Introduction to Linear Polynomials — Class 9 Maths Chapter 2 Notes

Example 1: Raju's Pen & Pencil Purchases

Raju buys \( x \) red boxes (with 4 pens each) and \( y \) blue boxes (with 5 pencils each), plus he gets 3 free pens. The total quantity is represented by:

• Terms: \( 4x \), \( 5y \), and \( 3 \) (the individual parts added together).
• Variables: \( x \) and \( y \).
• Coefficients: 4 (coefficient of \( x \)) and 5 (coefficient of \( y \)).
• Constant: 3 (does not change value).

1. Find the degrees of the following polynomials:
(i) \( 2x^2 - 5x + 3 \): Highest power of \( x \) is 2. Degree = 2 (Quadratic).
(ii) \( y^3 + 2y - 1 \): Highest power of \( y \) is 3. Degree = 3 (Cubic).
(iii) \( -9 \): Constant number, can be written as \( -9x^0 \). Degree = 0 (Constant).
(iv) \( 4z - 3 \): Highest power of \( z \) is 1. Degree = 1 (Linear).

2. Write polynomials of degrees 1, 2 and 3:
• Degree 1: \( 7x - 4 \)
• Degree 2: \( 3x^2 + 2x - 5 \)
• Degree 3: \( 2x^3 - x^2 + 8 \)

3. Coefficients of \( x^2 \) and \( x^3 \) in \( x^4 - 3x^3 + 6x^2 - 2x + 7 \):
• Coefficient of \( x^3 \) is \( -3 \).
• Coefficient of \( x^2 \) is \( 6 \).

4. Coefficient of \( z \) in \( 4z^3 + 5z^2 - 11 \):
Since there is no \( z \) term, it can be written as \( 0z \). Coefficient of \( z \) is \( 0 \).

5. Constant term of \( 9x^3 + 5x^2 - 8x - 10 \):
The standalone numerical term is \( -10 \).

1. Value of linear polynomial \( 5x - 3 \):
(i) \( x = 0 \implies 5(0) - 3 = \mathbf{-3} \)
(ii) \( x = -1 \implies 5(-1) - 3 = -5 - 3 = \mathbf{-8} \)
(iii) \( x = 2 \implies 5(2) - 3 = 10 - 3 = \mathbf{7} \)

2. Value of quadratic polynomial \( 7s^2 - 4s + 6 \):
(i) \( s = 0 \implies 7(0)^2 - 4(0) + 6 = \mathbf{6} \)
(ii) \( s = -3 \implies 7(-3)^2 - 4(-3) + 6 = 7(9) + 12 + 6 = 63 + 18 = \mathbf{81} \)
(iii) \( s = 4 \implies 7(4)^2 - 4(4) + 6 = 7(16) - 16 + 6 = 112 - 10 = \mathbf{102} \)

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

4. Integers with ratio \( 2:5 \) and difference 63:
Let integers be \( 2x \) and \( 5x \).
\( 5x - 2x = 63 \implies 3x = 63 \implies x = 21 \).
Integers are \( 2(21) = \mathbf{42} \) and \( 5(21) = \mathbf{105} \).

5. Ruby's Coins (₹2 and ₹5 coins total ₹88):
Let ₹5 coins = \( x \). Then ₹2 coins = \( 3x \).
Value: \( 5(x) + 2(3x) = 88 \implies 5x + 6x = 88 \implies 11x = 88 \implies x = 8 \).
₹5 coins = 8, ₹2 coins = 24. Total coins = 32 coins.

6. Fence cut into 4:1 ratio (Total 300 ft):
Shorter = \( x \), Longer = \( 4x \).
\( x + 4x = 300 \implies 5x = 300 \implies x = 60 \text{ ft} \).
Shorter piece = 60 ft, Longer piece = 240 ft.

7. Rectangle Dimensions (Length = \( 2w + 3 \), Perimeter = 24 cm):
\( 2(l + w) = 24 \implies (2w + 3) + w = 12 \implies 3w + 3 = 12 \implies 3w = 9 \implies w = 3 \text{ cm} \).
Length \( l = 2(3) + 3 = 9 \text{ cm} \). Dimensions: 9 cm × 3 cm.

1. Savings Bank Account (₹500 initial, ₹150/month pocket money):
At end of Month \( n \): Amount = \( 500 + 150n \) rupees.

2. Rally Member Dropout (120 initial, 9 drop out/hour):
Remaining members after \( n \) hours = \( 120 - 9n \).

3. Area of Rectangle (Length = 13 cm):
Formula: Area = \( 13 \times b \).
(i) \( b = 12 \text{ cm} \implies \mathbf{156 \text{ cm}^2} \)
(ii) \( b = 10 \text{ cm} \implies \mathbf{130 \text{ cm}^2} \)
(iii) \( b = 8 \text{ cm} \implies \mathbf{104 \text{ cm}^2} \)
Linear pattern for area: \( \text{Area}(b) = 13b \).

4. Volume of Box (Length 7 cm, Breadth 11 cm):
Base area = \( 7 \times 11 = 77 \text{ cm}^2 \). Volume = \( 77h \).
(i) \( h = 5 \text{ cm} \implies \mathbf{385 \text{ cm}^3} \)
(ii) \( h = 9 \text{ cm} \implies \mathbf{693 \text{ cm}^3} \)
(iii) \( h = 13 \text{ cm} \implies \mathbf{1001 \text{ cm}^3} \)
Linear pattern for volume: \( V(h) = 77h \).

5. Book Reading Pages Left (500 total, 20 pages/day):
Pages left after \( n \) days: \( L(n) = 500 - 20n \).
After 15 days: \( 500 - 20(15) = 500 - 300 = \mathbf{200 \text{ pages left}} \).

1. Plant Growth (Initial 1.75 ft, grows 0.5 ft/month):
(i) Height after 7 months = \( 1.75 + 0.5(7) = 1.75 + 3.5 = \mathbf{5.25 \text{ feet}} \).
(iii) Equation: \( h(t) = 1.75 + 0.5t \). It represents linear growth because the height increases by a constant 0.5 ft each month.

2. Mobile Phone Depreciation (₹10,000 cost, decreases ₹800/year):
(i) Value after 3 years = \( 10000 - 800(3) = \mathbf{₹7,600} \).
(iii) Equation: \( v(t) = 10000 - 800t \). It represents linear decay because the value decreases by a constant ₹800 each year.

3. Village Population (750 initial, 50 people/year added):
(i) Population after 6 years = \( 750 + 50(6) = \mathbf{1050 \text{ people}} \).
(iii) Equation: \( P(t) = 750 + 50t \). Represents linear growth because population increases by a constant 50 each year.

4. Telecom Balance (₹600 initial, reduces ₹15/day):
(i) Equation: \( b(x) = 600 - 15x \). Represents linear decay (reduces by constant ₹15/day).
(ii) Balance runs out when \( b(x) = 0 \implies 600 - 15x = 0 \implies x = \mathbf{40 \text{ days}} \).

1. Learning Platform Bill (10 modules = ₹400, 14 modules = ₹500):
Equations: \( 10a + b = 400 \) and \( 14a + b = 500 \).
Subtracting equations: \( 4a = 100 \implies \mathbf{a = 25} \).
Substituting: \( 10(25) + b = 400 \implies \mathbf{b = 150} \).
Linear relation: \( y = 25x + 150 \).

2. Gym Court Bill (10 hours = ₹800, 15 hours = ₹1100):
Equations: \( 10a + b = 800 \) and \( 15a + b = 1100 \).
Subtracting equations: \( 5a = 300 \implies \mathbf{a = 60} \).
Substituting: \( 10(60) + b = 800 \implies \mathbf{b = 200} \).
Linear relation: \( y = 60x + 200 \).

3. Celsius (\( ^\circ\text{C} \)) to Fahrenheit (\( ^\circ\text{F} \)) Conversion:
Form: \( ^\circ\text{C} = a^\circ\text{F} + b \).
• Ice melts: \( 0 = 32a + b \implies b = -32a \).
• Water boils: \( 100 = 212a + b \implies 100 = 212a - 32a = 180a \implies a = \frac{100}{180} = \mathbf{\frac{5}{9}} \).
• \( b = -32 \times \frac{5}{9} = \mathbf{-\frac{160}{9}} \).
Linear relation: \( ^\circ\text{C} = \frac{5}{9}(^\circ\text{F} - 32) \).

Problem 1: Write a polynomial of degree 3 in variable x with \( x^2 \) coefficient -7.

Answer: \( 2x^3 - 7x^2 + 5x - 1 \)

Problem 2: Find values of polynomials:

(i) \( 5x^2 - 3x + 7 \) at \( x = 1 \implies 5(1)^2 - 3(1) + 7 = 5 - 3 + 7 = \mathbf{9} \)
(ii) \( 4t^3 - t^2 + 6 \) at \( t = a \implies \mathbf{4a^3 - a^2 + 6} \)

Problem 3: If we multiply a number by \( \frac{5}{2} \) and add \( \frac{2}{3} \), we get \( -\frac{7}{12} \). Find the number.

Solution:
\( \frac{5}{2}x + \frac{2}{3} = -\frac{7}{12} \implies \frac{5}{2}x = -\frac{7}{12} - \frac{8}{12} = -\frac{15}{12} = -\frac{5}{4} \).
\( x = -\frac{5}{4} \times \frac{2}{5} = \mathbf{-\frac{1}{2}} \).

Problem 4: Positive number is 5 times another. Adding 21 to both makes one twice the other. Find numbers.

Solution: Let smaller be \( x \), larger be \( 5x \).
\( 5x + 21 = 2(x + 21) \implies 5x + 21 = 2x + 42 \implies 3x = 21 \implies x = 7 \).
Numbers are 7 and 35.

Problem 5: Initial ₹800 savings, saves ₹250/month:

Pattern: \( P(n) = 800 + 250n \).
(i) After 6 months: \( 800 + 250(6) = \) ₹2,300.
(ii) After 2 years (24 months): \( 800 + 250(24) = \) ₹6,800.

Problem 6*: Two-digit number digits differ by 3. Reversing and adding gives 143. Find numbers.

Solution: Let tens digit be \( x \), units be \( x - 3 \).
Original = \( 10x + (x - 3) = 11x - 3 \). Reversed = \( 10(x - 3) + x = 11x - 30 \).
Sum: \( (11x - 3) + (11x - 30) = 143 \implies 22x - 33 = 143 \implies 22x = 176 \implies x = 8 \).
Numbers are 85 or 58.

Problem 7*: Identify Slopes & Y-intercepts:

(i) \( y = -3x + 4 \implies \) Slope \( a = -3 \), y-intercept \( b = 4 \), cuts y-axis at \( (0, 4) \).
(ii) \( 2y = 4x + 7 \implies y = 2x + 3.5 \implies \) Slope \( a = 2 \), y-intercept \( b = 3.5 \), cuts y-axis at \( (0, 3.5) \).
(iii) \( 5y = 6x - 10 \implies y = \frac{6}{5}x - 2 \implies \) Slope \( a = \frac{6}{5} \), y-intercept \( b = -2 \).
(iv) \( 3y = 6x - 11 \implies y = 2x - \frac{11}{3} \implies \) Slope \( a = 2 \), y-intercept \( b = -\frac{11}{3} \).
Parallel Check: Lines (ii) and (iv) both have slope \( a = 2 \), so they are Parallel Lines!

Problem 8*: Temperature relation \( y = \frac{9}{5}(x - 273) + 32 \) (Kelvin \( x \) to Fahrenheit \( y \)):

(i) If \( x = 313 \text{ K} \implies y = \frac{9}{5}(313 - 273) + 32 = \frac{9}{5}(40) + 32 = 72 + 32 = \mathbf{104^\circ\text{F}} \).
(ii) If \( y = 158^\circ\text{F} \implies 158 - 32 = \frac{9}{5}(x - 273) \implies 126 \times \frac{5}{9} = x - 273 \implies 70 = x - 273 \implies x = \mathbf{343 \text{ K}} \).

Problem 9*: Work Done \( W = F \times d = 3d \):

When distance \( d = 2 \) units, Work done \( W = 3(2) = \mathbf{6 \text{ units}} \).

Problem 10*: Polynomial passing through \( (1, 5) \) and \( (3, 11) \):

Slope \( a = \frac{11 - 5}{3 - 1} = 3 \). Equation: \( y - 5 = 3(x - 1) \implies p(x) = 3x + 2 \).
Cuts y-axis at \( (0, 2) \), cuts x-axis at \( (-2/3, 0) \).

Problem 11*: Find \( p(x) = ax + b \) and \( q(x) = cx + d \):

Given \( p(0) = 5 \implies b = 5 \).
Given \( p(x) + q(x) = 6x + 4 \implies (a+c)x + (5+d) = 6x + 4 \implies a+c=6 \), \( d = -1 \).
Given \( p(x) - q(x) = (a-c)x + 6 \) cuts x-axis at \( (3, 0) \implies 3(a-c) + 6 = 0 \implies a-c = -2 \).
Solving \( a+c=6 \) and \( a-c=-2 \implies a = 2, c = 4 \).
Polynomials: \( p(x) = 2x + 5 \) and \( q(x) = 4x - 1 \).

Problem 12*: Matchstick Hexagon Pattern (\( 5n + 1 \)):

(iii) Rule for \( n \)-th stage = \( 5n + 1 \).
(iv) 15th stage matchsticks = \( 5(15) + 1 = \mathbf{76 \text{ matchsticks}} \).
(v) Can 200 matchsticks form a stage? Set \( 5n + 1 = 200 \implies 5n = 199 \implies n = 39.8 \). Since \( n \) is not an integer, No.

Problem 13*: Parallel lines \( p(x) \) through \( (2, 3), (6, 11) \) and \( q(x) \) through \( (4, -1) \):

Slope \( a = \frac{11-3}{6-2} = 2 \implies \mathbf{p(x) = 2x - 1} \).
Parallel \( q(x) \) has slope 2, passing through \( (4, -1) \implies -1 = 2(4) + d \implies d = -9 \implies \mathbf{q(x) = 2x - 9} \).
X-intercepts: \( p(x) \) cuts at \( (1/2, 0) \), \( q(x) \) cuts at \( (9/2, 0) \).

Problem 14: Common property of \( f(x) = ax + a, a > 0 \):

Set \( f(x) = 0 \implies a(x + 1) = 0 \implies x = -1 \). All such lines pass through the exact same x-intercept \( (-1, 0) \)!