• (i) 2x² - 5x + 3: The highest power of the variable x is 2, so the degree is 2.
• (ii) y³ + 2y - 1: The highest power of the variable y is 3, so the degree is 3.
• (iii) -9: This is a non-zero constant polynomial (can be written as -9x⁰). Its degree is 0.
• (iv) 4z - 3: The highest power of the variable z is 1, so the degree is 1.

Here are examples of polynomials with the specified degrees:

• Degree 1 (Linear): 5x + 2
• Degree 2 (Quadratic): 3y² - 4y + 7
• Degree 3 (Cubic): 2z³ + z² - 8z + 1

• The term containing x² is +6x², so the coefficient of x² is 6.
• The term containing x³ is -3x³, so the coefficient of x³ is -3.

The polynomial 4z³ + 5z² - 11 does not contain a term with z to the power of 1. It can be written as 4z³ + 5z² + 0z - 11. Therefore, the coefficient of z is 0.

The constant term is the term that does not contain any variable. In the polynomial 9x³ + 5x² - 8x - 10, the constant term is -10.

• (i) If x = 0: 5(0) - 3 = 0 - 3 = -3
• (ii) If x = -1: 5(-1) - 3 = -5 - 3 = -8
• (iii) If x = 2: 5(2) - 3 = 10 - 3 = 7

• (i) If s = 0: 7(0)² - 4(0) + 6 = 0 - 0 + 6 = 6
• (ii) If s = -3: 7(-3)² - 4(-3) + 6 = 7(9) + 12 + 6 = 63 + 12 + 6 = 81
• (iii) If s = 4: 7(4)² - 4(4) + 6 = 7(16) - 16 + 6 = 112 - 16 + 6 = 102

1. Let Salil's present age be x years.
2. His mother's present age is 3x years.
3. After 5 years, Salil's age will be x + 5, and his mother's age will be 3x + 5.
4. Their sum after 5 years is 70:
   (x + 5) + (3x + 5) = 70
   4x + 10 = 70
   4x = 60
   x = 15
5. Therefore, Salil's present age is 15 years, and his mother's present age is 45 years (3 × 15).

1. Let the two positive integers be 2x and 5x based on their ratio.
2. The difference between them is 63:
   5x - 2x = 63
   3x = 63
   x = 21
3. The two integers are:
   Smaller integer: 2 × 21 = 42
   Larger integer: 5 × 21 = 105

1. Let the number of five-rupee coins be x.
2. Then the number of two-rupee coins is 3x.
3. The total value of the coins is ₹88:
   (Value of five-rupee coins) + (Value of two-rupee coins) = 88
   5(x) + 2(3x) = 88
   5x + 6x = 88
   11x = 88
   x = 8
4. So, she has 8 five-rupee coins and 24 two-rupee coins (3 × 8).

1. Let the length of the shorter piece be x feet.
2. The length of the longer piece is 4x feet.
3. The total length is 300 feet:
   x + 4x = 300
   5x = 300
   x = 60
4. The lengths are:
   Shorter piece: 60 feet
   Longer piece: 4 × 60 = 240 feet

1. Let the width of the rectangle be w cm.
2. The length is 2w + 3 cm.
3. The perimeter of a rectangle is 2 × (Length + Width). Given the perimeter is 24 cm:
   2 × ((2w + 3) + w) = 24
   2 × (3w + 3) = 24
   6w + 6 = 24
   6w = 18
   w = 3
4. The dimensions are:
   Width: 3 cm
   Length: 2(3) + 3 = 9 cm

• At the end of the 1st month: ₹500 + (1 × ₹150) = ₹650
• At the end of the 2nd month: ₹500 + (2 × ₹150) = ₹800
• At the end of the 3rd month: ₹500 + (3 × ₹150) = ₹950, and so on.
• The linear expression for the amount in the n^th month is: 500 + 150n.

• After 1 hour: 120 - 9 = 111 members
• After 2 hours: 120 - 2(9) = 120 - 18 = 102 members
• After 3 hours: 120 - 3(9) = 120 - 27 = 93 members
• The linear expression for the members remaining at the end of the n^th hour is: 120 - 9n.

Area of a rectangle = Length × Breadth. Given Length = 13 cm.

• (i) If breadth is 12 cm: Area = 13 × 12 = 156 cm²
• (ii) If breadth is 10 cm: Area = 13 × 10 = 130 cm²
• (iii) If breadth is 8 cm: Area = 13 × 8 = 104 cm²
• The linear pattern representing the area as a function of the breadth b is: 13b.

Volume of a rectangular box = Length × Breadth × Height. Base Area = 7 × 11 = 77 cm².

• (i) If height is 5 cm: Volume = 77 × 5 = 385 cm³
• (ii) If height is 9 cm: Volume = 77 × 9 = 693 cm³
• (iii) If height is 13 cm: Volume = 77 × 13 = 1001 cm³
• The linear pattern representing the volume as a function of height h is: 77h.

• Pages read in 15 days: 15 × 20 = 300 pages
• Pages left after 15 days: 500 - 300 = 200 pages
• The linear pattern representing the number of pages left after d days is: 500 - 20d.

• (i) Height after 7 months: 1.75 + (7 × 0.5) = 1.75 + 3.5 = 5.25 feet.
• (ii) Table of values:

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

• (iii) Expression: h = 1.75 + 0.5t. It represents linear growth because the quantity (height) increases by a constant positive amount (0.5 feet) over equal intervals (each month).

• (i) Value after 3 years: 10000 - (3 × 800) = 10000 - 2400 = ₹7600.
• (ii) Table of values:

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

• (iii) Expression: v = 10000 - 800t. It represents linear decay because the value decreases by a constant fixed amount (₹800) over equal intervals (each year).

• (i) Population after 6 years: 750 + (6 × 50) = 750 + 300 = 1050.
• (ii) Table of values:

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

• (iii) Expression: P = 750 + 50t. It represents linear growth because the population increases by a constant amount (50 people) over equal intervals (each year).

• (i) Equation: b(x) = 600 - 15x. It represents linear decay because the balance decreases by a fixed amount (₹15) at constant intervals (every day).
• (ii) Balance runs out: Set b(x) = 0.
  600 - 15x = 0 ⇒ 15x = 600 ⇒ x = 40 days.
• (iii) Table of values:

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

1. We have two equations based on the information provided:
   400 = 10a + b
   500 = 14a + b
2. Subtract the first equation from the second to eliminate b:
   (500 - 400) = (14a + b) - (10a + b)
   100 = 4a
   a = 25
3. Substitute a = 25 back into the first equation to find b:
   400 = 10(25) + b
   400 = 250 + b
   b = 150
4. The relationship is y = 25x + 150.

1. Establish the two linear equations:
   800 = 10a + b
   1100 = 15a + b
2. Subtract the first from the second:
   300 = 5a
   a = 60
3. Substitute a = 60 into the first equation:
   800 = 10(60) + b
   800 = 600 + b
   b = 200
4. The relationship is y = 60x + 200.

1. Set up the equations based on the given points:
   0 = 32a + b
   100 = 212a + b
2. Subtract the first equation from the second:
   100 = 180a
   a = 100 / 180 = 5/9
3. Substitute a = 5/9 into the first equation:
   0 = 32(5/9) + b
   b = -160/9
4. The relationship is °C = (5/9)°F - 160/9 (or °C = (5/9)(°F - 32)).

Reflections on the parameters 'a' (slope) and 'b' (y-intercept):

• (i) y = 3x - 1, y = 3x, y = 3x + 1: Here, the slope a is constant (3), so the three lines are parallel. The y-intercept b changes (-1, 0, 1), shifting the parallel lines vertically up and down the y-axis.
• (ii) y = 4x, y = 2x, y = x: Here, the y-intercept b is fixed at 0 (all pass through the origin). The slope a changes (4, 2, 1). As a increases, the lines become progressively steeper (closer to the y-axis).
• (iii) y = -6x, y = -3x, y = -x: All pass through the origin. Since slopes are negative, they represent linear decay. As the magnitude of a increases (from 1 to 6), the lines get steeper in the negative direction.
• (iv) y = 5x, y = -5x: Both pass through the origin. They have the same steepness but opposite directions; y = 5x represents growth (rising left to right), while y = -5x represents decay (falling left to right). They are reflections of each other across the y-axis.
• (v) y = -2x - 3, y = -2x, y = 2x + 3: The first two lines (y = -2x - 3 and y = -2x) are parallel to each other with a negative slope, but with different y-intercepts (-3 and 0). The third line (y = 2x + 3) has a positive slope (intersecting the others) and cuts the y-axis at +3.

A possible example is: 5x³ - 7x² + 2x + 1. (Any polynomial where the highest power is x³ and the term containing x² is -7x² is correct).

• (i) If x = 1: 5(1)² - 3(1) + 7 = 5 - 3 + 7 = 9
• (ii) If t = a: Simply replace t with a to get 4a³ - a² + 6

1. Let the number be x.
2. Set up the linear equation: (5/2)x + 2/3 = -7/12
3. Subtract 2/3 from both sides:
   (5/2)x = -7/12 - 2/3
   (5/2)x = -7/12 - 8/12
   (5/2)x = -15/12
   (5/2)x = -5/4
4. Multiply both sides by 2/5 to isolate x:
   x = (-5/4) × (2/5)
   x = -1/2

1. Let the smaller number be x, so the larger number is 5x.
2. After adding 21 to both, the new numbers are x + 21 and 5x + 21.
3. Since 5x was already 5 times larger than x, 5x + 21 will be the larger of the two new numbers. So, it must be twice the smaller new number:
   5x + 21 = 2(x + 21)
4. Solve for x:
   5x + 21 = 2x + 42
   3x = 21
   x = 7
5. The numbers are:
   Smaller number: 7
   Larger number: 5 × 7 = 35

• (i) After 6 months: 800 + (6 × 250) = 800 + 1500 = ₹2300
• (ii) After 2 years (24 months): 800 + (24 × 250) = 800 + 6000 = ₹6800
• Linear Pattern: The total amount A after m months is expressed as A(m) = 800 + 250m.

1. Let the tens digit be t and the units digit be u. The original number is 10t + u.
2. The interchanged number is 10u + t.
3. Their sum is 143:
   (10t + u) + (10u + t) = 143
   11t + 11u = 143
   t + u = 13
4. We are also given that the digits differ by 3, so either t - u = 3 or u - t = 3.
5. Case 1: t - u = 3
   Adding the equations: (t + u) + (t - u) = 13 + 3 ⇒ 2t = 16 ⇒ t = 8
   If t = 8, then u = 5. The number is 85.
6. Case 2: u - t = 3
   Adding the equations: (u + t) + (u - t) = 13 + 3 ⇒ 2u = 16 ⇒ u = 8
   If u = 8, then t = 5. The number is 58.
7. The two possible numbers are 58 and 85.

Convert each to the form y = ax + b:

• (i) y = -3x + 4:
  Slope = -3, y-intercept = 4, Coordinates = (0, 4).
• (ii) 2y = 4x + 7 ⇒ y = 2x + 3.5:
  Slope = 2, y-intercept = 3.5, Coordinates = (0, 3.5).
• (iii) 5y = 6x - 10 ⇒ y = (6/5)x - 2:
  Slope = 6/5 (or 1.2), y-intercept = -2, Coordinates = (0, -2).
• (iv) 3y = 6x - 11 ⇒ y = 2x - 11/3:
  Slope = 2, y-intercept = -11/3, Coordinates = (0, -11/3).

Parallel lines: Lines (ii) and (iv) both have a slope of 2, so they are parallel to each other.

• (i) From Kelvin to Fahrenheit:
  Substitute x = 313 into the equation:
  y = 9/5(313 - 273) + 32
  y = 9/5(40) + 32
  y = 9(8) + 32 = 72 + 32 = 104 °F
• (ii) From Fahrenheit to Kelvin:
  Substitute y = 158 into the equation:
  158 = 9/5(x - 273) + 32
  126 = 9/5(x - 273)
  126 × (5/9) = x - 273
  70 = x - 273
  x = 70 + 273 = 343 K

1. The relationship is Work = Force × distance.
2. Given the constant force is 3 units, the linear equation is: w = 3d.
3. When distance d = 2 units, the work done is: w = 3(2) = 6 units.
4. (To verify on a graph: Plot the line passing through (0,0), (1,3), and (2,6). Reading the y-value at x=2 confirms it is 6).

1. (i) Find the polynomial p(x):
   Let p(x) = ax + b. We have:
   5 = a(1) + b ⇒ a + b = 5
   11 = a(3) + b ⇒ 3a + b = 11
   Subtracting the first from the second gives 2a = 6 ⇒ a = 3.
   Substitute a = 3 into the first: 3 + b = 5 ⇒ b = 2.
   The polynomial is p(x) = 3x + 2.
2. (ii) Coordinates where it cuts the axes:
   y-axis intercept (set x = 0): p(0) = 3(0) + 2 = 2. Coordinates: (0, 2).
   x-axis intercept (set p(x) = 0): 0 = 3x + 2 ⇒ x = -2/3. Coordinates: (-2/3, 0).

1. From (i), p(0) = 5 ⇒ a(0) + b = 5 ⇒ b = 5.
2. From (iii), p(x) + q(x) = (a + c)x + (b + d) = 6x + 4.
   This means a + c = 6 and b + d = 4.
   Since b = 5, then 5 + d = 4 ⇒ d = -1.
3. From (ii), p(3) - q(3) = 0 ⇒ p(3) = q(3).
   3a + b = 3c + d
   Substitute b = 5 and d = -1:
   3a + 5 = 3c - 1 ⇒ 3a - 3c = -6 ⇒ a - c = -2.
4. We now have a system for a and c:
   a + c = 6
   a - c = -2
   Adding them: 2a = 4 ⇒ a = 2.
   Substituting back: 2 + c = 6 ⇒ c = 4.
5. The polynomials are: p(x) = 2x + 5 and q(x) = 4x - 1.

• (i) & (ii) Table Completion:
  Stage 1: 6 matchsticks.
  Stage 2: 6 + 5 = 11 matchsticks (shares 1 side).
  Stage 3: 11 + 5 = 16 matchsticks.

  Stage	1	2	3	4	5	n
  Number of matchsticks	6	11	16	21	26	5n + 1

• (iii) Rule: The pattern increases by 5 each time. The rule is 5n + 1.
• (iv) 15th Stage: 5(15) + 1 = 75 + 1 = 76 matchsticks.
• (v) Can 200 matchsticks form a stage?
  Set 5n + 1 = 200 ⇒ 5n = 199 ⇒ n = 199/5 = 39.8.
  No, because n must be a whole number integer (a specific stage), and 199 is not perfectly divisible by 5.

1. Find p(x):
   3 = 2a + b
   11 = 6a + b
   Subtracting gives 4a = 8 ⇒ a = 2. Substituting back gives b = -1.
   So, p(x) = 2x - 1.
2. Find q(x):
   Because q(x) is parallel to p(x), its slope c must equal a. So, c = 2.
   q(x) = 2x + d. It passes through (4, -1), so:
   -1 = 2(4) + d ⇒ -1 = 8 + d ⇒ d = -9.
   So, q(x) = 2x - 9.
3. X-axis intercepts (where y = 0):
   For p(x): 2x - 1 = 0 ⇒ x = 1/2. Coordinates: (1/2, 0).
   For q(x): 2x - 9 = 0 ⇒ x = 9/2. Coordinates: (9/2, 0).

Given f(x) = ax + a, where a > 0. We can factor this as f(x) = a(x + 1).

1. Positive Slope: The slope is a. Since a > 0, all lines have a positive slope and rise from left to right (linear growth).
2. Equal Slope and Y-intercept: When x = 0, f(0) = a. The y-intercept is (0, a), which is above the origin since a > 0.
3. Common X-intercept: To find where the line cuts the x-axis, set f(x) = 0:
   a(x + 1) = 0
   Since a ≠ 0, we have x + 1 = 0 ⇒ x = -1.
   Therefore, all such lines pass through the identical fixed point (-1, 0) on the x-axis.