Introduction to Linear Polynomials

Chapter 2 • Class 9 • Ganita Manjri

Table of Contents

Imp

What is an algebraic expression?

An algebraic expression is a combination of numbers, variables, and operation signs. In this chapter, expressions with one variable are especially important.

Examples: 4x + 5y + 3, 10x − x², 3z + 7

Terms are the separate parts joined by + or −.
Variables are letters like x, y, and z.
Coefficients are the numbers attached to variables.
Constant is the number without a variable.

Imp

What is a polynomial?

A polynomial in one variable is an algebraic expression made using one variable and whole-number powers of that variable.

5y³ + y² + 2y − 1

The highest power of the variable is called the degree.
Degree 0: constant polynomial.
Degree 1: linear polynomial.
Degree 2: quadratic polynomial.
Degree 3: cubic polynomial.

Degree of a polynomial

The degree tells us the highest power of the variable in the expression.

Polynomial	Highest power	Name
8	0	Constant polynomial
3z + 7	1	Linear polynomial
x² + 5x + 1	2	Quadratic polynomial
5y³ + y² + 2y − 1	3	Cubic polynomial

Imp idea about linear polynomials

A linear polynomial has degree 1. Its general form is shown below.

ax + b

Here, a is the coefficient of x and b is the constant term. Examples include 4x, 2x + 3, 50m + 200, and 2n − 1.

Linear patterns

Constant difference

In a linear pattern, the difference between consecutive terms stays the same.

1, 3, 5, 7, 9, ...

Since the difference is always 2, this is a linear pattern.

Nth term rule

If a pattern starts 1, 3, 5, 7, ... then the nth term can be written as:

2n − 1

This is a linear polynomial because the highest power of n is 1.

Tile pattern

The number of tiles increases by 2 at each new stage.

Input-output process

Put in a value of x, apply the rule, and get the output.

Linear growth and decay

A rising straight line shows growth, while a falling straight line shows decay.

Linear equations and graphs

Imp

From polynomial to equation

When a linear polynomial is made equal to a constant, we get a linear equation.

2x + 10 = 64

Such an equation can be solved to find the value of the variable.

Imp

Linear relationship

The general form of a linear relationship between x and y is:

y = ax + b

a is the slope of the line.
b is the y-intercept.
If b = 0, the line passes through the origin.
Same slope but different y-intercepts give parallel lines.

Graph of y = 2x + 1

Any two correct points are enough to draw a straight line.

Parallel lines

These lines have the same slope, so they are parallel.

Remember: In y = ax + b, changing a changes the steepness, while changing b shifts the line up or down.

Exercise set 2.1

1. Degrees of the given polynomials

(i) 2x² − 5x + 3 → degree 2
(ii) y³ + 2y − 1 → degree 3
(iii) −9 → degree 0
(iv) 4z − 3 → degree 1

2. Write one polynomial each of degrees 1, 2 and 3

Degree 1: 3x + 2
Degree 2: x² − 4x + 1
Degree 3: 2x³ + x − 5

3. Coefficients of x² and x³ in x⁴ − 3x³ + 6x² − 2x + 7

Coefficient of x² = 6
Coefficient of x³ = −3

4. Coefficient of z in 4z³ + 5z² − 11

There is no z-term, so the coefficient of z is 0.

5. Constant term of 9x³ + 5x² − 8x − 10

Constant term = −10

Exercise set 2.2

1. Find the value of 5x − 3

(i) x = 0 → 5(0) − 3 = −3
(ii) x = −1 → 5(−1) − 3 = −8
(iii) x = 2 → 10 − 3 = 7

2. Find the value of 7s² − 4s + 6

(i) s = 0 → 6
(ii) s = −3 → 7(9) + 12 + 6 = 81
(iii) s = 4 → 7(16) − 16 + 6 = 102

3. Present ages of Salil and his mother

Let Salil's present age be x years.
Mother's age = 3x years.
After 5 years: (x + 5) + (3x + 5) = 70
4x + 10 = 70
4x = 60
x = 15
Answer: Salil = 15 years, Mother = 45 years

4. Two positive integers with difference 63 and ratio 2:5

Let the integers be 2k and 5k.
5k − 2k = 63
3k = 63
k = 21
Answer: The integers are 42 and 105

5. Ruby's coins

Let the number of ₹5 coins be x.
Then the number of ₹2 coins = 3x.
Total value = 5x + 2(3x) = 88
11x = 88
x = 8
Answer: ₹5 coins = 8 and ₹2 coins = 24

6. Fence cut into two pieces

Let the shorter piece be x feet.
Longer piece = 4x feet.
x + 4x = 300
5x = 300
x = 60
Answer: Shorter piece = 60 ft, Longer piece = 240 ft

7. Dimensions of the rectangle

Let width = w cm.
Length = 2w + 3 cm.
Perimeter = 24 cm, so 2(l + w) = 24
l + w = 12
(2w + 3) + w = 12
3w = 9
w = 3
l = 2(3) + 3 = 9
Answer: Width = 3 cm and Length = 9 cm

Exercise set 2.3

1. Savings account with pocket money

Amount after n months = 500 + 150n
2nd month = 800
3rd month = 950
4th month = 1100
Linear expression: A = 500 + 150n

2. Rally members after n hours

Members left after n hours = 120 − 9n
After 1 hour = 111
After 2 hours = 102
After 3 hours = 93
Linear expression: M = 120 − 9n

3. Area of a rectangle when length = 13 cm

Area = 13 × breadth = 13b
(i) b = 12 → 156 cm²
(ii) b = 10 → 130 cm²
(iii) b = 8 → 104 cm²
Linear pattern: A = 13b

4. Volume of a box when length = 7 cm and breadth = 11 cm

Volume = 7 × 11 × h = 77h
(i) h = 5 → 385 cm³
(ii) h = 9 → 693 cm³
(iii) h = 13 → 1001 cm³
Linear pattern: V = 77h

5. Pages left after 15 days

Pages left after d days = 500 − 20d
After 15 days = 500 − 20(15) = 200
Linear pattern: P = 500 − 20d

Exercise set 2.4

1. Plant growth

Height after t months = h = 1.75 + 0.5t
(i) After 7 months: h = 1.75 + 0.5(7) = 5.25 feet

(ii) Table for t = 0 to 10:

t	0	1	2	3	4	5	6	7	8	9	10
h (ft)	1.75	2.25	2.75	3.25	3.75	4.25	4.75	5.25	5.75	6.25	6.75

(iii) It is linear growth because the height increases by 0.5 ft every month.

2. Mobile phone value

Value after t years = v = 10000 − 800t
(i) After 3 years: v = 7600

(ii) Table for t = 0 to 8:

t	0	1	2	3	4	5	6	7	8
v (₹)	10000	9200	8400	7600	6800	6000	5200	4400	3600

(iii) It is linear decay because the value decreases by ₹800 every year.

3. Population of a village

Population after t years = P = 750 + 50t
(i) After 6 years: P = 1050

(ii) Table for t = 0 to 10:

t	0	1	2	3	4	5	6	7	8	9	10
P	750	800	850	900	950	1000	1050	1100	1150	1200	1250

(iii) It is linear growth because the population increases by 50 every year.

4. Recharge balance

(i) Remaining balance after x days:

b(x) = 600 − 15x

This is linear decay because the balance reduces by the same amount each day.

(ii) Balance becomes zero when 600 − 15x = 0
15x = 600
x = 40
Answer: The balance runs out after 40 days.

(iii) Table for x = 1 to 10:

x	1	2	3	4	5	6	7	8	9	10
b(x)	585	570	555	540	525	510	495	480	465	450

Exercise set 2.5

1. Learning platform bill

Given points: (10, 400) and (14, 500)
a = (500 − 400) ÷ (14 − 10) = 100 ÷ 4 = 25
400 = 25(10) + b
b = 150
Answer: a = 25, b = 150
Equation: y = 25x + 150

2. Gym and badminton court bill

Given points: (10, 800) and (15, 1100)
a = (1100 − 800) ÷ (15 − 10) = 300 ÷ 5 = 60
800 = 60(10) + b
b = 200
Answer: a = 60, b = 200
Equation: y = 60x + 200

3. Celsius and Fahrenheit relation

Given relation: C = aF + b
When F = 32, C = 0
When F = 212, C = 100
So:
0 = 32a + b
100 = 212a + b
Subtracting gives 100 = 180a, so a = 5/9
Then b = −160/9
Answer: a = 5/9 and b = −160/9
Equation:

C = (5/9)F − 160/9 = (5/9)(F − 32)

Exercise set 2.6

1. What the graphs show

(i) y = 4x, y = 2x, y = x
All pass through the origin. Greater positive slope gives a steeper rising line.

(ii) y = −6x, y = −3x, y = −x
All pass through the origin. Greater magnitude of negative slope gives a steeper falling line.

(iii) y = 5x, y = −5x
Both are equally steep, but one rises and the other falls.

(iv) y = 3x − 1, y = 3x, y = 3x + 1
All have the same slope 3, so they are parallel.

(v) y = −2x − 3, y = −2x, y = 2x + 3
The first two are parallel because both have slope −2. The third is not parallel because its slope is positive.

End-of-chapter exercises

1. One degree-3 polynomial with coefficient of x² equal to −7

One example is:

x³ − 7x² + 2x + 1

2. Values of the given polynomials

(i) 5x² − 3x + 7 at x = 1
= 5(1) − 3(1) + 7 = 9

(ii) 4t³ − t² + 6 at t = a
= 4a³ − a² + 6

3. Find the number

Let the number be x.
(5/2)x + 2/3 = −7/12
(5/2)x = −7/12 − 2/3
(5/2)x = −7/12 − 8/12 = −15/12 = −5/4
x = (−5/4) × (2/5) = −1/2
Answer: The number is −1/2

4. Two positive numbers

Let the smaller number be x.
Then the larger number is 5x.
After adding 21 to both, one new number becomes twice the other:
5x + 21 = 2(x + 21)
5x + 21 = 2x + 42
3x = 21
x = 7
Answer: The numbers are 7 and 35

5. Savings after 6 months and 2 years

Amount after n months = 800 + 250n
(i) After 6 months = 800 + 250(6) = 2300
(ii) After 2 years = 24 months
Amount = 800 + 250(24) = 6800
Linear pattern: A = 800 + 250n

6. Two-digit numbers with digits differing by 3

Let the tens digit be x and the ones digit be y.
Original number + reversed number = 143
(10x + y) + (10y + x) = 143
11(x + y) = 143
x + y = 13
Since the digits differ by 3, the digits are 8 and 5.
Answer: The two numbers are 85 and 58

7. Slopes, y-intercepts, and parallel lines

(i) y = −3x + 4
Slope = −3, y-intercept = 4, cuts y-axis at (0, 4)

(ii) 2y = 4x + 7 → y = 2x + 7/2
Slope = 2, y-intercept = 7/2, cuts y-axis at (0, 7/2)

(iii) 5y = 6x − 10 → y = (6/5)x − 2
Slope = 6/5, y-intercept = −2, cuts y-axis at (0, −2)

(iv) 3y = 6x − 11 → y = 2x − 11/3
Slope = 2, y-intercept = −11/3, cuts y-axis at (0, −11/3)

Parallel lines: (ii) and (iv), because both have slope 2.

8. Kelvin and Fahrenheit

Given:

y = (9/5)(x − 273) + 32

(i) If x = 313 K:
y = (9/5)(40) + 32 = 72 + 32 = 104°F

(ii) If y = 158°F:
158 = (9/5)(x − 273) + 32
126 = (9/5)(x − 273)
70 = x − 273
x = 343 K
Answers: 104°F and 343 K

9. Work and distance

Work = force × distance
If force = 3 units, then:

w = 3d

This is the required linear equation.
When d = 2, work done = 3(2) = 6 units.

10. Find the polynomial from two points

The graph passes through (1, 5) and (3, 11).
Slope = (11 − 5) ÷ (3 − 1) = 6 ÷ 2 = 3
So p(x) = 3x + b
Using (1, 5): 5 = 3(1) + b, so b = 2
(i) p(x) = 3x + 2
(ii) y-axis cut = (0, 2)
x-axis cut: 3x + 2 = 0 → x = −2/3, so the point is (−2/3, 0)

11. Find p(x) and q(x)

Let p(x) = ax + b and q(x) = cx + d.
Given p(0) = 5, so b = 5.
Also p(x) + q(x) = 6x + 4, so:
a + c = 6 and b + d = 4
5 + d = 4, so d = −1
p(x) − q(x) cuts the x-axis at (3, 0), so p(3) − q(3) = 0
(3a + 5) − (3c − 1) = 0
3a − 3c + 6 = 0
a − c = −2
Solving: a + c = 6 and a − c = −2
a = 2 and c = 4
Answer: p(x) = 2x + 5 and q(x) = 4x − 1

12. Growing pattern of hexagons with matchsticks

Each new hexagon shares one side with the previous one, so each new stage adds 5 matchsticks.

(i) Stage 4 = 21 matchsticks, Stage 5 = 26 matchsticks

(ii) Table:

Stage number	1	2	3	4	5	n
Matchsticks	6	11	16	21	26	5n + 1

(iii) Rule: M = 5n + 1
(iv) 15th stage: M = 5(15) + 1 = 76
(v) For 200 matchsticks:
5n + 1 = 200
5n = 199
n = 199/5, which is not a whole number.
So 200 matchsticks cannot form a stage in this pattern.

13. Two parallel linear polynomials

For p(x), points are (2, 3) and (6, 11).
Slope = (11 − 3) ÷ (6 − 2) = 8 ÷ 4 = 2
So p(x) = 2x + b
Using (2, 3): 3 = 2(2) + b, so b = −1
p(x) = 2x − 1

q(x) is parallel to p(x), so it has the same slope 2.
q(x) passes through (4, −1):
−1 = 2(4) + d
d = −9
q(x) = 2x − 9

x-axis cuts:
For p(x): 2x − 1 = 0 → x = 1/2, so point is (1/2, 0)
For q(x): 2x − 9 = 0 → x = 9/2, so point is (9/2, 0)

14. What is common in f(x) = ax + a, where a > 0?

Since f(x) = a(x + 1), all such lines have these common properties:

They are increasing lines because a > 0.
The slope and the y-intercept are both equal to a.
All cut the x-axis at the same point: (−1, 0).