 2) and (2,
−2), giving (−2 − 2)/(2 − (−2)) = −1; in (iii) they are (−2, −1) and (1, 1), giving (1 −
(−1))/(1 − (−2)) = 2/3. In (i) l is the line y = 1.5 and in (iv) it is x = −1.

Tip: line (ii) passes through the origin and goes down to the right, as y = −x
does; line (iii) goes up to the right, so its slope must be positive.

Q4 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?

Picture with Q4 (page 103)
An illustration of a short flight of steps beside a curved brick wall, with a long,
gently sloping ramp with a handrail running alongside the steps. A person in a
wheelchair is going up the ramp.

The picture printed with this question, described here; see the textbook for the picture itself.

The required slope is rise/run = 1/12. The ramp must rise 18 cm, so

Page 21 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

18/run = 1/12
run = 18 × 12 = 216 cm

The horizontal length of the ramp should be 216 cm (2.16 m).

Why: a slope of 1/12 means every 1 cm of height needs 12 cm of horizontal
length. 18 cm of height therefore needs 18 × 12 cm.

Check it yourself: 216 ÷ 18 = 12, so the rise-to-run ratio is 1 : 12 ✔.

Did you know? a gentle slope like 1 : 12 lets a wheelchair user go up without too
much effort and come down safely. That is why the ramp is much longer than
the staircase it runs beside.

Think and Reflect — Page 104
Section 13.4 Pair of Linear Equations in Two Variables
THINK AND REFLECT

Q1 The cost of one paratha and 2 bowls of dahi is ₹90. Also, the cost of 2 parathas and 7
bowls of dahi is ₹260. Can you figure out the cost of one paratha and one bowl of dahi
and explain your reasoning?

Picture on page 103
A two-part menu board. Left panel: “₹90/-” above “1 + 2” (Paratha, Dahi), with a
picture of one paratha and two bowls of dahi. Right panel: “₹260/-” above “2 +
7” (Paratha, Dahi), with two parathas and seven bowls of dahi.

The picture printed with this question, described here; see the textbook for the picture itself.

Let one paratha cost ₹p and one bowl of dahi ₹d.

Page 22 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

p + 2d = 90 ... (1)
2p + 7d = 260 ... (2)

Double (1): 2p + 4d = 180 ... (3)
(2) − (3): 3d = 80 ⇒ d = 80/3 ≈ 26.67
From (1): p = 90 − 160/3 = 110/3 ≈ 36.67

A paratha costs ₹110/3 ≈ ₹36.67 and a bowl of dahi costs ₹80/3 ≈ ₹26.67.
Reasoning: two of the first order (2 parathas + 4 dahi) cost ₹180. The second order has the
same 2 parathas but 3 more bowls of dahi, and costs ₹80 more. So 3 bowls of dahi cost ₹80.

Check it yourself: 110/3 + 2 × 80/3 = 270/3 = 90 ✔ and 2 × 110/3 + 7 × 80/3 =
780/3 = 260 ✔.

About the book: the book’s numbers (₹90 and ₹260, as also printed on the
menu board) give prices that are not whole rupees: 3 bowls of dahi cost ₹80, so
one bowl is ₹26.666…. The answers ₹36.67 and ₹26.67 are rounded to the
nearest paisa; the exact values are ₹110/3 and ₹80/3.

Why: comparing two orders that contain the same number of parathas isolates
the cost of the extra dahi. This is the idea of the elimination method that you
meet in Section 13.5.

In-text Questions — Page 104

Page 23 of 77

as e
Class 9 Maths Chapter 13 Two Variables, One Line
ag l AglaSem · NCERT Solutions

Section 13.4 Pair of Linear Equations in Two Variables
co m
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emthe amount paid by two different groups.
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Group B: 3 adult tickets and 2 child tickets for ₹700
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How much do individual adult and child tickets cost? If the cost of one adult ticket is ₹x

em we seek. Can you see why? ag
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m
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m a s
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la
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agl Check it yourself: Group A: 2(180) + 3(80) = 360 + 240 = 600 ✔. Group B:
3(180) + 2(80) = 540 + 160 = 700 ✔.
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Exercise Set 13.4 — Page 105.com agl
se m
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co m
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m as e
.co

ag l Page 24 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Section 13.4 Pair of Linear Equations in Two Variables

Q1 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 ₹850.
(ii) On the second day, the family buys 4 movie tickets and 1 snack box for ₹1100.

Cost = (number of tickets × ₹x) + (number of snack boxes × ₹y).

First day: 2x + 3y = 850
Second day: 4x + y = 1100

The pair is 2x + 3y = 850 and 4x + y = 1100.

Check it yourself: solving the pair gives x = 245, y = 120 (a ticket ₹245, a snack
box ₹120): 490 + 360 = 850 ✔ and 980 + 120 = 1100 ✔.

Tip: in standard form: 2x + 3y − 850 = 0 and 4x + y − 1100 = 0.

Q2 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 ₹122.
(ii) Meena travelled 8 km and paid a total fare of ₹160.

Fare = fixed charge + (distance × charge per km).

Sahil: x + 6y = 122
Meena: x + 8y = 160

Page 25 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

The pair is x + 6y = 122 and x + 8y = 160.

Check it yourself: subtracting, 2y = 38, so y = 19 and x = 122 − 114 = 8. Fixed
charge ₹8, ₹19 per km: 8 + 6 × 19 = 122 ✔, 8 + 8 × 19 = 160 ✔.

Common mistake: writing 6x + y = 122. The fixed charge is paid once per trip,
so x has coefficient 1; it is the per-km charge that is multiplied by the distance.

Q3 In a sports meet, tickets for adults cost ₹150 each and tickets for children cost ₹100
each. A total of 200 people attended the meet, and the total amount collected from
ticket sales was ₹25,000. (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.

One equation counts people, the other counts money.

Number of people: x + y = 200
Money collected: 150x + 100y = 25000

The pair is x + y = 200 and 150x + 100y = 25000 (the second can be divided by 50 to give 3x
+ 2y = 500).

Check it yourself: solving: 3x + 2y = 500 and 2x + 2y = 400 give x = 100, y =
100. 100 adults and 100 children: 15000 + 10000 = 25000 ✔.

Why: dividing an equation by a non-zero number does not change its solutions
(Exercise 13.2, Q7), so either form of the money equation is correct.

Page 26 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Think and Reflect — Page 106
Section 13.5.1 Algebraic Methods for Solving a Pair of Linear Equations
THINK AND REFLECT

Q1 Solve the puzzle and discuss your strategy with your friends. Can you determine the
value represented by each shape?

28

30

18

20

? 30 23 22
The shape puzzle (page 106). Each row total is on the right and each column total below; the
first column total is unknown.

Let T, R, H, C be the values of a triangle, rectangle, hexagon and circle.

Page 27 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Row 1: 2T + 2R = 28 ⇒ T + R = 14
Column 4: T + R + 2C = 22 ⇒ 14 + 2C = 22 ⇒ C = 4
Row 3: 3C + T = 18 ⇒ 12 + T = 18 ⇒ T = 6
So R = 14 − 6 = 8

Row 2: 2H + 2R = 30 ⇒ H + 8 = 15 ⇒ H = 7
Column 1: T + H + 2C = 6 + 7 + 8 = 21

Triangle = 6, rectangle = 8, hexagon = 7, circle = 4, and the missing column total is 21.

Check it yourself: Row 4: 3C + R = 12 + 8 = 20 ✔. Column 2: 3R + T = 24 + 6 =
30 ✔. Column 3: R + H + 2C = 8 + 7 + 8 = 23 ✔.

Why: the strategy is to look for a total that contains a group already known.
Row 1 tells us T + R as a block, and column 4 contains that same block, so
substituting it leaves one unknown. This is the substitution idea the section
goes on to develop.

About the book: the text below the puzzle says “T, S and C”; the letter for the
rectangle should be R, since it writes T + R = 14.

Think and Reflect — Page 107
Section 13.5.1 Substitution Method (Example 7)
THINK AND REFLECT

Q1 In Example 7 the pair 7x − 15y = 2 … (1) and x + 2y = 3 … (2) was solved by writing x = 3
− 2y. What if we had expressed y in terms of x, i.e., take Equation (2) and write y =
(1/2)(3 − x). Would we still get the same solution?

Yes. Substitute y = (3 − x)/2 in (1):

Page 28 of 77

ase
Class 9 Maths Chapter 13 Two Variables, One Line
ag l AglaSem · NCERT Solutions

co m
e m.
7x − 15(3 − x)/2 = 2
m l as
.co
Multiply by 2: 14x − 45 + 15x = 4
m ag
la se
29x = 49 ⇒ x = 49/29
agy = (3 − 49/29)/2 = (87/29 − 49/29)/2 = 38/58 = 19/29

co m
em
. ag
s
We get x = 49/29, y = 19/29, exactly as before.
g la
− Explanation a

. c om
s em
Why: a pair of intersecting lines has only one common point. Whichever
m a
.co we choose to express first, we are finding the same
variable
a l
gcommon point, so
a semanswer cannot change. Only the amount of work changes: here writing x =
the
ag l 3 − 2y avoids fractions, so it is the easier choice.

m a s
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com
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Think .and agl
a s em13.5.1 Elimination Method
agl
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The method used in solvingethis m .cproblem (Example 9: 9x − 4y = 2000 … (1), 7x − 3y = ag
Q1
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aElimination
a g
2000 … (2)) is called the Method because we eliminate one of the variables
to obtain a linear equation in the other variable. In the example above, we eliminated y.
m
Try the same problem by eliminating x instead of y. Also try to solve the two equations
using the Substitution Method. Assess the pros and cons of each .comethod.
m
a se
. com ag l
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agl Eliminating x: c
m .
m a s e
. co agl
a sem
agl

co m
m .
m as e
.co

ag l Page 29 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

(1) × 7: 63x − 28y = 14000
(2) × 9: 63x − 27y = 18000
Subtract: y = 4000

In (1): 9x = 2000 + 16000 ⇒ x = 2000

Substitution: from (2), y = (7x − 2000)/3. Put in (1):

9x − 4(7x − 2000)/3 = 2000
27x − 28x + 8000 = 6000 ⇒ x = 2000
y = (14000 − 2000)/3 = 4000

All methods give x = 2000, y = 4000 (incomes ₹18,000 and ₹14,000). Elimination avoids
fractions but here needs larger multipliers for x (7 and 9) than for y (3 and 4). Substitution is
quick when a coefficient is 1, but here it brings fractions.

Why: both methods reduce the pair to one equation in one unknown. Which is
easier depends on the coefficients: look for a coefficient 1 (substitution) or for
coefficients that are already equal or small (elimination).

Tip: eliminate the variable whose coefficients have the smaller LCM. For y the
LCM of 4 and 3 is 12; for x the LCM of 9 and 7 is 63.

Q2 Does a pair of linear equations always have a unique solution? Can you analyse this
using the Elimination Method? While subtracting the equations to eliminate a particular
variable, if the other variable remains, then clearly there is a unique solution. But what
if the second variable gets eliminated too while eliminating the first variable? Can you
think of a situation where this happens?

No, not always. Both variables vanish together when one equation’s left side is a multiple of the
other’s. For example:

Page 30 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

x − y = 10 and 10x − 10y = 100
10 × first: 10x − 10y = 100

Subtract: 0 = 0 (always true)

Here the equations are the same, so there are infinitely many solutions. But with

x − y = 10 and 10x − 10y = 101
Subtract: 0 = 1 (never true)

there is no solution. So when both variables are eliminated, the result is either “0 = 0” (infinitely
many) or “0 = a non-zero number” (none).

Why: multiplying makes the x-coefficients equal. If that same multiplier also
makes the y-coefficients equal, then a1/a2 = b1/b2, and subtracting removes y
as well. Only the constants are left to compare.

Did you know? in pictures: “0 = 0” means the two lines are the same line; “0 =
1” means the lines are parallel and never meet.

Think and Reflect — Page 110

Page 31 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Section 13.5.1 Elimination Method — pairs with infinitely many solutions
THINK AND REFLECT

Q1 How many solutions are there to the equations 10x − 5y = 20 and 4x − 2y = 8? Give 3
more examples of pairs of equations that lead to 0 = 0 after subtraction, and therefore
have infinitely many solutions. Can you find a simple rule to check when this happens?

(1) × 2: 20x − 10y = 40
(2) × 5: 20x − 10y = 40
Subtract: 0 = 0

Both are 2x − y = 4 in disguise (divide by 5 and by 2), so there are infinitely many solutions,
e.g. (0, −4), (2, 0), (3, 2).
Three more examples:

x + 2y = 3 and 2x + 4y = 6

3x − y = 5 and 9x − 3y = 15
x + y = 7 and 5x + 5y = 35

Rule: one equation is a non-zero multiple of the other, i.e. a1/a2 = b1/b2 = c1/c2.

Check it yourself: for the first pair: 10/4 = −5/−2 = 20/8 = 5/2 ✔ (the constants
are on the right here, but the ratio of the constants is the same whichever side
they are written on).

Why: if every coefficient and the constant are multiplied by the same k, the
second equation is just k times the first, and Exercise 13.2 Q7 showed such
equations have exactly the same solutions.

Page 32 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Q2 The rule for infinitely many solutions was stated as a1/a2 = b1/b2 = c1/c2 = k, for a2, b2
and c2 each not equal to 0. What if one of a2, b2, c2 is zero?

Then that ratio cannot be written (division by 0), so use the rule in its multiple form instead: a1
= ka2, b1 = kb2, c1 = kc2 with k ≠ 0. A zero in one equation must then match a zero in the same
place in the other.

2x + 3y = 0 and 4x + 6y = 0: k = 1/2, both c's are 0 → infinitely many
2x + 3y + 5 = 0 and 4x + 6y = 0: c1 = 5 ≠ k × 0 → no solution

y − 2 = 0 and 3y − 6 = 0 (a1 = a2 = 0): k = 1/3 → same line y = 2

So the rule still works; just compare by multiples, or compare only the ratios that exist and
check the zero terms directly.

Why: “a1/a2 = k” is only a short way of writing a1 = k a2. The second form
makes sense even when a2 = 0 (then a1 must be 0 too).

Tip: another way that never divides: the lines are parallel or the same exactly
when a1b2 = a2b1 (cross-multiplying the first two ratios).

Q3 Does a pair of linear equations always have either a unique solution or infinitely many
solutions?

No. A pair can also have no solution. Example:

x − y = 10 and x − y = 11

The same expression x − y cannot equal both 10 and 11, so no pair (x, y) satisfies both.
Elimination gives 0 = 1, which is false. Graphically the two lines are parallel (same slope 1,
different intercepts) and never meet.

Page 33 of 77

as e
Class 9 Maths Chapter 13 Two Variables, One Line
ag l AglaSem · NCERT Solutions

co m
m.

o m l ase
Why:mthere ag they cross at
.c are exactly three possibilities for two lines in a plane:
la sepoint (unique solution), they are the same line (infinitely many), or they are
ag
one
parallel and distinct (no solution). The book’s next example, x − y = 10 and 10x −
10y = 101, is of the third kind.
com
m . ag
la se
ag
Think and Reflect — Page 111 co m
em.
c om
Section 13.5.1 Elimination Method — pairs with no solution
gl as
.
m REFLECT a
se
THINK AND
a
agl
Give 3 more examples of pairs of equations that have no solution. Can you find a
s
Q1
m a
.co agl
simple rule to check when this will happen?

se m
la
ag

co m
.
x + y = 2 and x + y = 5
e m
om
2x − 3y = 4 and 4x − 6y = 7
c gl as
. a
a sexm+ 2y = 1 and 3x + 6y = 10
agl
In each, multiplying the first equation makes the left sides identical but the right sides differ
m a s em
.co
(e.g. 4x − 6y = 8 against 4x − 6y = 7), and subtracting gives a false statement such as 0 = 1. So
a gl
there is no solution.
a s em
glvariable parts are in the same ratio but the constants are not.
Rule: a1/a2 = b1/b2 ≠ c1/c2: the
a
co m
m .
om la se
em ag 4/7 ≠ 1/2. Third pair: 1/3
.c Check it yourself: second pair: 2/4 = −3/−6 = 1/2, but
a s
agl
= 2/6 = 1/3, but 1/10 ≠ 1/3.
.c
s e m
m a
. co agl
a sem
agl

co m
m .
m as e
.co

ag l Page 34 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Why: equal ratios of a and b mean the lines have the same slope; a different
ratio of constants means they cut the axes at different places. Same slope,
different line: parallel lines.

Think and Reflect — Page 112
Section 13.5.1 Summary of cases
THINK AND REFLECT

Q1 The book summarises: 1. when a1/a2 ≠ b1/b2, the pair has a unique solution; 2. when
a1/a2 = b1/b2 = c1/c2, the pair has an infinite number of solutions; and 3. when a1/a2 =
b1/b2 ≠ c1/c2, the pair has no solution. Are the converses of the above statements true?

Yes, all three converses are true (with a2, b2, c2 non-zero):
1. If the pair has a unique solution, then a1/a2 ≠ b1/b2.
2. If it has infinitely many solutions, then all three ratios are equal.
3. If it has no solution, then a1/a2 = b1/b2 ≠ c1/c2.
Reason: the three conditions cover every possible pair, and no pair satisfies two of them. The
three outcomes also cannot happen together. So, for example, if a pair has a unique solution it
cannot be in case 2 or 3 (those give infinitely many or none); it must be in case 1.

Why: this “process of elimination” works because each case leads to a different
outcome. Suppose a pair has no solution. Case 1 would give exactly one
solution and case 2 infinitely many, so neither can hold; only case 3 is left.

Did you know? so the ratio test is a complete check: it tells you the number of
solutions, and the number of solutions tells you which ratio pattern you have.

Page 35 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Q2 Can there be methods other than elimination and substitution to reduce a pair of linear
equations in two variables to a linear equation in one variable? If so, describe them.

Yes. Two examples:
1. Comparison (equating) method. Make the same variable the subject of both equations and
set the two expressions equal. For x + y = 25 and x − y = 11:

y = 25 − x and y = x − 11
25 − x = x − 11 ⇒ 2x = 36 ⇒ x = 18, y = 7

2. Adding and subtracting. When the coefficients are swapped, as in 2x + 3y = 600, 3x + 2y =
700, adding gives x + y = 260 and subtracting gives x − y = 100, two very simple equations.

Why: each method produces, in one step, an equation with a single unknown,
which is the aim stated in Section 13.5.

Did you know? the cross-multiplication method solves a1x + b1y + c1 = 0, a2x
+ b2y + c2 = 0 directly by a formula: x = (b1c2 − b2c1)/(a1b2 − a2b1) and y =
(c1a2 − c2a1)/(a1b2 − a2b1). It is elimination done once in general. The graphical
method of Section 13.6 is a further, non-algebraic way to find the solution.

In-text Questions — Page 113

Page 36 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Section 13.6 Graphical Method for Solving a Pair of Linear Equations in Two Variables

Q1 Consider the following pairs of linear equations in two variables.
Pair 1: 2x + y = 6 and x − y = 2
Pair 2: x + y = 4 and 2x + 2y = 8
Pair 3: 3x − 2y = 6 and 6x − 4y = 12
Pair 4: x − 2y = 4 and 2x − 4y = 6
For each pair, prepare a table of values (with at least two ordered pairs). Plot the points
on the Cartesian plane. Draw the straight lines.

EQUATION Points (x, y)

2x + y = 6 (0, 6), (3, 0), (1, 4)

x−y=2 (0, −2), (2, 0), (4, 2)

x+y=4 (0, 4), (4, 0), (2, 2)

2x + 2y = 8 (0, 4), (4, 0), (1, 3)

3x − 2y = 6 (0, −3), (2, 0), (4, 3)

6x − 4y = 12 (0, −3), (2, 0), (4, 3)

x − 2y = 4 (0, −2), (4, 0), (2, −1)

2x − 4y = 6 (0, −1.5), (3, 0), (1, −1)

Page 37 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Pair 1 Pair 2
y y
6 2x + y = 6 6
2x + 2y = 8
4 4 x+y=4

2 2

(8/3, 2/3) x x
−2 0 2 4 6 −2 0 2 4 6

−2 −2
x−y=2
−4 −4
Pair 3 Pair 4
y y
6 6x − 4y = 12 6
3x − 2y = 6
4 4

2x − 4y = 6
2 2

x x
−2 0 2 4 6 −2 0 2 4 6

−2 −2

x − 2y = 4
−4 −4
The four pairs drawn on the same scale. Dashed red lines lie exactly on top of the blue ones
(coincident lines).

Pair 1: the lines cross at one point, (8/3, 2/3). Pairs 2 and 3: the two lines coincide (one line).
Pair 4: the lines are parallel and never meet.

Page 38 of 77

ase
Class 9 Maths Chapter 13 Two Variables, One Line
ag l AglaSem · NCERT Solutions

co m
m.

o m l ase
m .itcyourself: Pair 1: adding the equations gives 3x = 8, soagx = 8/3 and y =
se− 2 = 2/3, matching the graph (about (2.7, 0.7)).
Check
la
ag
8/3

.
Why: in Pairs 2 and 3 the second equation
comis 2 times the first, so it has the ag
same solutions and the same line. eInmPair 4 the left side of the second is 2 times
glas2 gives x − 2y = 3, a line with the same slope
a
the first but 6 ≠ 2 × 4; dividing by
1/2 but a different intercept.
co m
em.
m l as
.co ag
a sem
agl
Think and Reflect — Page 113
Section 13.6 Graphical Method for Solving a Pair of Linear Equations in Two Variables
m a s
.co agl
THINK AND REFLECT

a s em
l the graph? What can we say about the number of
agfrom
How do we find solutions
Q1
solutions when the two lines (i) intersect at a point, (ii) are parallel, (iii) are coincident?
co m
m .
om l ase
g lies on both lines. We
.c of the pair must satisfy both equations, so it is a pointathat
m
se off the common point(s) of the two lines.
A solution
la
ag
read
(i) Intersect at a point: exactly one common point, so a unique solution, e.g. (8/3, 2/3) for Pair
1.
m a s em
.co (Pair 4). a gl
m is common, so infinitely many solutions (Pairs 2 and 3).
(ii) Parallel: no common point, so no solution
(iii) Coincident: every point of theseline
a
agl

co m
m .
om
Why:
la seevery solution is a
every point on a line is a solution of its equation, and
g
.c point on the line. So the solutions of both equationsatogether
em
are exactly the
a s
agl
points the two lines share.
.c
s e m
omonly approximately when it is not at a grid a
c agl
Tip: a graph can show a solution
.
emabout (2.7, 0.7); use algebra to get the exact value.
point. (8/3, 2/3) looks like
a s
agl

co m
m .
m as e
.co

ag l Page 39 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Think and Reflect — Page 115
Section 13.6 Graphical Method (Example 12: parallel lines)
THINK AND REFLECT

Q1 Can you prove that two lines of equal slope are parallel? (Hint: Consider the equations
of the two lines to be y = mx + d1 and y = mx + d2.)

Take two different lines with the same slope: y = mx + d1 and y = mx + d2, with d1 ≠ d2.
Suppose they had a common point (p, q). Then

q = mp + d1 and q = mp + d2

⇒ mp + d1 = mp + d2 ⇒ d1 = d2

This contradicts d1 ≠ d2. So the lines have no common point: they are parallel. (If d1 = d2, they
are the same line.)

Why: at every value of x, the two lines are separated vertically by the same
amount d1 − d2, which is never 0. A gap that never closes means the lines never
meet.

Did you know? vertical lines such as x = 2 and x = 5 have undefined slope and
are also parallel, since no point can have x = 2 and x = 5 at once.

Check it yourself: Example 12: x + 2y − 4 = 0 is y = −x/2 + 2 and 2x + 4y − 12 =
0 is y = −x/2 + 3. Equal slopes −1/2, different intercepts: parallel, as Fig. 13.15
shows.

Exercise Set 13.5 — Pages 116–118

Page 40 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Sections 13.5 Finding Solutions to a Pair of Linear Equations and 13.6 Graphical Method

Q1 (i) 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.

Let the integers be x and y, with difference x − y.

x + y = 5 ... (1)

x − y = −21 ... (2)
Add: 2x = −16 ⇒ x = −8
From (1): y = 5 − (−8) = 13

The numbers are −8 and 13.

Check it yourself: −8 + 13 = 5 ✔ and −8 − 13 = −21 ✔.

Why: a negative difference simply means the first number is smaller than the
second. Adding the equations eliminates y at once because its coefficients are
+1 and −1.

Q1 (ii) (ii) The difference between two numbers is 26 and one number is three times the
other. Find the numbers.

Let the larger number be x and the smaller y.

Page 41 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

x − y = 26 ... (1)
x = 3y ... (2)
Substitute: 3y − y = 26 ⇒ 2y = 26 ⇒ y = 13
x = 3 × 13 = 39

The numbers are 39 and 13.

Check it yourself: 39 − 13 = 26 ✔ and 39 = 3 × 13 ✔.

Why: equation (2) already gives x in terms of y, so substitution is the natural
method.

Tip: if negative numbers are allowed there is a second answer: −13 and −39.
Here −13 − (−39) = 26 and −39 = 3 × (−13). With y = 3x instead of x = 3y, the
equations give exactly this pair.

Q1 (iii) (iii) The coach of a cricket team buys 7 bats and 6 balls for ₹8880. Later, she buys 3
bats and 5 balls for ₹4000. Find the cost of each bat and each ball.

Let a bat cost ₹x and a ball ₹y.

7x + 6y = 8880 ... (1)
3x + 5y = 4000 ... (2)
(1) × 5: 35x + 30y = 44400
(2) × 6: 18x + 30y = 24000
Subtract: 17x = 20400 ⇒ x = 1200

In (2): 3600 + 5y = 4000 ⇒ y = 80

A bat costs ₹1200 and a ball costs ₹80.

Page 42 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Check it yourself: 7 × 1200 + 6 × 80 = 8400 + 480 = 8880 ✔; 3 × 1200 + 5 ×
80 = 3600 + 400 = 4000 ✔.

Tip: here y was eliminated by making both coefficients 30 (the LCM of 6 and 5).
Eliminating x instead, using 21 (the LCM of 7 and 3), works equally well and
gives the same answer.

Q1 (iv) (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
₹155 and for a journey of 15 km, the total amount paid is ₹220. 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?

Let the fixed charge be ₹x and the charge per km ₹y.

x + 10y = 155 ... (1)
x + 15y = 220 ... (2)
(2) − (1): 5y = 65 ⇒ y = 13
x = 155 − 130 = 25
For 25 km: 25 + 25 × 13 = 25 + 325 = 350

Fixed charge ₹25, ₹13 per km; a 25 km journey costs ₹350.

Check it yourself: 25 + 15 × 13 = 25 + 195 = 220 ✔.

Why: the fare grows linearly with distance: fare = 25 + 13 × (km). The fixed
charge is the y-intercept and ₹13 per km is the slope of this line.

Page 43 of 77

as e
Class 9 Maths Chapter 13 Two Variables, One Line
ag l AglaSem · NCERT Solutions

comthe
(v) A fraction becomes equal to 9/11 if 2 is added to both the numerator.and
emit becomes
Q1 (v)
m as
denominator. If 3 is added to both the numerator and the denominator,
m .coto 5/6. Find the fraction.
equal agl
la se
ag

m
Let the fraction be x/y.
. co ag
se m
la
ag
(x + 2)/(y + 2) = 9/11 ⇒ 11x + 22 = 9y + 18 ⇒ 11x − 9y = −4 ... (1)

(x + 3)/(y + 3) = 5/6 ⇒ 6x + 18 = 5y + 15 ⇒ 6x − 5y = −3 ... (2)

co m
m.
(1) × 5: 55x − 45y = −20

m as e
.co
(2) × 9: 54x − 45y = −27
agl
a sem x = 7
agl
Subtract:
In (2): 42 − 5y = −3 ⇒ y = 9

m a s
m .co agl
se
The fraction is 7/9.
la
− Explanation ag

c o m
Check it yourself: (7 + 2)/(9 + 2) = 9/11 ✔ and (7 + 3)/(9 + 3) =m .
s e 10/12 = 5/6 ✔.
. com agla
a s em
agl
Why: cross-multiplying turns each condition into a linear equation in x and y;
after that it is an ordinary pair.
se m
com g l a
m . a
e
a g las
Q1 (vi) (vi) If we add 1 to the numerator and subtract 1 from the denominator, a fraction
o m
reduces to 1. It becomes equal to 1/2 if we add 1 only to the denominator.
c
What is
the fraction? m. se
m la
.co
mANSWER ag
s e
ag la Let the fraction be x/y.
.c
s e m
m a
. co agl
a sem
agl

co m
m .
m as e
.co

ag l Page 44 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

(x + 1)/(y − 1) = 1 ⇒ x + 1 = y − 1 ⇒ x − y = −2 ... (1)
x/(y + 1) = 1/2 ⇒ 2x = y + 1 ⇒ 2x − y = 1 ... (2)
(2) − (1): x = 3
y=x+2=5

The fraction is 3/5.

Check it yourself: (3 + 1)/(5 − 1) = 4/4 = 1 ✔ and 3/(5 + 1) = 3/6 = 1/2 ✔.

Q1 (vii) (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 years and Sonu’s y years.

Five years ago: x − 5 = 3(y − 5) ⇒ x − 3y = −10 ... (1)
Ten years later: x + 10 = 2(y + 10) ⇒ x − 2y = 10 ... (2)
(2) − (1): y = 20

x = 10 + 2(20) = 50

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

Check it yourself: five years ago: 45 and 15, and 45 = 3 × 15 ✔. Ten years later:
60 and 30, and 60 = 2 × 30 ✔.

Common mistake: subtracting 5 from only one of the ages. Five years ago both
were 5 years younger.

Page 45 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Q1 (viii) (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 the tens digit be x and the units digit y. The number is 10x + y; reversed it is 10y + x.

x + y = 9 ... (1)
9(10x + y) = 2(10y + x) ⇒ 90x + 9y = 20y + 2x ⇒ 88x = 11y ⇒ y = 8x ... (2)
In (1): x + 8x = 9 ⇒ x = 1, y = 8

The number is 18.

Check it yourself: 1 + 8 = 9 ✔; 9 × 18 = 162 and 2 × 81 = 162 ✔.

Q1 (ix) (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 her get x notes of ₹50 and y notes of ₹100.

x + y = 25 ... (1)

50x + 100y = 2000 ⇒ x + 2y = 40 ... (2)
(2) − (1): y = 15
x = 25 − 15 = 10

10 notes of ₹50 and 15 notes of ₹100.

Page 46 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Check it yourself: 10 + 15 = 25 notes ✔; 10 × 50 + 15 × 100 = 500 + 1500 =
₹2000 ✔.

Tip: dividing the money equation by 50 first keeps the numbers small.

Q1 (x) (x) A lending library has a fixed charge for the first three days and an additional
charge for each day thereafter. Saritha paid ₹27 for a book kept for seven days,
while Susy paid ₹21 for the book she kept for five days. Find the fixed charge and
the charge for each extra day.

Let the fixed charge be ₹x and the charge per extra day ₹y. Seven days = 3 + 4 extra days; five
days = 3 + 2 extra days.

x + 4y = 27 ... (1)
x + 2y = 21 ... (2)
(1) − (2): 2y = 6 ⇒ y = 3
x = 21 − 6 = 15

Fixed charge ₹15; ₹3 for each extra day.

Check it yourself: Saritha: 15 + 4 × 3 = 27 ✔; Susy: 15 + 2 × 3 = 21 ✔.

Common mistake: using 7y and 5y. The extra charge starts only after the first
three days, so the extra days are 7 − 3 = 4 and 5 − 3 = 2.

Page 47 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Q2 Form a pair of linear equations and find their common solutions graphically.
10 students of Grade 9 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.

Let the number of boys be x and of girls y.

x + y = 10 ... (1)

y = x + 4 ... (2)

x 0 5 10

y = 10 − x 10 5 0

x 0 2 3

y=x+4 4 6 7

Page 48 of 77

as e
Class 9 Maths Chapter 13 Two Variables, One Line
ag l AglaSem · NCERT Solutions

co m
e m.
my l as
.co ag
a s em10
agl y=x+4
com
m . ag
8
la se
ag
(3, 7)
co m
6 em.
m l as
m .co g
x + y =a 10
la se
ag 4
m a s
m .co agl
la se
2 ag

co m
m . x
com 0 ase
. 2 4 6 ag8l 10
a se m
agl
se m
m y = x + 4 (x = boys, y = girls).
co
Graphs of x + y = 10 and
g l a
m . a
l aseand 7 girls took part.
The lines meet at (3, 7). So 3gboys
a
co m
m .
om la se substituting (2)
em
.c Check it yourself: 3 + 7 = 10 ✔ and 7 = 3 + 4 ✔.
ag
Algebraically,
a s in (1): 2x + 4 = 10, so x = 3.
agl c
m .
a s e
om7) is one. agl
Why: the common point is the only pair that fits both facts. Only whole-number
points make sense here, and.c(3,
se m
g l a
a

co m
m .
m as e
.co

ag l Page 49 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Q3 Use the ratios a1/a2, b1/b2, c1/c2 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 + 9 = 0

(i) a1/a_2 = 5/7, b1/b_2 = −4/6 = −2/3

5/7 ≠ −2/3
(ii) 9/18 = 1/2, 3/6 = 1/2, 12/24 = 1/2

(iii) 6/2 = 3, −3/−1 = 3, 10/9

(i) Intersect at a point (a1/a2 ≠ b1/b2).
(ii) Coincident (all three ratios equal 1/2).
(iii) Parallel (a1/a2 = b1/b2 = 3, but c1/c2 = 10/9 ≠ 3).

Why: equal a and b ratios mean equal slopes. Then the c ratio decides: equal
means the same line, unequal means parallel lines.

Check it yourself: in (i) the slopes are 5/4 and −7/6, clearly different. In (ii) the
second equation is exactly 2 × the first. In (iii) the first equation is 3(2x − y) + 10
= 0 and the second 2x − y + 9 = 0; the same 2x − y cannot equal both −10/3
and −9.

Page 50 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Q4 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) 1/2 = 1/2 = 5/10 → coincident
(ii) 1/3 = 1/3, but 8/16 = 1/2 ≠ 1/3 → parallel

(iii) 2/4 = 1/2, 1/(−2) = −1/2 → intersect
(iv) 2/4 = 1/2 = 1/2, but 2/5 ≠ 1/2 → parallel

(i) has infinitely many solutions: every point of the line x + y = 5, e.g. (0, 5), (2, 3), (5, 0).
(ii) No solution.
(iii) Unique solution x = 2, y = 2, where the lines through (0, 6), (3, 0) and (0, −2), (1, 0) meet.
(iv) No solution.

(i) Coincident lines (iii) Lines meet at (2, 2)
6 y y 2x + y = 6
5 6
2x + 2y = 10
4
x+y=5 4
3
2
1 2 (2, 2)
x
0 1 2 3 4 5 6 x
0 1 2 3 4

−2
4x − 2y = 4

Graphs for Q4 (i) and (iii).

Page 51 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Check it yourself: (iii): 2(2) + 2 − 6 = 0 ✔ and 4(2) − 2(2) − 4 = 0 ✔.

Why: for (ii), dividing the second equation by 3 gives x − y = 16/3, while the first
says x − y = 8. For (iv), the equations reduce to x − y = 1 and x − y = 5/4. In
each case the same expression would need two different values, so the lines
are parallel.

Q5 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 the length be l m and the width w m. Half the perimeter is l + w.

l + w = 36 ... (1)
l − w = 4 ... (2)
Add: 2l = 40 ⇒ l = 20
w = 36 − 20 = 16

The garden is 20 m long and 16 m wide.

Check it yourself: 20 − 16 = 4 ✔; perimeter 2(20 + 16) = 72 m, half of which is
36 m ✔.

Common mistake: using 2(l + w) = 36. The 36 m is half the perimeter, which is l
+ w.

Page 52 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Q6 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: 3x + 2y − 7 = 0, since 2/3 ≠ 3/2.
(ii) Parallel: 4x + 6y − 9 = 0, since 2/4 = 3/6 = 1/2 but −8/−9 = 8/9 ≠ 1/2.
(iii) Coincident: 4x + 6y − 16 = 0, since 2/4 = 3/6 = −8/−16 = 1/2.

Why: (i) needs a, b not in the same ratio; (ii) needs a and b multiplied by the
same number but c by a different one; (iii) needs the whole equation multiplied
by one non-zero number.

Your own answer: many answers are correct. For example, x − y = 0 also
intersects it (at (8/5, 8/5)), 2x + 3y = 0 is also parallel, and 6x + 9y − 24 = 0 is
also coincident. Test your answer with the three ratios.

Q7 Here is a problem that was posed by Mahāvīrāchārya in Gaṇita sāra saṅgraha (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.
(Hint: The given problem can be modelled as 9x + 7y = 107, 7x + 9y = 101. Can you
figure out a way of solving these equations without directly using elimination or the
substitution method? What is special in this pair of equations? The x and y coefficients
are interchanged in the 2 equations. What will happen if we add the pair of equations?
What will happen if we subtract the pair of equations? Can the resulting equations be
solved to find the values of x and y?)

The coefficients 9 and 7 are swapped, so adding and subtracting give very simple equations.

Page 53 of 77

as e
Class 9 Maths Chapter 13 Two Variables, One Line
ag l AglaSem · NCERT Solutions

com
m.
Add: 16x + 16y = 208 ⇒ x + y = 13
m as e
.co
Subtract: 2x − 2y = 6 ⇒ x − y = 3
ag l
se m
la
Add these: 2x = 16 ⇒ x = 8; then y = 13 − 8 = 5
ag
m
A citron costs 8 and a fragrant wood-apple costs 5 (in the units of money of the problem).
co
m . ag
la se
ag
Check it yourself: 9 × 8 + 7 × 5 = 72 + 35 = 107 ✔; 7 × 8 + 9 × 5 = 56 +m45 =
.co
em
101 ✔.
m as
.co
em adding makes both coefficients 9 + 7 = 16, so we can divide by 16. agl
a s
agl Subtracting makes them 9 − 7 = 2 and −2. Each new equation is very easy, and
Why:

s
together they give x and y at once. This works for any pair ax + by = p, bx + ay
com a
. agl
= q with a ≠ b.
e m
glas
Tip: the problem namesano currency, so give the answer as plain numbers.

co m
m .
om l ase
.c a g
m
e 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost
sQ8
a
agl ₹46. Find the cost of each pencil and pen.

se m
com g l a
m . a
Let a pencil cost ₹x and a pen ₹y.se
a
agl
m
5x + 7y = 50 ... (1)
. co
7x + 5y = 46 ... (2)
se m
com12x + 12y = 96 ⇒ x + y = 8 la
.Add:
em (1) − (2): −2x + 2y = 4 ⇒ y − x = 2 ag
a s
agl c
m .
e
So 2y = 10 ⇒ y = 5, x = 3
m a s
.
em₹5.
co agl
a s
A pencil costs ₹3 and a pen costs
agl

co m
m .
m as e
.co

ag l Page 54 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Check it yourself: 5 × 3 + 7 × 5 = 15 + 35 = 50 ✔; 7 × 3 + 5 × 5 = 21 + 25 = 46
✔.

Why: this is the same trick as Q7: the coefficients are interchanged, so adding
and subtracting give x + y and y − x directly.

Q9 Find the height of the stool.

85 cm 25 cm

Q9 (page 118), redrawn. Left: the cat sits on the stool; from the floor to the top of the cat’s
head is 85 cm. Right: the same cat sits on the floor beside the stool; the top of the stool is 25
cm above the top of the cat’s head.

Let the height of the stool be h cm and the height of the sitting cat be c cm.

Page 55 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Cat on the stool: h + c = 85 ... (1)
Cat on the floor: h − c = 25 ... (2)
Add: 2h = 110 ⇒ h = 55

c = 85 − 55 = 30

The stool is 55 cm high (and the cat is 30 cm tall).

Why: in the left picture the heights add up (stool + cat). In the right picture the
25 cm is how much taller the stool is than the cat, a difference. A sum and a
difference of the same two numbers can always be added to find the larger one.

Check it yourself: 55 + 30 = 85 ✔ and 55 − 30 = 25 ✔.

End-of-Chapter Exercises — Pages 119–121

Page 56 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Whole chapter

Q1 The graph of the line y = 3x, passing through (0, 0) and (2, 6) is given below.

y-axis
8
7
6 (2, 6)
5
4
3
y = 3x
2
1
(0, 0)
−3 −2 −1 0 1 2 3x-axis
4
−1
−2
−3
−4

Page 57 of 77

Class 9 Maths Chapter 13 Two Variables, One Line AglaSem · NCERT Solutions

Graph for Q1, page 119 — the line y = 3x through (0, 0) and (2, 6).

(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) slope = (6 − 0)/(2 − 0) = 6/2 = 3

(i) Slope = 3: from (0, 0) to (2, 6) the run is 2 and the rise is 6.
(ii) y-intercept = 0: the line meets the y-axis at the origin (0, 0). For the tank, the water level is
0 cm at time 0, i.e. the tank starts empty.
(iii)(a) y = 3 × 5 = 15 cm. (b) 21 = 3x ⇒ x = 7 minutes. (The level rises 3 cm every minute.)
(iv) No. At x = 4, y = 3 × 4 = 12, not 10, so (4, 10) does not satisfy y = 3x.
(v) (3, 9). Every point of x = 3 has x-coordinate 3; on y = 3x that gives y = 9.

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Class 9 Maths Chapter 13 Two Variables, One Line
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