Chapter 2: Introduction to Linear Polynomials

The degree of a polynomial is the highest power of the variable in the polynomial.

(i) 2x² – 5x + 3

• The terms are: 2x² (power 2), –5x (power 1), 3 (power 0)
• Highest power = 2
• ∴ Degree = 2

(ii) y³ + 2y – 1

• The terms are: y³ (power 3), 2y (power 1), –1 (power 0)
• Highest power = 3
• ∴ Degree = 3

(iii) –9

• This is a non-zero constant polynomial. It can be written as –9x⁰.
• Highest power = 0
• ∴ Degree = 0

(iv) 4z – 3

• The terms are: 4z (power 1), –3 (power 0)
• Highest power = 1
• ∴ Degree = 1

Key Idea: Any polynomial $p(x) = a_n x^n + \cdots + a_0$ with $a_n \neq 0$ has $\deg = n$.

Reference: Chapter 2, $\text{Page 18}$

$\text{Degree} = \begin{cases} 1 \ (\text{Linear}): & 2x + 5 \\ 2 \ (\text{Quadratic}): & x^2 + 3x + 1 \\ 3 \ (\text{Cubic}): & x^3 - 2x^2 + x + 4 \end{cases}$

$\underbrace{2x+5}_{\deg = 1} \xrightarrow{\text{add higher power}} \underbrace{x^2+3x+1}_{\deg = 2} \xrightarrow{\text{add higher power}} \underbrace{x^3-2x^2+x+4}_{\deg = 3}$

Real fact: Degree-1 models straight lines (simple interest), degree-2 models parabolas (projectile motion), and degree-3 models inflection curves (population growth with limits).

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The coefficient of a term is the numerical factor multiplied with the variable part.

Given polynomial: x⁴ – 3x³ + 6x² – 2x + 7

Identify each term:

Term	Variable Part	Coefficient
x⁴	x⁴	1
–3x³	x³	–3
6x²	x²	6
–2x	x	–2
7	(constant)	7

• Coefficient of x² = 6
• Coefficient of x³ = –3

Given polynomial: 4z³ + 5z² – 11

We need to find the coefficient of z (i.e., the term with z¹).

• Looking at the polynomial: 4z³ + 5z² + 0·z – 11
• There is no term containing z (z to the power 1) in this polynomial.
• This means the coefficient of z is 0.

∴ The coefficient of z = 0

The constant term of a polynomial is the term that does not contain any variable (i.e., the term with x⁰).

Given polynomial: 9x³ + 5x² – 8x – 10

Identify each term:

• 9x³ → contains variable x
• 5x² → contains variable x
• –8x → contains variable x
• –10 → does not contain any variable ✓

∴ The constant term = –10

We substitute each value of x into the polynomial p(x) = 5x – 3.

(i) x = 0

p(0) = 5(0) – 3 = 0 – 3 = –3

(ii) x = –1

p(–1) = 5(–1) – 3 = –5 – 3 = –8

(iii) x = 2

p(2) = 5(2) – 3 = 10 – 3 = 7

We substitute each value of s into the polynomial p(s) = 7s² – 4s + 6.

(i) s = 0

p(0) = 7(0)² – 4(0) + 6 = 0 – 0 + 6 = 6

(ii) s = –3

p(–3) = 7(–3)² – 4(–3) + 6 = 7 × 9 + 12 + 6 = 63 + 12 + 6 = 81

(iii) s = 4

p(4) = 7(4)² – 4(4) + 6 = 7 × 16 – 16 + 6 = 112 – 16 + 6 = 102

Let Salil's present age = x years

Then, Salil's mother's present age = 3x years

After 5 years:

• Salil's age = x + 5
• Mother's age = 3x + 5

Setting up the equation:

(x + 5) + (3x + 5) = 70

4x + 10 = 70

4x = 70 – 10 = 60

x = 60 ÷ 4 = 15

Therefore:

• Salil's present age = 15 years
• Salil's mother's present age = 3 × 15 = 45 years

Verification: After 5 years, Salil = 20, Mother = 50. Sum = 70 ✓

Let the two positive integers be in the ratio 2:5.

So, let the integers be 2k and 5k for some positive integer k.

Given: The difference between the two integers is 63.

5k – 2k = 63

3k = 63

k = 63 ÷ 3 = 21

Therefore:

• Smaller integer = 2 × 21 = 42
• Larger integer = 5 × 21 = 105

Verification: 105 – 42 = 63 ✓ and 42:105 = 2:5 ✓

Let the number of five-rupee coins = x

Then, the number of two-rupee coins = 3x

Total amount:

Value from five-rupee coins = 5x

Value from two-rupee coins = 2 × 3x = 6x

Setting up the equation:

5x + 6x = 88

11x = 88

x = 88 ÷ 11 = 8

Therefore:

• Number of five-rupee coins = 8
• Number of two-rupee coins = 3 × 8 = 24

Verification: 8 × 5 + 24 × 2 = 40 + 48 = ₹88 ✓

Let the length of the shorter piece = x feet

Then, the length of the longer piece = 4x feet

Setting up the equation:

x + 4x = 300

5x = 300

x = 300 ÷ 5 = 60

Therefore:

• Length of the shorter piece = 60 feet
• Length of the longer piece = 4 × 60 = 240 feet

Verification: 60 + 240 = 300 ✓ and 240 = 4 × 60 ✓

Let the width of the rectangle = w cm

Then, the length = (2w + 3) cm

Perimeter of a rectangle = 2(length + width)

2(2w + 3 + w) = 24

2(3w + 3) = 24

6w + 6 = 24

6w = 24 – 6 = 18

w = 18 ÷ 6 = 3

Therefore:

• Width = 3 cm
• Length = 2(3) + 3 = 6 + 3 = 9 cm

Verification: Perimeter = 2(9 + 3) = 2 × 12 = 24 cm ✓

Step-by-Step Solution

Given Information:

• Initial amount = ₹500
• Monthly pocket money = ₹150

Building the pattern:

Month (n)	Calculation	Amount (₹)
1	500 + 1 × 150	650
2	500 + 2 × 150	800
3	500 + 3 × 150	950
4	500 + 4 × 150	1100

Finding the pattern:

At the end of the nth month, the amount = Initial amount + (n × monthly pocket money)

$A(n) = 500 + 150n$

Linear Expression: A(n) = 500 + 150n

This is a linear growth pattern, where the amount increases by a fixed value of ₹150 each month.

Step-by-Step Solution

Given Information:

• Initial members = 120
• Members dropping out each hour = 9

Building the pattern:

Hour (n)	Calculation	Members remaining
1	120 − 1 × 9	111
2	120 − 2 × 9	102
3	120 − 3 × 9	93
4	120 − 4 × 9	84

Finding the pattern:

At the end of the nth hour, the members remaining = Initial members − (n × members dropping per hour)

$M(n) = 120 - 9n$

Linear Expression: M(n) = 120 − 9n

This is a linear decay pattern, where the number of members decreases by a fixed value of 9 each hour.

Step-by-Step Solution

Given Information:

• Length = 13 cm (fixed)
• Breadth varies: 12 cm, 10 cm, 8 cm
• Area of rectangle = Length × Breadth

Calculating the areas:

(i) Breadth = 12 cm: $\text{Area} = 13 \times 12 = 156 \text{ cm}^2$

(ii) Breadth = 10 cm: $\text{Area} = 13 \times 10 = 130 \text{ cm}^2$

(iii) Breadth = 8 cm: $\text{Area} = 13 \times 8 = 104 \text{ cm}^2$

Observing the pattern:

Breadth (b) cm	8	10	12
Area (cm²)	104	130	156

As the breadth decreases by 2 cm, the area decreases by 26 cm².

Linear Pattern:

Let b = breadth of the rectangle.

$A(b) = 13b$

This is a linear expression in b. As the breadth increases by 1 cm, the area increases by 13 cm² (a constant amount), confirming it is a linear pattern.

Step-by-Step Solution

Given Information:

• Length = 7 cm (fixed)
• Breadth = 11 cm (fixed)
• Base area = 7 × 11 = 77 cm²
• Volume = Length × Breadth × Height = 77 × Height

Calculating the volumes:

(i) Height = 5 cm: $V = 7 \times 11 \times 5 = 77 \times 5 = 385 \text{ cm}^3$

(ii) Height = 9 cm: $V = 7 \times 11 \times 9 = 77 \times 9 = 693 \text{ cm}^3$

(iii) Height = 13 cm: $V = 7 \times 11 \times 13 = 77 \times 13 = 1001 \text{ cm}^3$

Observing the pattern:

Height (h) cm	5	9	13
Volume (cm³)	385	693	1001

As height increases by 4 cm, volume increases by 308 cm³.

Linear Pattern:

Let h = height of the rectangular box.

$V(h) = 77h$

This is a linear expression in h. As height increases by 1 cm, volume increases by a fixed amount of 77 cm³, confirming this is a linear growth pattern.

Step-by-Step Solution

Given Information:

• Total pages = 500
• Pages read per day = 20

Building the linear pattern:

Let d = number of days elapsed.

Pages read after d days = 20d

Pages remaining after d days: $P(d) = 500 - 20d$

Calculating pages left after specific days:

Day (d)	Pages Remaining
1	500 − 20 = 480
2	500 − 40 = 460
5	500 − 100 = 400
10	500 − 200 = 300
15	500 − 300 = 200

Pages left after 15 days: $P(15) = 500 - 20 \times 15 = 500 - 300 = \mathbf{200 \text{ pages}}$

Linear Pattern: P(d) = 500 − 20d

This is a linear decay pattern — as each day passes, the number of remaining pages decreases by a constant 20 pages.

Solution

Part (i): Height after 7 months

Given:

• Initial height = 1.75 feet
• Growth rate = 0.5 feet per month
• Time, t = 7 months

Calculation: $h = 1.75 + 0.5 \times t$ $h = 1.75 + 0.5 \times 7$ $h = 1.75 + 3.5 = \mathbf{5.25 \text{ feet}}$

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Part (ii): Table of values (t = 0 to 10)

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

• The height increases by a constant 0.5 feet every month.

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Part (iii): Expression relating h and t

$\boxed{h = 0.5t + 1.75}$

Why it represents linear growth:

• This is of the form h = at + b, where a = 0.5 and b = 1.75.
• The variable t appears with power 1 (no t², no t³, etc.), making it a linear polynomial.
• The height increases by a fixed (constant) amount of 0.5 feet each month — this constant rate of change is the defining feature of linear growth.
• On a graph, this would produce a straight line with slope 0.5 and y-intercept 1.75.

Solution

Part (i): Value after 3 years

Given:

• Initial value = ₹10,000
• Depreciation rate = ₹800 per year
• Time, t = 3 years

Calculation: $v = 10000 - 800 \times t$ $v = 10000 - 800 \times 3$ $v = 10000 - 2400 = \mathbf{₹7600}$

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Part (ii): Table of values (t = 0 to 8)

Year (t)	Value v (₹)
0	10,000
1	9,200
2	8,400
3	7,600
4	6,800
5	6,000
6	5,200
7	4,400
8	3,600

• The value decreases by a constant ₹800 every year.

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Part (iii): Expression relating v and t

$\boxed{v = 10000 - 800t}$

Why it represents linear decay:

• This is of the form v = at + b, where a = −800 and b = 10,000.
• The variable t has power 1, making it a linear polynomial.
• The value decreases by a fixed (constant) amount of ₹800 each year — this constant rate of decrease is the defining feature of linear decay.
• On a graph, this produces a straight line with a negative slope (−800), showing a steady decrease over time.

Solution

Part (i): Population after 6 years

Given:

• Initial population = 750
• Rate of increase = 50 people per year
• Time, t = 6 years

Calculation: $P = 750 + 50 \times t$ $P = 750 + 50 \times 6$ $P = 750 + 300 = \mathbf{1050}$

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Part (ii): Table of values (t = 0 to 10)

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

• The population increases by a constant 50 people every year.

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Part (iii): Expression relating P and t

$\boxed{P = 50t + 750}$

Why it represents linear growth:

• This is of the form P = at + b, where a = 50 and b = 750.
• The variable t appears with power 1, making this a linear polynomial.
• The population increases by a fixed (constant) amount of 50 people each year — this constant rate of change is the hallmark of linear growth.
• On a graph, this produces a straight line with slope 50 and y-intercept 750.

Solution

Part (i): Equation for remaining balance b(x)

Given:

• Initial balance = ₹600
• Daily reduction = ₹15
• x = number of days

$\boxed{b(x) = 600 - 15x}$

Why it represents linear decay:

• This is of the form b = ax + c, where a = −15 and c = 600.
• The variable x has power 1, so this is a linear polynomial.
• The balance decreases by a fixed (constant) amount of ₹15 every day — a constant rate of decrease is the defining property of linear decay.
• On a graph, this is a straight line with a negative slope (−15).

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Part (ii): Number of days for balance to run out

The balance runs out when b(x) = 0: $600 - 15x = 0$ $15x = 600$ $x = \frac{600}{15} = \mathbf{40 \text{ days}}$

The balance will run out after 40 days.

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Part (iii): Table of values (x = 1 to 10)

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

• The balance decreases by ₹15 each day, confirming the constant rate of linear decay.

Solution

We are given that the monthly bill follows the relation y = ax + b.

Step 1: Set up the two equations using the given information.

• When x = 10, y = 400: $10a + b = 400 \quad \text{...(1)}$

• When x = 14, y = 500: $14a + b = 500 \quad \text{...(2)}$

Step 2: Subtract equation (1) from equation (2) to eliminate b.

$(14a + b) - (10a + b) = 500 - 400$ $4a = 100$ $a = 25$

Step 3: Substitute a = 25 into equation (1) to find b.

$10(25) + b = 400$ $250 + b = 400$ $b = 150$

Step 4: Write the final relation.

$y = 25x + 150$

Conclusion:

• a = 25 (cost per module = ₹25)
• b = 150 (fixed monthly fee = ₹150)

Solution

We are given that the monthly bill follows the relation y = ax + b.

Step 1: Set up the two equations using the given information.

• When x = 10, y = 800: $10a + b = 800 \quad \text{...(1)}$

• When x = 15, y = 1100: $15a + b = 1100 \quad \text{...(2)}$

Step 2: Subtract equation (1) from equation (2) to eliminate b.

$(15a + b) - (10a + b) = 1100 - 800$ $5a = 300$ $a = 60$

Step 3: Substitute a = 60 into equation (1) to find b.

$10(60) + b = 800$ $600 + b = 800$ $b = 200$

Step 4: Write the final relation.

$y = 60x + 200$

Conclusion:

• a = 60 (cost per hour for badminton court = ₹60)
• b = 200 (fixed monthly fee = ₹200)

Solution

We are given that °C = a·°F + b.

Step 1: Set up the two equations using the given conditions.

• When °C = 0, °F = 32 (ice melts): $0 = 32a + b \quad \text{...(1)}$

• When °C = 100, °F = 212 (water boils): $100 = 212a + b \quad \text{...(2)}$

Step 2: Subtract equation (1) from equation (2) to eliminate b.

$100 - 0 = 212a - 32a$ $100 = 180a$ $a = \frac{100}{180} = \frac{5}{9}$

Step 3: Substitute a = 5/9 into equation (1) to find b.

$0 = 32 \times \frac{5}{9} + b$ $0 = \frac{160}{9} + b$ $b = -\frac{160}{9}$

Step 4: Write the final relation.

$°C = \frac{5}{9} \cdot °F - \frac{160}{9}$

This can also be written as: $°C = \frac{5}{9}(°F - 32)$

Verification:

• When °F = 32: °C = (5/9)(32 − 32) = 0 ✓
• When °F = 212: °C = (5/9)(212 − 32) = (5/9)(180) = 100 ✓

Conclusion:

• a = 5/9
• b = −160/9

Drawing the Graphs of y = 4x, y = 2x, y = x**

Step 1: Prepare a table of values for each line.

Since all three lines pass through the origin (0, 0), we choose a few values of x:

x	y = x	y = 2x	y = 4x
0	0	0	0
1	1	2	4
2	2	4	8
-1	-1	-2	-4

Step 2: Plot the points and draw each line.

• Plot points for y = x: (0,0), (1,1), (2,2), (-1,-1) → draw a straight line.
• Plot points for y = 2x: (0,0), (1,2), (2,4), (-1,-2) → draw a straight line.
• Plot points for y = 4x: (0,0), (1,4), (2,8), (-1,-4) → draw a straight line.

Step 3: Observe the role of 'a' in y = ax.

• All three lines pass through the origin (0, 0).
• The value of 'a' is the slope (steepness) of the line.
• As 'a' increases (from 1 to 2 to 4), the line becomes steeper (closer to the y-axis).
• Since all values of 'a' are positive, all lines slope upward from left to right.
• Conclusion: In y = ax, the coefficient 'a' controls the slope/steepness of the line. Larger positive 'a' → steeper line.

Drawing the Graphs of y = –6x, y = –3x, y = –x

Step 1: Prepare a table of values for each line.

x	y = –x	y = –3x	y = –6x
0	0	0	0
1	-1	-3	-6
2	-2	-6	-12
-1	1	3	6

Step 2: Plot the points and draw each line.

• y = –x: (0,0), (1,–1), (–1,1) → line slopes downward.
• y = –3x: (0,0), (1,–3), (–1,3) → steeper downward slope.
• y = –6x: (0,0), (1,–6), (–1,6) → even steeper downward slope.

Step 3: Observe the role of 'a'.

• All lines pass through the origin (0, 0).
• Since 'a' is negative, all lines slope downward from left to right.
• As the magnitude of 'a' increases (1 → 3 → 6), the line becomes steeper.
• Conclusion: A negative value of 'a' gives a downward-sloping line. The larger the magnitude of negative 'a', the steeper the line.

Drawing the Graphs of y = 5x and y = –5x

Step 1: Prepare a table of values.

x	y = 5x	y = –5x
0	0	0
1	5	-5
2	10	-10
-1	-5	5
-2	-10	10

Step 2: Plot the points and draw each line.

• y = 5x: passes through (0,0), (1,5), (–1,–5) → slopes steeply upward.
• y = –5x: passes through (0,0), (1,–5), (–1,5) → slopes steeply downward.

Step 3: Observe the role of 'a'.

• Both lines pass through the origin and have the same steepness (|a| = 5).
• They are mirror images of each other about the x-axis (or y-axis).
• y = 5x rises from left to right (positive slope).
• y = –5x falls from left to right (negative slope).
• Conclusion: Changing the sign of 'a' reflects the line about the x-axis while keeping the steepness the same.

Drawing the Graphs of y = 3x – 1, y = 3x, y = 3x + 1

Step 1: Prepare a table of values.

x	y = 3x – 1	y = 3x	y = 3x + 1
-1	-4	-3	-2
0	-1	0	1
1	2	3	4
2	5	6	7

Step 2: Plot the points and draw each line.

• y = 3x – 1: crosses y-axis at (0, –1).
• y = 3x: passes through origin (0, 0).
• y = 3x + 1: crosses y-axis at (0, 1).

Step 3: Observe the role of 'b'.

• All three lines have the same slope (a = 3), so they are parallel to each other.
• The value of 'b' is the y-intercept — the point where the line crosses the y-axis.
• Increasing 'b' shifts the line upward (parallel shift).
• Decreasing 'b' shifts the line downward (parallel shift).
• Conclusion: In y = ax + b, the value of 'b' shifts the line up or down without changing its slope. Parallel lines have the same 'a' but different 'b'.

Drawing the Graphs of y = –2x – 3, y = –2x, y = 2x + 3

Step 1: Prepare a table of values.

x	y = –2x – 3	y = –2x	y = 2x + 3
-2	1	4	-1
-1	-1	2	1
0	-3	0	3
1	-5	-2	5
2	-7	-4	7

Step 2: Plot the points and draw each line.

• y = –2x: passes through (0,0), slopes downward with slope –2.
• y = –2x – 3: parallel to y = –2x but shifted down by 3 (y-intercept = –3).
• y = 2x + 3: slopes upward with slope 2, y-intercept = 3.

Step 3: Observe the role of 'a' and 'b'.

• y = –2x and y = –2x – 3 are parallel (same slope a = –2, different b).
  • 'b' = –3 shifts the line down by 3 units.
• y = –2x – 3 and y = 2x + 3 have opposite slopes (–2 vs +2) and are reflections of each other about the y-axis in terms of direction.
  • Note that y = 2x + 3 is actually the reflection of y = –2x – 3 in a sense — same magnitude of slope but opposite sign, and 'b' changes from –3 to +3.
• Role of 'a': Determines the direction and steepness of the slope.
• Role of 'b': Determines the y-intercept (vertical position) of the line.
• Conclusion: 'a' controls slope direction and steepness; 'b' controls the vertical shift. Lines with the same 'a' are parallel. Changing the sign of 'a' reflects the line's direction.

A polynomial of degree 3 has the highest power of x equal to 3.

The general form of a degree-3 polynomial is: $p(x) = ax^3 + bx^2 + cx + d$ where $a \neq 0$.

Given condition: The coefficient of $x^2$ is $-7$, so $b = -7$.

Choosing simple values for the other coefficients (e.g., $a = 1$, $c = 2$, $d = 5$):

$\boxed{p(x) = x^3 - 7x^2 + 2x + 5}$

Note: Many correct answers are possible. Any polynomial of the form $ax^3 - 7x^2 + cx + d$ where $a \neq 0$ is acceptable.

(i) Finding the value of 5x² – 3x + 7 at x = 1:

Substitute $x = 1$:

$p(1) = 5(1)^2 - 3(1) + 7$

$= 5(1) - 3 + 7$

$= 5 - 3 + 7$

$= \boxed{9}$

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(ii) Finding the value of 4t³ – t² + 6 at t = a:

Substitute $t = a$:

$p(a) = 4(a)^3 - (a)^2 + 6$

$= \boxed{4a^3 - a^2 + 6}$

Let the number be $x$.

According to the problem: $\frac{5}{2} \cdot x + \frac{2}{3} = -\frac{7}{12}$

Step 1: Isolate the term with $x$: $\frac{5}{2}x = -\frac{7}{12} - \frac{2}{3}$

Step 2: Find a common denominator for the right side (LCD = 12): $-\frac{7}{12} - \frac{2}{3} = -\frac{7}{12} - \frac{8}{12} = -\frac{15}{12} = -\frac{5}{4}$

Step 3: Solve for $x$: $\frac{5}{2}x = -\frac{5}{4}$ $x = -\frac{5}{4} \times \frac{2}{5} = -\frac{10}{20} = -\frac{1}{2}$

Verification: $\frac{5}{2} \times \left(-\frac{1}{2}\right) + \frac{2}{3} = -\frac{5}{4} + \frac{2}{3} = -\frac{15}{12} + \frac{8}{12} = -\frac{7}{12}$ ✓

$\boxed{x = -\dfrac{1}{2}}$

Let the smaller number be $x$.

Then the larger number is $5x$ (since it is 5 times the other).

After adding 21 to both:

• Smaller number becomes: $x + 21$
• Larger number becomes: $5x + 21$

Setting up the equation:

One of the new numbers becomes twice the other. Since the larger number is $5x + 21$, it is reasonable that: $5x + 21 = 2(x + 21)$

Solving: $5x + 21 = 2x + 42$ $5x - 2x = 42 - 21$ $3x = 21$ $x = 7$

So the two numbers are:

• Smaller number: $x = 7$
• Larger number: $5x = 35$

Verification:

• After adding 21: $7 + 21 = 28$ and $35 + 21 = 56$
• Is $56 = 2 \times 28$? Yes! ✓

$\boxed{\text{The two numbers are } 7 \text{ and } 35.}$

Setting up the linear pattern:

Let $n$ = number of months, and $A$ = total amount after $n$ months.

You start with ₹800 and save ₹250 every month, so: $A = 800 + 250n$

This is a linear equation in $n$ with slope 250 (monthly savings) and y-intercept 800 (initial amount).

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(i) After 6 months ($n = 6$): $A = 800 + 250 \times 6 = 800 + 1500 = ₹2300$

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(ii) After 2 years ($n = 24$ months): $A = 800 + 250 \times 24 = 800 + 6000 = ₹6800$

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Linear Pattern Table:

Month (n)	Amount (₹)
0	800
1	1050
2	1300
6	2300
24	6800

$\boxed{A = 800 + 250n}$

Setting up the problem:

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

• Original number = $10x + y$
• Reversed number = $10y + x$

Condition 1: The digits differ by 3: $x - y = 3 \quad \text{...(1)}$ (assuming $x > y$; we'll check the other case too)

Condition 2: Sum of original and reversed numbers = 143: $(10x + y) + (10y + x) = 143$ $11x + 11y = 143$ $x + y = 13 \quad \text{...(2)}$

Solving equations (1) and (2):

Adding (1) and (2): $2x = 16 \Rightarrow x = 8$

Substituting in (2): $8 + y = 13 \Rightarrow y = 5$

The two numbers:

• Original number: $10(8) + 5 = 85$
• Reversed number: $10(5) + 8 = 58$

Verification: $85 + 58 = 143$ ✓ and $8 - 5 = 3$ ✓

If $y > x$, then $y - x = 3$ gives $y = 8, x = 5$, so the original number is 58 and the reversed number is 85 — the same pair.

$\boxed{\text{The two numbers are } 85 \text{ and } 58.}$

To graph each line, we convert it to the slope-intercept form $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

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(i) $y = -3x + 4$

Already in slope-intercept form.

• Slope $m = -3$
• y-intercept $b = 4$ → cuts y-axis at (0, 4)

Points for graph:

x	y = –3x + 4
0	4
1	1
2	–2

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(ii) $2y = 4x + 7$

Divide both sides by 2: $y = 2x + \frac{7}{2}$

• Slope $m = 2$
• y-intercept $b = \frac{7}{2} = 3.5$ → cuts y-axis at $\left(0, \dfrac{7}{2}\right)$

Points for graph:

x	y = 2x + 3.5
0	3.5
1	5.5
–1	1.5

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(iii) $5y = 6x – 10$

Divide both sides by 5: $y = \frac{6}{5}x - 2$

• Slope $m = \dfrac{6}{5}$
• y-intercept $b = -2$ → cuts y-axis at (0, –2)

Points for graph:

x	y = (6/5)x – 2
0	–2
5	4
–5	–8

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(iv) $3y = 6x – 11$

Divide both sides by 3: $y = 2x - \frac{11}{3}$

• Slope $m = 2$
• y-intercept $b = -\dfrac{11}{3}$ → cuts y-axis at $\left(0, -\dfrac{11}{3}\right)$

Points for graph:

x	y = 2x – 11/3
0	–3.67
2	0.33
3	2.33

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Summary Table:

Equation	Slope	y-intercept	y-axis point
y = –3x + 4	–3	4	(0, 4)
2y = 4x + 7	2	7/2	(0, 7/2)
5y = 6x – 10	6/5	–2	(0, –2)
3y = 6x – 11	2	–11/3	(0, –11/3)

Are any lines parallel?

Lines (ii) and (iv) both have slope = 2 but different y-intercepts ($\frac{7}{2}$ and $-\frac{11}{3}$).

$\boxed{\text{Lines (ii) } 2y = 4x + 7 \text{ and (iv) } 3y = 6x - 11 \text{ are parallel.}}$

Given: $y = \dfrac{9}{5}(x - 273) + 32$

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(i) Find y (°F) when x = 313 K:

$y = \frac{9}{5}(313 - 273) + 32$ $= \frac{9}{5}(40) + 32$ $= \frac{360}{5} + 32$ $= 72 + 32$ $= \boxed{104 \text{ °F}}$

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(ii) Find x (K) when y = 158 °F:

$158 = \frac{9}{5}(x - 273) + 32$

Step 1: Subtract 32 from both sides: $158 - 32 = \frac{9}{5}(x - 273)$ $126 = \frac{9}{5}(x - 273)$

Step 2: Multiply both sides by $\frac{5}{9}$: $x - 273 = 126 \times \frac{5}{9} = \frac{630}{9} = 70$

Step 3: Solve for x: $x = 273 + 70 = \boxed{343 \text{ K}}$

Setting up the linear equation:

Work done = Force × Distance $w = F \cdot d$

With constant force $F = 3$ units: $\boxed{w = 3d}$

This is a linear equation in two variables $w$ and $d$, passing through the origin with slope 3.

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Table of values for the graph:

d (distance)	w = 3d (work)
0	0
1	3
2	6
3	9
4	12

Graph: Plot $d$ on the x-axis and $w$ on the y-axis. The graph is a straight line passing through the origin (0, 0) with slope 3.

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Work done when d = 2 units: $w = 3 \times 2 = \boxed{6 \text{ units of work}}$

Verification from graph: The point (2, 6) lies on the line $w = 3d$. ✓

Let $p(x) = ax + b$.

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(i) Finding p(x):

Using point (1, 5): $a(1) + b = 5 \Rightarrow a + b = 5 \quad \text{...(1)}$

Using point (3, 11): $a(3) + b = 11 \Rightarrow 3a + b = 11 \quad \text{...(2)}$

Subtracting (1) from (2): $2a = 6 \Rightarrow a = 3$

Substituting in (1): $3 + b = 5 \Rightarrow b = 2$

$\boxed{p(x) = 3x + 2}$

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(ii) Finding where p(x) cuts the axes:

y-axis (x = 0): $p(0) = 3(0) + 2 = 2$ → Cuts y-axis at (0, 2)

x-axis (p(x) = 0): $3x + 2 = 0 \Rightarrow x = -\frac{2}{3}$ → Cuts x-axis at $\left(-\dfrac{2}{3}, 0\right)$

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(iii) Graph verification:

Plot the following points and draw the line:

x	p(x) = 3x + 2
–1	–1
0	2
1	5
3	11

• The line passes through (1, 5) and (3, 11) ✓
• The line cuts y-axis at (0, 2) ✓
• The line cuts x-axis at (–2/3, 0) ✓

Given:

• $p(x) = ax + b$, $q(x) = cx + d$

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From condition (i): $p(0) = 5$ $a(0) + b = 5 \Rightarrow b = 5$

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From condition (iii): $p(x) + q(x) = 6x + 4$ $(ax + b) + (cx + d) = 6x + 4$ $(a + c)x + (b + d) = 6x + 4$

Comparing coefficients:

• $a + c = 6 \quad \text{...(1)}$
• $b + d = 4 \quad \text{...(2)}$

Since $b = 5$, from (2): $5 + d = 4 \Rightarrow d = -1$

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From condition (ii): $p(x) - q(x)$ cuts the x-axis at $(3, 0)$

$p(x) - q(x) = (ax + b) - (cx + d) = (a - c)x + (b - d)$

At x = 3, this equals 0: $(a - c)(3) + (b - d) = 0$ $3(a - c) + (5 - (-1)) = 0$ $3(a - c) + 6 = 0$ $a - c = -2 \quad \text{...(3)}$

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Solving (1) and (3):

Adding: $2a = 4 \Rightarrow a = 2$

Substituting in (1): $2 + c = 6 \Rightarrow c = 4$

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The polynomials are: $\boxed{p(x) = 2x + 5 \quad \text{and} \quad q(x) = 4x - 1}$

Verification:

• $p(0) = 5$ ✓
• $p(x) - q(x) = (2x + 5) - (4x - 1) = -2x + 6$; at $x = 3$: $-2(3) + 6 = 0$ ✓
• $p(x) + q(x) = (2x + 5) + (4x - 1) = 6x + 4$ ✓

Understanding the pattern:

• Stage 1: 1 hexagon → 6 matchsticks
• Stage 2: 2 hexagons → when a new hexagon shares 1 side, we add 5 matchsticks (6 – 1 shared)
• Stage 3: 3 hexagons → add 5 more matchsticks

So the pattern is: start with 6, then add 5 for each new hexagon.

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(i) Next two stages:

• Stage 4: 4 hexagons in a row — draw a chain of 4 hexagons.
  • Matchsticks = 6 + 3 × 5 = 6 + 15 = 21
• Stage 5: 5 hexagons in a row — draw a chain of 5 hexagons.
  • Matchsticks = 6 + 4 × 5 = 6 + 20 = 26

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(ii) Completed table:

Stage Number (n)	1	2	3	4	5	...	n
Matchsticks	6	11	16	21	26	...	5n + 1

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(iii) Rule for the nth stage:

• Stage 1: $6 = 5(1) + 1$
• Stage 2: $11 = 5(2) + 1$
• Stage 3: $16 = 5(3) + 1$

$\boxed{\text{Number of matchsticks} = 5n + 1}$

This is a linear polynomial in $n$.

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(iv) Matchsticks for Stage 15: $= 5(15) + 1 = 75 + 1 = \boxed{76 \text{ matchsticks}}$

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(v) Can 200 matchsticks form a stage?

We need to check if $5n + 1 = 200$ has a whole number solution: $5n = 199 \Rightarrow n = \frac{199}{5} = 39.8$

Since $n = 39.8$ is not a natural number, 200 matchsticks cannot form a stage in this pattern.

The nearest stages would use $5(39) + 1 = 196$ or $5(40) + 1 = 201$ matchsticks.

Finding p(x):

Let $p(x) = ax + b$.

Using point (2, 3): $2a + b = 3 \quad \text{...(1)}$

Using point (6, 11): $6a + b = 11 \quad \text{...(2)}$

Subtracting (1) from (2): $4a = 8 \Rightarrow a = 2$

Substituting in (1): $4 + b = 3 \Rightarrow b = -1$

$\boxed{p(x) = 2x - 1}$

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Finding q(x):

Since $q(x)$ is parallel to $p(x)$, they have the same slope: $c = 2$

So $q(x) = 2x + d$.

Using point (4, –1): $2(4) + d = -1$ $8 + d = -1$ $d = -9$

$\boxed{q(x) = 2x - 9}$

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Where do they meet the x-axis?

For p(x) = 0: $2x - 1 = 0 \Rightarrow x = \frac{1}{2}$ → p(x) meets x-axis at $\left(\dfrac{1}{2}, 0\right)$

For q(x) = 0: $2x - 9 = 0 \Rightarrow x = \frac{9}{2}$ → q(x) meets x-axis at $\left(\dfrac{9}{2}, 0\right)$

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Summary:

• $p(x) = 2x - 1$ meets the x-axis at $\left(\frac{1}{2}, 0\right)$
• $q(x) = 2x - 9$ meets the x-axis at $\left(\frac{9}{2}, 0\right)$
• Both lines have slope 2 and are parallel to each other.

Analysing f(x) = ax + a, where a > 0:

We can factor: $f(x) = a(x + 1)$

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Key observations:

1. Common x-intercept (zero of the function):

Set $f(x) = 0$: $a(x + 1) = 0$ Since $a > 0$, we have $a \neq 0$, so: $x + 1 = 0 \Rightarrow x = -1$

All such lines pass through the point $(-1, 0)$ on the x-axis.

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2. Slope: The slope of $f(x) = ax + a$ is $a$, which is positive (since $a > 0$).

Different values of $a$ give different slopes, so these are not parallel in general.

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3. Verification with examples:

a	f(x)	At x = –1
1	x + 1	0 ✓
2	2x + 2	–2 + 2 = 0 ✓
3	3x + 3	–3 + 3 = 0 ✓
5	5x + 5	–5 + 5 = 0 ✓

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$\boxed{\text{All lines of the form } f(x) = ax + a \text{ (}a > 0\text{) pass through the common point } (-1, 0).}$

They all share the same x-intercept at x = –1, meaning they all meet the x-axis at the same point (–1, 0).