Introduction to Linear Polynomials

Can you identify the terms, variables and coefficients of the algebraic expression 200l + 160w + 50lw?

Answer

The given algebraic expression is 200l + 160w + 50lw.

The terms are separated by + and - signs.

Terms: 200l, 160w and 50lw

Variables: l and w

Coefficients:

Coefficient of l = 200

Coefficient of w = 160

Coefficient of lw = 50

How is 200l + 160w + 50lw different from the algebraic expression 4x + 5y + 3?

Answer

Let us compare the two algebraic expressions term by term:

Expression 1: 200l + 160w + 50lw

Expression 2: 4x + 5y + 3

Key differences:

1. Linearity: In the expression 4x + 5y + 3, each term has degree at most 1 (the variables x and y appear only to the first power and are not multiplied together). Hence, it is a linear expression. On the other hand, 200l + 160w + 50lw contains the term 50lw, which is a product of two variables l and w, giving it degree 2. Hence, 200l + 160w + 50lw is not a linear expression.

2. Product of variables: The expression 200l + 160w + 50lw contains the term 50lw, where two variables are multiplied together. The expression 4x + 5y + 3 has no such term — its variables appear only individually.

3. Constant term: The expression 4x + 5y + 3 has a constant term, namely 3. The expression 200l + 160w + 50lw has no constant term.

Hence, the main difference is that 4x + 5y + 3 is a linear expression in two variables, while 200l + 160w + 50lw is not linear because of the term 50lw, which is a product of two variables.

Can you identify the terms, variables and coefficients of the algebraic expression 10x – x²?

Answer

The given algebraic expression is 10x – x².

The terms are separated by + and - signs.

Terms: 10x and -x²

Variable: x

Coefficients:

Coefficient of x = 10

Coefficient of x² = -1

Can you point out any similarity or difference between the algebraic expressions 4x + 5y + 3 and 10x – x²?

Answer

Let us examine the two algebraic expressions term by term:

Expression 1: 4x + 5y + 3

Expression 2: 10x − x²

Similarities:

1. Both are algebraic expressions made up of more than one term.

2. Both contain the variable x, with a term in which x appears to the first power (4x in the first expression and 10x in the second).

Differences:

1. Number of variables: The expression 4x + 5y + 3 contains two variables (x and y), whereas 10x − x² contains only one variable (x).

2. Linearity: In 4x + 5y + 3, each term has degree at most 1, so it is a linear expression. In 10x − x², the term −x² has degree 2, so it is not a linear expression (it is a quadratic expression).

3. Constant term: The expression 4x + 5y + 3 has a constant term (3), while 10x − x² has no constant term.

Hence, although both expressions share the variable x with a first-degree term, 4x + 5y + 3 is a linear expression in two variables with a constant term, whereas 10x − x² is a quadratic (non-linear) expression in one variable with no constant term.

Find the degrees of the following polynomials:
(i) 2x² – 5x + 3
(ii) y³ + 2y – 1
(iii) – 9
(iv) 4z – 3

Answer

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

(i) 2x² – 5x + 3

The highest power of x is 2.

∴ Degree = 2

(ii) y³ + 2y – 1

The highest power of y is 3.

∴ Degree = 3

(iii) – 9

The constant -9 can be written as -9x⁰, in which the power of the variable x is 0.

∴ Degree = 0

(iv) 4z – 3

The highest power of z is 1.

∴ Degree = 1

Write polynomials of degrees 1, 2 and 3.

Answer

A polynomial of degree 1 is one in which the highest power of the variable is 1.

Polynomial of degree 1: 2x + 5

A polynomial of degree 2 is one in which the highest power of the variable is 2.

Polynomial of degree 2: 3x² – 4x + 1

A polynomial of degree 3 is one in which the highest power of the variable is 3.

Polynomial of degree 3: x³ + 2x² – x + 7

What are the coefficients of x² and x³ in the polynomial x⁴ – 3x³ + 6x² – 2x + 7?

Answer

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

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

The term containing x² is 6x².

The term containing x³ is -3x³.

∴ Coefficient of x² is 6 and the coefficient of x³ is -3.

What is the coefficient of z in the polynomial 4z³ + 5z² – 11?

Answer

Given polynomial: 4z³ + 5z² – 11

In this polynomial, there is no term containing z (i.e., z¹).

The polynomial can be written as 4z³ + 5z² + 0z – 11.

∴ Coefficient of z is 0.

What is the constant term of the polynomial 9x³ + 5x² – 8x – 10?

Answer

Given polynomial: 9x³ + 5x² – 8x – 10

The constant term is the term that does not contain any variable.

In the given polynomial, the term without any variable is -10.

∴ Constant term = -10

Find the perimeter of squares with sides 1 cm, 1.5 cm, 2 cm, 2.5 cm and 3 cm. What will happen to the perimeters if the sides increase by 0.5 cm?

Answer

The perimeter of a square is given by P = 4 × side.

For side = 1 cm:

Perimeter = 4 × 1 = 4 cm

For side = 1.5 cm:

Perimeter = 4 × 1.5 = 6 cm

For side = 2 cm:

Perimeter = 4 × 2 = 8 cm

For side = 2.5 cm:

Perimeter = 4 × 2.5 = 10 cm

For side = 3 cm:

Perimeter = 4 × 3 = 12 cm

The perimeters are: 4 cm, 6 cm, 8 cm, 10 cm and 12 cm.

The difference between consecutive perimeters is constant (= 2 cm).

Hence, when the side of the square increases by 0.5 cm, the perimeter increases by 2 cm each time.

A chess club charges a joining fee of ₹200 plus ₹50 for every match played. The following table shows the amount a player will have to pay as the number of matches varies.

If a player paid ₹750, how many matches did he play?

Answer

From the table, the amount paid by a player for m matches is given by:

Amount = ₹(200 + 50m)

Given:

Amount paid = ₹750

So,

200 + 50m = 750

⇒ 50m = 750 - 200

⇒ 50m = 550

⇒ m = $\dfrac{550}{50}$

⇒ m = 11

Hence, the player played 11 matches.

We have learnt that to evaluate the value of an algebraic expression, we substitute a value of the variable in the given expression. Consider Example 3, where the wire is bent to form a rectangle. Here, the area of the rectangle, 10x – x², is a function of x. Can you interpret this as an input-output process? What value does the expression take when x = 6 cm?

Answer

Yes, the expression 10x – x² can be interpreted as an input-output process.

Here, the input is the value of x (the length of the rectangle), and the output is the value of 10x – x² (the area of the rectangle).

For every value of x, there is a corresponding value of the expression 10x – x².

Substituting x = 6 in the expression 10x – x²:

= 10(6) – (6)²

= 60 – 36

= 24

Hence, when x = 6 cm, the value of the expression is 24 cm².

Find the value of the linear polynomial 5x – 3 if:
(i) x = 0
(ii) x = –1
(iii) x = 2

Answer

Given linear polynomial: 5x – 3

(i) x = 0

Substituting x = 0 in the polynomial:

= 5(0) – 3

= 0 – 3

= -3

∴ Value of the polynomial when x = 0 is -3.

(ii) x = –1

Substituting x = -1 in the polynomial:

= 5(-1) – 3

= -5 – 3

= -8

∴ Value of the polynomial when x = -1 is -8.

(iii) x = 2

Substituting x = 2 in the polynomial:

= 5(2) – 3

= 10 – 3

= 7

∴ Value of the polynomial when x = 2 is 7.

Find the value of the quadratic polynomial 7s² – 4s + 6 if:
(i) s = 0
(ii) s = –3
(iii) s = 4

Answer

Given quadratic polynomial: 7s² – 4s + 6

(i) s = 0

Substituting s = 0 in the polynomial:

= 7(0)² – 4(0) + 6

= 0 – 0 + 6

= 6

∴ Value of the polynomial when s = 0 is 6.

(ii) s = -3

Substituting s = -3 in the polynomial:

= 7(-3)² – 4(-3) + 6

= 7(9) + 12 + 6

= 63 + 12 + 6

= 81

∴ Value of the polynomial when s = -3 is 81.

(iii) s = 4

Substituting s = 4 in the polynomial:

= 7(4)² – 4(4) + 6

= 7(16) – 16 + 6

= 112 – 16 + 6

= 102

∴ Value of the polynomial when s = 4 is 102.

The present age of Salil's mother is three times Salil's present age. After 5 years, their ages will add up to 70 years. Find their present ages.

Answer

Let Salil's present age be x years.

Then, his mother's present age = 3x years.

After 5 years:

Salil's age = (x + 5) years

Mother's age = (3x + 5) years

According to the question, the sum of their ages after 5 years will be 70 years.

So,

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

⇒ x + 5 + 3x + 5 = 70

⇒ 4x + 10 = 70

⇒ 4x = 70 - 10

⇒ 4x = 60

⇒ x = $\dfrac{60}{4}$

⇒ x = 15

So, Salil's present age = 15 years.

Mother's present age = 3 × 15 = 45 years.

Hence, Salil's present age is 15 years and his mother's present age is 45 years.

The difference between two positive integers is 63. The ratio of the two integers is 2 : 5. Find the two integers.

Answer

Let the two positive integers be 2x and 5x (since their ratio is 2 : 5).

Note that 5x > 2x, so the difference is 5x - 2x.

According to the question, the difference between them is 63.

So,

5x – 2x = 63

⇒ 3x = 63

⇒ x = $\dfrac{63}{3}$

⇒ x = 21

So, the two integers are:

First integer = 2x = 2 × 21 = 42

Second integer = 5x = 5 × 21 = 105

Hence, the two integers are 42 and 105.

Ruby has 3 times as many two-rupee coins as she has five rupee-coins. If she has a total ₹88, how many coins does she have of each type?

Answer

Let the number of five-rupee coins Ruby has be x.

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

Total amount from five-rupee coins = ₹5x

Total amount from two-rupee coins = ₹2(3x) = ₹6x

According to the question, the total amount is ₹88.

So,

5x + 6x = 88

⇒ 11x = 88

⇒ x = $\dfrac{88}{11}$

⇒ x = 8

So,

Number of five-rupee coins = 8

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

Hence, Ruby has 24 two-rupee coins and 8 five-rupee coins.

A farmer cuts a 300 feet fence into two pieces of different sizes. The longer piece is four times as long as the shorter piece. How long are the two pieces?

Answer

Let the length of the shorter piece be x feet.

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

Total length of the fence = 300 feet.

So,

x + 4x = 300

⇒ 5x = 300

⇒ x = $\dfrac{300}{5}$

⇒ x = 60

So,

Length of the shorter piece = 60 feet

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

Hence, the two pieces are 60 feet and 240 feet long.

If the length of a rectangle is three more than twice its width and its perimeter is 24 cm, what are the dimensions of the rectangle?

Answer

Let the width of the rectangle be x cm.

Then, the length of the rectangle = (2x + 3) cm.

Perimeter of a rectangle = 2(length + width)

According to the question, the perimeter is 24 cm.

So,

2[(2x + 3) + x] = 24

⇒ 2[3x + 3] = 24

⇒ 6x + 6 = 24

⇒ 6x = 24 - 6

⇒ 6x = 18

⇒ x = $\dfrac{18}{6}$

⇒ x = 3

So,

Width of the rectangle = 3 cm

Length of the rectangle = 2(3) + 3 = 6 + 3 = 9 cm

Hence, the dimensions of the rectangle are: length = 9 cm and width = 3 cm.

Predict the number of squares in the next three stages of the pattern and write the sequence of numbers up to Stage 7 of the pattern.

Answer

From the pattern, we observe that:

Stage 1 has 1 square.

Stage 2 has 3 squares.

Stage 3 has 5 squares.

Stage 4 has 7 squares.

Each stage has 2 more squares than the previous stage.

So, the next three stages will have:

Stage 5 = 7 + 2 = 9 squares

Stage 6 = 9 + 2 = 11 squares

Stage 7 = 11 + 2 = 13 squares

Hence, the sequence of the number of squares up to Stage 7 is: 1, 3, 5, 7, 9, 11, 13.

Using the expression 2n – 1, can you find out how many tiles will be there in the 15th stage and the 26th stage of the pattern? Also, which stage will contain 21 tiles and 47 tiles?

Answer

The expression for the number of tiles at the n^th stage is 2n – 1.

For the 15th stage:

Substituting n = 15:

= 2(15) – 1

= 30 – 1

= 29

So, the 15th stage has 29 tiles.

For the 26th stage:

Substituting n = 26:

= 2(26) – 1

= 52 – 1

= 51

So, the 26th stage has 51 tiles.

To find which stage contains 21 tiles:

2n – 1 = 21

⇒ 2n = 21 + 1

⇒ 2n = 22

⇒ n = $\dfrac{22}{2}$

⇒ n = 11

So, the 11th stage contains 21 tiles.

To find which stage contains 47 tiles:

2n – 1 = 47

⇒ 2n = 47 + 1

⇒ 2n = 48

⇒ n = $\dfrac{48}{2}$

⇒ n = 24

So, the 24th stage contains 47 tiles.

Hence, the 15th stage has 29 tiles, the 26th stage has 51 tiles, the 11th stage contains 21 tiles, and the 24th stage contains 47 tiles.

Bela has ₹100 for pocket money. She spends 5 every day. What amount will be left on the 15th day? How many days will it take for the entire amount to be spent?

Answer

Let n be the number of days and A be the amount left (in ₹) after n days.

Since Bela starts with ₹100 and spends ₹5 every day, the amount left after n days is:

A = 100 − 5n

Amount left on the 15th day:

Substituting n = 15 in A = 100 − 5n:

A = 100 − 5 × 15

⇒ A = 100 − 75

⇒ A = 25

So, ₹25 will be left on the 15th day.

Number of days for the entire amount to be spent:

The entire amount is spent when A = 0.

Substituting A = 0 in A = 100 − 5n:

100 − 5n = 0

⇒ 5n = 100

⇒ n = $\dfrac{100}{5}$

⇒ n = 20

So, it will take 20 days to spend the entire amount.

Hence, ₹25 will be left on the 15th day, and it will take 20 days for the entire amount to be spent.

An auto-rikshaw fare starts at ₹25 and remains the same for the initial 2 km. Then it increases by ₹15 per km. For how many km will the fare be ₹130?

Answer

Let x be the total distance travelled (in km) and F be the fare (in ₹).

For the initial 2 km, the fare is ₹25. After 2 km, the fare increases by ₹15 for every additional km. So for x km (where x ≥ 2), the fare is:

F = 25 + 15(x − 2)

Given that the fare is ₹130, we have:

130 = 25 + 15(x − 2)

⇒ 15(x − 2) = 130 − 25

⇒ 15(x − 2) = 105

⇒ x − 2 = $\dfrac{105}{15}$

⇒ x − 2 = 7

⇒ x = 7 + 2

⇒ x = 9

Hence, the fare will be ₹130 for a travel of 9 km.

A student has ₹500 in her savings bank account. She gets ₹150 every month as pocket money. How much money will she have at the end of every month from the second month onwards? Find a linear expression to represent the amount she will have in the n^th month.

Answer

Initial amount in the savings bank account = ₹500

Pocket money received every month = ₹150

Amount at the end of each month:

After 1st month = 500 + 150 × 1 = ₹650

After 2nd month = 500 + 150 × 2 = ₹800

After 3rd month = 500 + 150 × 3 = ₹950

After 4th month = 500 + 150 × 4 = ₹1100

In general, after n months, the amount will be:

Amount after n months = ₹(500 + 150n)

Hence, the linear expression representing the amount in the n^th month is ₹(500 + 150n).

A rally starts with 120 members. Each hour, 9 members drop out of the group. How many members will remain after 1, 2, 3, … hours? Find a linear expression to represent the number of members at the end of the n^th hour.

Answer

Initial number of members = 120

Members dropping out every hour = 9

Number of members remaining:

After 1 hour = 120 - 9 × 1 = 111

After 2 hours = 120 - 9 × 2 = 102

After 3 hours = 120 - 9 × 3 = 93

After 4 hours = 120 - 9 × 4 = 84

In general, after n hours, the number of members remaining will be:

Number of members after n hours = 120 - 9n

Hence, the linear expression representing the number of members at the end of the n^th hour is (120 - 9n).

Suppose the length of a rectangle is 13 cm. Find the area if the breadth is (i) 12 cm, (ii) 10 cm, (iii) 8 cm. Find the linear pattern representing the area of the rectangle.

Answer

Length of the rectangle = 13 cm

Area of a rectangle = length × breadth

(i) Breadth = 12 cm

Area = 13 × 12 = 156 cm²

(ii) Breadth = 10 cm

Area = 13 × 10 = 130 cm²

(iii) Breadth = 8 cm

Area = 13 × 8 = 104 cm²

In general, if the breadth of the rectangle is b cm, the area will be 13b cm².

Hence, the linear pattern representing the area of the rectangle is 13b, where b is the breadth in cm.

Suppose the length of a rectangular box is 7 cm and breadth is 11 cm. Find the volume if the height is (i) 5 cm, (ii) 9 cm, (iii) 13 cm. Find the linear pattern representing the volume of the rectangular box.

Answer

Length of the rectangular box = 7 cm

Breadth of the rectangular box = 11 cm

Volume of a rectangular box = length × breadth × height

= 7 × 11 × height

= 77 × height

(i) Height = 5 cm

Volume = 77 × 5 = 385 cm³

(ii) Height = 9 cm

Volume = 77 × 9 = 693 cm³

(iii) Height = 13 cm

Volume = 77 × 13 = 1001 cm³

In general, if the height of the box is h cm, the volume will be 77h cm³.

Hence, the linear pattern representing the volume of the rectangular box is 77h, where h is the height in cm.

Sarita is reading a book of 500 pages. She reads 20 pages every day. How many pages will be left after 15 days? Express this as a linear pattern.

Answer

Total number of pages in the book = 500

Pages read every day = 20

Pages read after 15 days = 20 × 15 = 300

Pages left after 15 days = 500 - 300 = 200

In general, after n days, the number of pages left will be:

Number of pages left = 500 - 20n

Hence, after 15 days, 200 pages will be left, and the linear pattern representing the number of pages left is (500 - 20n), where n is the number of days.

The cost of a journey is given by the linear function C(d) = 100 + 60d where C indicates total cost in rupees and d the distance travelled in km. What is the cost for travelling 15 km? For how many kilometres will the cost of the journey be ₹700?

Answer

The cost of the journey is given by the linear function:

C(d) = 100 + 60d

where C is the total cost (in ₹) and d is the distance travelled (in km).

Cost for travelling 15 km:

Substituting d = 15 in C(d) = 100 + 60d:

C(15) = 100 + 60 × 15

⇒ C(15) = 100 + 900

⇒ C(15) = 1000

So, the cost for travelling 15 km is ₹1000.

Distance for which the cost is ₹700:

Substituting C(d) = 700 in C(d) = 100 + 60d:

100 + 60d = 700

⇒ 60d = 700 − 100

⇒ 60d = 600

⇒ d = $\dfrac{600}{60}$

⇒ d = 10

So, the cost will be ₹700 for a distance of 10 km.

Hence, the cost for travelling 15 km is ₹1000, and the cost of the journey will be ₹700 for a distance of 10 km.

The height of water in a cylindrical tank is 3 m at the start of summer. The height h m at the end of t months is given by the linear function h(t) = 3 – 0.5t. What will be the height of the water at the end of 5 months?

Answer

The height of water in the tank is given by the linear function:

h(t) = 3 − 0.5t

where h is the height of water (in m) and t is the time elapsed (in months).

Height of water at the end of 5 months:

Substituting t = 5 in h(t) = 3 − 0.5t:

h(5) = 3 − 0.5 x 5

⇒ h(5) = 3 − 2.5

⇒ h(5) = 0.5

Hence, the height of the water at the end of 5 months will be 0.5 m.

Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month.
(i) Find the height after 7 months.
(ii) Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month.
(iii) Find an expression that relates h and t, and explain why it represents linear growth.

Answer

Initial height of the plant = 1.75 feet

Growth per month = 0.5 feet

(i) Height after 7 months = 1.75 + 0.5 × 7

= 1.75 + 3.5

= 5.25 feet

∴ The height of the plant after 7 months is 5.25 feet.

(ii) The height of the plant at the end of t months is given by h = 1.75 + 0.5t.

(iii) The expression relating h and t is:

h = 1.75 + 0.5t

This represents linear growth because as t (months) increases by 1, the height h increases by a constant value of 0.5 feet. The change in h for every unit change in t is the same, which is the characteristic feature of linear growth.

A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year.
(i) Find the value of the phone after 3 years.
(ii) Make a table of values for t varying from 0 to 8 years and show how the value of the phone, v, depreciates with time.
(iii) Find an expression that relates v and t, and explain why it represents linear decay.

Answer

Initial value of the mobile phone = ₹10,000

Decrease in value per year = ₹800

(i) Value of the phone after 3 years = 10000 - 800 × 3

= 10000 - 2400

= ₹7600

∴ The value of the phone after 3 years is ₹7600.

(ii) The value of the phone at the end of t years is given by v = 10000 - 800t.

(iii) The expression relating v and t is:

v = 10000 - 800t

This represents linear decay because as t (years) increases by 1, the value v decreases by a constant amount of ₹800. The change in v for every unit change in t is the same (a fixed decrease), which is the characteristic feature of linear decay.

The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.
(i) Find the population of the village after 6 years.
(ii) Make a table of values for t varying from 0 to 10 years and show how the population, P, increases every year.
(iii) Find an expression that relates P and t, and explain why it represents linear growth.

Answer

Initial population of the village = 750

Increase in population per year = 50

(i) Population of the village after 6 years = 750 + 50 × 6

= 750 + 300

= 1050

∴ The population of the village after 6 years is 1050.

(ii) The population of the village at the end of t years is given by P = 750 + 50t.

(iii) The expression relating P and t is:

P = 750 + 50t

This represents linear growth because as t (years) increases by 1, the population P increases by a constant amount of 50 people. The change in P for every unit change in t is the same, which is the characteristic feature of linear growth.

A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the recharge.
(i) Write an equation that models the remaining balance b(x) after using the scheme for x days. Explain why it represents linear decay.
(ii) After how many days will the balance run out?
(iii) Make a table of values for x varying from 1 to 10 days and show how the balance b(x), reduces with time.

Answer

Initial balance = ₹600

Reduction per day = ₹15

(i) The remaining balance after x days is given by:

b(x) = 600 - 15x

This represents linear decay because as x (days) increases by 1, the balance b(x) decreases by a constant amount of ₹15. The change in b(x) for every unit change in x is a fixed decrease, which is the characteristic feature of linear decay.

(ii) For the balance to run out, b(x) = 0.

So,

600 - 15x = 0

⇒ 15x = 600

⇒ x = $\dfrac{600}{15}$

⇒ x = 40

∴ The balance will run out after 40 days.

(iii) The balance after x days is given by b(x) = 600 - 15x.

A telecom company charges a fixed monthly fee and an additional cost per GB of the internet data used. A student observes that when she used 10 GB, her bill was ₹350. When she used 20 GB, her bill was ₹550. If the monthly bill y depends on the amount of data used, x (in GB), according to the relation y = ax + b, Can you guess what the numbers 20 and 150 in the equation y = 20x + 150 represent?

Answer

The linear relationship is given by y = 20x + 150,

where x is the number of GB of internet data used and y is the total monthly bill in rupees.

The number 20 represents the additional cost per GB of internet data used. That is, for every additional GB of data used, the bill increases by ₹20.

The number 150 represents the fixed monthly fee charged by the telecom company. This is the amount the user must pay even if no data is used (i.e., when x = 0, y = 150).

Hence, 20 represents the cost per GB of data used, and 150 represents the fixed monthly fee.

A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes that when she accessed 10 modules, her bill was ₹400. When she accessed 14 modules, her bill was ₹500. If the monthly bill y depends on the number of modules accessed, x, according to the relation y = ax + b, find the values of a and b.

Answer

Given:

Number of modules accessed (x) and monthly bill (y) follow the relation y = ax + b.

When x = 10, y = 400.

When x = 14, y = 500.

Substituting these values into y = ax + b, we get the following two equations:

400 = 10a + b ...(i)

500 = 14a + b ...(ii)

From equation (i): b = 400 - 10a

Substituting this into equation (ii):

500 = 14a + (400 - 10a)

⇒ 500 = 14a + 400 - 10a

⇒ 500 = 4a + 400

⇒ 4a = 500 - 400

⇒ 4a = 100

⇒ a = $\dfrac{100}{4}$

⇒ a = 25

Substituting a = 25 in b = 400 - 10a:

b = 400 - 10(25)

= 400 - 250

= 150

Hence, a = 25 and b = 150.

A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hours, her bill was ₹800. When she used it for 15 hours, her bill was ₹1100. If the monthly bill y depends on the hours of the use of the badminton court, x, according to the relation y = ax + b, find the values of a and b.

Answer

Given:

Number of hours of court use (x) and monthly bill (y) follow the relation y = ax + b.

When x = 10, y = 800.

When x = 15, y = 1100.

Substituting these values into y = ax + b, we get the following two equations:

800 = 10a + b ...(i)

1100 = 15a + b ...(ii)

From equation (i): b = 800 - 10a

Substituting this into equation (ii):

1100 = 15a + (800 - 10a)

⇒ 1100 = 15a + 800 - 10a

⇒ 1100 = 5a + 800

⇒ 5a = 1100 - 800

⇒ 5a = 300

⇒ a = $\dfrac{300}{5}$

⇒ a = 60

Substituting a = 60 in b = 800 - 10a:

b = 800 - 10(60)

= 800 - 600

= 200

Hence, a = 60 and b = 200.

Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by °C = a °F + b. Find a and b, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit. (Hint: When °C = 0, °F = 32 and when °C = 100, °F = 212. Use this information to find a and b, and thus, the linear relationship between °C and °F.)

Answer

Given:

The relationship between °C and °F is °C = a(°F) + b.

When °C = 0, °F = 32.

When °C = 100, °F = 212.

Substituting these values into °C = a(°F) + b, we get the following two equations:

0 = 32a + b ...(i)

100 = 212a + b ...(ii)

From equation (i): b = -32a

Substituting this into equation (ii):

100 = 212a + (-32a)

⇒ 100 = 212a - 32a

⇒ 100 = 180a

⇒ a = $\dfrac{100}{180}$

⇒ a = $\dfrac{5}{9}$

Substituting a = $\dfrac{5}{9}$ in b = -32a:

b = -32 × $\dfrac{5}{9}$

= $-\dfrac{160}{9}$

The linear relationship is:

°C = $\dfrac{5}{9}$ (°F) - $\dfrac{160}{9}$ = $\dfrac{5}{9}$ (°F - 32)

Hence, a = $\dfrac{5}{9}$, b = $-\dfrac{160}{9}$ and Linear relationship = $\dfrac{5}{9}$ (°F - 32)

Identify other points on the line y = 2x + 1 by completing the following table.

Answer

The equation of the line is y = 2x + 1.

Substituting each value of x in the equation:

For x = 1: y = 2(1) + 1 = 3

For x = 2: y = 2(2) + 1 = 5

For x = 5: y = 2(5) + 1 = 11

For x = 7: y = 2(7) + 1 = 15

For x = 9: y = 2(9) + 1 = 19

For x = 12: y = 2(12) + 1 = 25

For x = 20: y = 2(20) + 1 = 41

The completed table is:

Hence, the points (1, 3), (2, 5), (5, 11), (7, 15), (9, 19), (12, 25) and (20, 41) all lie on the line y = 2x + 1.

Differentiate between the graphs of the equations y = 3x + 1, and y = –3x + 1.

Answer

Both equations are of the form y = ax + b.

To draw the graph, we identify suitable points on each line.

Equation 1 : y = 3x + 1

x = 1, y = 4

x = 2, y = 7

x = -1, y = -2

Equation 2 : y = –3x + 1

x = 0, y = 1

x = 2, y = -5

x = -2, y = 7

Similarities:

Both lines have the same y-intercept, b = 1, so both lines cut the y-axis at the same point (0, 1).

Differences:

The line y = 3x + 1 has a positive slope (a = 3), so it rises from the bottom-left to the top-right of the coordinate plane. It represents linear growth.

The line y = -3x + 1 has a negative slope (a = -3), so it falls from the top-left to the bottom-right of the coordinate plane. It represents linear decay.

Hence, the two lines pass through the same point (0, 1) on the y-axis but have slopes opposite in sign, so one rises and the other falls.

Does this help you to conclude anything about the linear equation y = ax + b when a is fixed but b varies?

Answer

Looking at the graphs of y = 2x – 1, y = 2x + 1 and y = 2x + 5, we observe:

The slope (a = 2) is the same for all three lines, so they are equally inclined and have the same direction.

The y-intercept (b) is different for each line:

For y = 2x – 1, the line cuts the y-axis at (0, -1).

For y = 2x + 1, the line cuts the y-axis at (0, 1).

For y = 2x + 5, the line cuts the y-axis at (0, 5).

The lines do not intersect each other; they are parallel to one another.

Hence, when a is fixed but b varies in the linear equation y = ax + b, the lines have the same slope but different y-intercepts, and they are parallel to each other.

Draw the graphs of the following sets of lines. In each case, reflect on the role of 'a' and 'b'.
(i) y = 4x, y = 2x, y = x
(ii) y = – 6x, y = – 3x, y = – x
(iii) y = 5x, y = –5x
(iv) y = 3x – 1, y = 3x, y = 3x + 1
(v) y = –2x – 3, y = –2x, y = 2x + 3

Answer

To draw the graphs, we identify two points for each line by substituting values of x and finding the corresponding values of y.

(i) y = 4x, y = 2x, y = x

For y = 4x: when x = 0, y = 0; when x = 1, y = 4. Points: (0, 0) and (1, 4).

For y = 2x: when x = 0, y = 0; when x = 1, y = 2. Points: (0, 0) and (1, 2).

For y = x: when x = 0, y = 0; when x = 1, y = 1. Points: (0, 0) and (1, 1).

Observation:

All three lines pass through the origin (0, 0) since b = 0 in each.

The slopes are different (4, 2, 1). As 'a' increases, the line becomes steeper. The line y = 4x is the steepest, while y = x is the least steep.

(ii) y = -6x, y = -3x, y = -x

For y = -6x: when x = 0, y = 0; when x = 1, y = -6. Points: (0, 0) and (1, -6).

For y = -3x: when x = 0, y = 0; when x = 1, y = -3. Points: (0, 0) and (1, -3).

For y = -x: when x = 0, y = 0; when x = 1, y = -1. Points: (0, 0) and (1, -1).

Observation:

All three lines pass through the origin (0, 0) since b = 0 in each.

All three lines have negative slopes, so they fall from top-left to bottom-right. As the magnitude of 'a' increases, the line becomes steeper. The line y = -6x is the steepest, while y = -x is the least steep.

(iii) y = 5x, y = -5x

For y = 5x: when x = 0, y = 0; when x = 1, y = 5. Points: (0, 0) and (1, 5).

For y = -5x: when x = 0, y = 0; when x = 1, y = -5. Points: (0, 0) and (1, -5).

Observation:

Both lines pass through the origin (0, 0) since b = 0 in each.

The slopes are equal in magnitude but opposite in sign. The line y = 5x rises from bottom-left to top-right, while y = -5x falls from top-left to bottom-right. The two lines are reflections of each other across the x-axis (and the y-axis).

(iv) y = 3x - 1, y = 3x, y = 3x + 1

For y = 3x - 1: when x = 0, y = -1; when x = 1, y = 2. Points: (0, -1) and (1, 2).

For y = 3x: when x = 0, y = 0; when x = 1, y = 3. Points: (0, 0) and (1, 3).

For y = 3x + 1: when x = 0, y = 1; when x = 1, y = 4. Points: (0, 1) and (1, 4).

Observation:

All three lines have the same slope (a = 3), so they are parallel to each other.

The y-intercepts are different (-1, 0, 1). The lines cut the y-axis at (0, -1), (0, 0) and (0, 1), respectively.

When 'a' is fixed and 'b' varies, the lines are parallel and shift along the y-axis.

(v) y = -2x - 3, y = -2x, y = 2x + 3

For y = -2x - 3: when x = 0, y = -3; when x = 1, y = -5. Points: (0, -3) and (1, -5).

For y = -2x: when x = 0, y = 0; when x = 1, y = -2. Points: (0, 0) and (1, -2).

For y = 2x + 3: when x = 0, y = 3; when x = 1, y = 5. Points: (0, 3) and (1, 5).

Observation:

The lines y = -2x - 3 and y = -2x have the same slope (a = -2), so they are parallel to each other. Their y-intercepts are -3 and 0, respectively.

The line y = 2x + 3 has a positive slope (a = 2), so it is not parallel to the other two lines. Its y-intercept is 3.

This shows that lines with the same slope (regardless of the y-intercept) are parallel, while lines with different slopes are not parallel.

Write a polynomial of degree 3 in the variable x, in which the coefficient of the x² term is –7.

Answer

A polynomial of degree 3 has its highest power of the variable as 3.

The general form of such a polynomial is ax³ + bx² + cx + d, where a ≠ 0.

Here, the coefficient of x² is given to be -7.

So, b = -7.

We may choose any non-zero value for a (the coefficient of x³) and any values for c and d.

For example, taking a = 1, c = 5 and d = 2, we get:

Polynomial: x³ - 7x² + 5x + 2

(Note: Many other polynomials are also possible, such as 2x³ - 7x² + 4, 3x³ - 7x² - x + 1, etc.)

Find the values of the following polynomials at the indicated values of the variables.
(i) 5x² – 3x + 7 if x = 1
(ii) 4t³ – t² + 6 if t = a

Answer

(i) 5x² – 3x + 7 if x = 1

Substituting x = 1 in the polynomial:

= 5(1)² – 3(1) + 7

= 5(1) – 3 + 7

= 5 – 3 + 7

= 9

∴ Value of the polynomial when x = 1 is 9.

(ii) 4t³ – t² + 6 if t = a

Substituting t = a in the polynomial:

= 4(a)³ – (a)² + 6

= 4a³ – a² + 6

∴ Value of the polynomial when t = a is 4a³ – a² + 6.

If we multiply a number by $\dfrac{5}{2}$ and add $\dfrac{2}{3}$ to the product, we get $\dfrac{-7}{12}$. Find the number.

Answer

Let the number be x.

According to the question:

$\dfrac{5}{2}x + \dfrac{2}{3} = \dfrac{-7}{12}$

$\Rightarrow \dfrac{5}{2}x = \dfrac{-7}{12} - \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{5}{2}x = \dfrac{-7}{12} - \dfrac{2 \times 4}{3 \times 4} \quad\text{[Taking LCM of 12 and 3 as 12]}\\[1em] \Rightarrow \dfrac{5}{2}x = \dfrac{-7}{12} - \dfrac{8}{12} \\[1em] \Rightarrow \dfrac{5}{2}x = \dfrac{-7 - 8}{12} \\[1em] \Rightarrow \dfrac{5}{2}x = \dfrac{-15}{12} \\[1em] \Rightarrow x = \dfrac{-15}{12} \times \dfrac{2}{5} \\[1em] \Rightarrow x = \dfrac{-15 \times 2}{12 \times 5} \\[1em] \Rightarrow x = \dfrac{-30}{60} \\[1em] \Rightarrow x = -\dfrac{1}{2}$

Hence, the number is $-\dfrac{1}{2}$.

A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Answer

Let the smaller number be x.

Then, the larger number = 5x.

After adding 21 to both:

New smaller number = x + 21

New larger number = 5x + 21

According to the question, after adding 21, one of the new numbers becomes twice the other.

Since 5x + 21 > x + 21, the larger new number is twice the smaller new number.

So,

5x + 21 = 2(x + 21)

⇒ 5x + 21 = 2x + 42

⇒ 5x - 2x = 42 - 21

⇒ 3x = 21

⇒ x = $\dfrac{21}{3}$

⇒ x = 7

So,

Smaller number = 7

Larger number = 5 × 7 = 35

Hence, the two numbers are 7 and 35.

If you have ₹800 and you save ₹250 every month, find the amount you have after (i) 6 months (ii) 2 years. Express this as a linear pattern.

Answer

Initial amount = ₹800

Amount saved every month = ₹250

After n months, the total amount will be:

Amount = 800 + 250n

(i) After 6 months:

Amount = 800 + 250 × 6

= 800 + 1500

= ₹2300

∴ The amount after 6 months is ₹2300.

(ii) After 2 years:

2 years = 2 × 12 = 24 months

Amount = 800 + 250 × 24

= 800 + 6000

= ₹6800

∴ The amount after 2 years is ₹6800.

The linear pattern representing the amount after n months is (800 + 250n).

The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. Find both the numbers.

Answer

Let the digit at the tens place be x and the digit at the units place be y.

Then the original two-digit number = 10x + y.

When the digits are interchanged, the new number = 10y + x.

According to the question, the digits differ by 3.

So, |x - y| = 3

This gives us either x - y = 3 or y - x = 3.

Also, the sum of the original number and the interchanged number is 143.

So,

(10x + y) + (10y + x) = 143

⇒ 11x + 11y = 143

⇒ 11(x + y) = 143

⇒ x + y = $\dfrac{143}{11}$

⇒ x + y = 13

Case 1: x - y = 3

Adding x + y = 13 and x - y = 3:

2x = 16

⇒ x = 8

Substituting x = 8 in x + y = 13:

8 + y = 13

⇒ y = 5

Original number = 10(8) + 5 = 85

Interchanged number = 10(5) + 8 = 58

Case 2: y - x = 3

Adding x + y = 13 and y - x = 3:

2y = 16

⇒ y = 8

Substituting y = 8 in x + y = 13:

x + 8 = 13

⇒ x = 5

Original number = 10(5) + 8 = 58

Interchanged number = 10(8) + 5 = 85

Hence, the two numbers are 58 and 85.

Draw the graph of the following equations, and identify their slopes and y-intercepts. Also, find the coordinates of the points where these lines cut the y-axis.
(i) y = –3x + 4
(ii) 2y = 4x + 7
(iii) 5y = 6x – 10
(iv) 3y = 6x – 11
Are any of the lines parallel?

Answer

In a linear equation of the form y = ax + b, 'a' is the slope and 'b' is the y-intercept. The line cuts the y-axis at the point (0, b).

(i) y = -3x + 4

This is already in the form y = ax + b with a = -3 and b = 4.

To draw the graph, we need two points:

When x = 0, y = -3(0) + 4 = 4. Point: (0, 4).

When x = 1, y = -3(1) + 4 = 1. Point: (1, 1).

Slope = -3, y-intercept = 4. The line cuts the y-axis at (0, 4).

(ii) 2y = 4x + 7

Dividing both sides by 2:

y = $\dfrac{4x + 7}{2} = 2x + \dfrac{7}{2}$

So, a = 2 and b = $\dfrac{7}{2}$ = 3.5

To draw the graph:

When x = 0, y = 3.5. Point: (0, 3.5).

When x = 1, y = 2(1) + 3.5 = 5.5. Point: (1, 5.5).

Slope = 2, y-intercept = $\dfrac{7}{2}$ (or 3.5). The line cuts the y-axis at $\left(0, \dfrac{7}{2}\right)$.

(iii) 5y = 6x - 10

Dividing both sides by 5:

y = $\dfrac{6x - 10}{5} = \dfrac{6}{5}x - 2$

So, a = $\dfrac{6}{5}$ and b = -2.

To draw the graph:

When x = 0, y = -2. Point: (0, -2).

When x = 5, y = $\dfrac{6}{5}(5)$ - 2 = 6 - 2 = 4. Point: (5, 4).

Slope = $\dfrac{6}{5}$, y-intercept = -2. The line cuts the y-axis at (0, -2).

(iv) 3y = 6x - 11

Dividing both sides by 3:

y = $\dfrac{6x - 11}{3} = 2x - \dfrac{11}{3}$

So, a = 2 and b = $-\dfrac{11}{3}$.

To draw the graph:

When x = 0, y = $-\dfrac{11}{3} = -3.6$. Point: (0, -3.6)

When x = 3, y = 2(3) - $\dfrac{11}{3} = 6 - \dfrac{11}{3} = \dfrac{18 - 11}{3} = \dfrac{7}{3} = 2.3$. Point: (3, 2.3)

Slope = 2, y-intercept = $-\dfrac{11}{3}$ or (-3.6). The line cuts the y-axis at $\left(0, -\dfrac{11}{3}\right)$.

Are any of the lines parallel?

Comparing the slopes:

Slope of (i) = -3

Slope of (ii) = 2

Slope of (iii) = $\dfrac{6}{5}$

Slope of (iv) = 2

Lines (ii) and (iv) have the same slope (2), but different y-intercepts ($\dfrac{7}{2}$ and $-\dfrac{11}{3}$).

Hence, the lines 2y = 4x + 7 and 3y = 6x - 11 are parallel to each other.

If the temperature of a liquid can be measured in Kelvin units as x K and in Fahrenheit units as y °F, the relation between the two systems of measurement of temperature is given by the linear equation y = $\dfrac{9}{5}$(x – 273) + 32.
(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 K.
(ii) If the temperature is 158 °F, then find the temperature in Kelvin.

Answer

The given relation is y = $\dfrac{9}{5}$(x - 273) + 32.

(i) When x = 313 K:

Substituting x = 313 in the equation:

$y = \left(\dfrac{9}{5}(313 - 273) + 32\right) \\[1em] = \left(\dfrac{9}{5}(40) + 32\right) \\[1em] = \left(\dfrac{9 \times 40}{5} + 32\right) \\[1em] = \left(\dfrac{360}{5} + 32\right) \\[1em] = (72 + 32) \\[1em] = 104 \text{°F}$

∴ The temperature in Fahrenheit is 104 °F.

(ii) When y = 158 °F:

Substituting y = 158 in the equation:

$\phantom{=}158 = \left(\dfrac{9}{5}(x - 273) + 32\right) \\[1em] \Rightarrow 158 - 32 = \left(\dfrac{9}{5}(x - 273)\right) \\[1em] \Rightarrow 126 = \left(\dfrac{9}{5}(x - 273)\right) \\[1em] \Rightarrow x - 273 = \left(\dfrac{126 \times 5}{9}\right) \\[1em] \Rightarrow x - 273 = \left(\dfrac{630}{9}\right) \\[1em] \Rightarrow x - 273 = 70 \\[1em] \Rightarrow x = (70 + 273) \\[1em] \Rightarrow x = 343 \text{ K}$

∴ The temperature in Kelvin is 343 K.

The work done by a body on the application of a constant force is the product of the constant force and the distance travelled by the body in the direction of the force. Express this in the form of a linear equation in two variables (work w and distance d), and draw its graph by taking the constant force as 3 units. What is the work done when the distance travelled is 2 units? Verify it by plotting it on the graph.

Answer

Given:

Work done = constant force × distance travelled

Let the constant force be F units, work be w units and distance be d units.

So, w = F × d.

When the constant force is 3 units, the equation becomes:

w = 3d

This is a linear equation in two variables w and d.

To draw the graph, we identify the below points:

When d = 0, w = 3(0) = 0. Point: (0, 0).

When d = 1, w = 3(1) = 3. Point: (1, 3).

When d = 2, w = 3(2) = 6. Point: (2, 6).

Work done when d = 2:

Substituting d = 2 in w = 3d:

w = 3(2) = 6 units

∴ The work done when the distance travelled is 2 units is 6 units.

This is verified by the graph, since the point (2, 6) lies on the line w = 3d.

The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11).
(i) Find the polynomial p(x).
(ii) Find the coordinates where the graph of p(x) cuts the axes.
(iii) Draw the graph of p(x) and verify your answers.

Answer

(i) Let the linear polynomial be p(x) = ax + b.

Since the graph passes through (1, 5):

5 = a(1) + b

⇒ a + b = 5 ...(i)

Since the graph passes through (3, 11):

11 = a(3) + b

⇒ 3a + b = 11 ...(ii)

Subtracting equation (i) from equation (ii):

(3a + b) - (a + b) = 11 - 5

⇒ 2a = 6

⇒ a = 3

Substituting a = 3 in equation (i):

3 + b = 5

⇒ b = 2

∴ The polynomial is p(x) = 3x + 2.

(ii) To find where the graph cuts the y-axis, put x = 0:

p(0) = 3(0) + 2 = 2

So, the graph cuts the y-axis at (0, 2).

To find where the graph cuts the x-axis, put p(x) = 0 (i.e., y = 0):

3x + 2 = 0

⇒ 3x = -2

⇒ x = $-\dfrac{2}{3}$

So, the graph cuts the x-axis at $\left(-\dfrac{2}{3}, 0\right)$.

∴ The graph cuts the y-axis at (0, 2) and the x-axis at $\left(-\dfrac{2}{3}, 0\right)$.

(iii) To draw the graph, we use the points already found: (1, 5), (3, 11), (0, 2) and $\left(-\dfrac{2}{3}, 0\right)$.

The graph passes through all the four points, verifying our answers.

Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:
(i) p(0) = 5.
(ii) The polynomial p(x) – q(x) cuts the x-axis at (3, 0).
(iii) The sum p(x) + q(x) is equal to 6x + 4 for all real x.

Find the polynomials p(x) and q(x).

Answer

Given:

p(x) = ax + b and q(x) = cx + d

Using condition (i): p(0) = 5

p(0) = a(0) + b

5 = b

⇒ b = 5

Using condition (iii): p(x) + q(x) = 6x + 4

p(x) + q(x) = (ax + b) + (cx + d)

= (a + c)x + (b + d)

Comparing with 6x + 4:

a + c = 6 ...(i)

b + d = 4

Substituting b = 5:

5 + d = 4

⇒ d = -1

Using 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)

Substituting b = 5 and d = -1:

p(x) - q(x) = (a - c)x + (5 - (-1)) = (a - c)x + 6

Since the graph cuts the x-axis at (3, 0), substituting x = 3, p(x) - q(x) = 0:

(a - c)(3) + 6 = 0

⇒ 3(a - c) = -6

⇒ a - c = -2 ...(ii)

Solving (i) and (ii):

a + c = 6

a - c = -2

Adding:

2a = 4

⇒ a = 2

Substituting value of a in a + c = 6:

2 + c = 6

⇒ c = 4

So, a = 2, b = 5, c = 4, d = -1.

Hence, p(x) = 2x + 5 and q(x) = 4x - 1.

Look at the first three stages of a growing pattern of hexagons made using matchsticks. A new hexagon gets added at every stage which shares a side with the last hexagon of the previous stage.

(i) Draw the next two stages of the pattern. How many matchsticks will be required at these stages?
(ii) Complete the following table.

(iii) Find a rule to determine the number of matchsticks required for the n^th stage.
(iv) How many matchsticks will be required for the 15th stage of the pattern?
(v) Can 200 matchsticks form a stage in this pattern? Justify your answer.

Answer

From the pattern:

Stage 1 has 1 hexagon, requiring 6 matchsticks.

Stage 2 has 2 hexagons; the second hexagon shares one side with the first, so it adds 5 matchsticks. Total = 6 + 5 = 11 matchsticks.

Stage 3 has 3 hexagons; each new hexagon adds 5 matchsticks. Total = 11 + 5 = 16 matchsticks.

So, each new stage adds 5 more matchsticks to the previous stage.

(i) Stage 4 has 4 hexagons. Total matchsticks = 16 + 5 = 21

Stage 5 has 5 hexagons. Total matchsticks = 21 + 5 = 26

Hence, the 4th stage requires 21 matchsticks and the 5th stage requires 26 matchsticks.

(ii) Stage 1: 6 matchsticks

Stage 2: 11 matchsticks

Stage 3: 16 matchsticks

Stage 4: 21 matchsticks

Stage 5: 26 matchsticks

The completed table is:

(iii) Each stage has 5 more matchsticks than the previous stage, starting with 6 in stage 1.

Stage 1: 6 = 5(1) + 1

Stage 2: 11 = 5(2) + 1

Stage 3: 16 = 5(3) + 1

Stage 4: 21 = 5(4) + 1

Stage 5: 26 = 5(5) + 1

In general, at the n^th stage, the number of matchsticks = 5n + 1.

Hence, the rule is: Number of matchsticks at n^th stage = 5n + 1.

(iv) Number of matchsticks at the 15th stage:

= 5(15) + 1

= 75 + 1

= 76

∴ The 15th stage requires 76 matchsticks.

(v) For 200 matchsticks to form a stage, we need:

5n + 1 = 200

⇒ 5n = 200 - 1

⇒ 5n = 199

⇒ n = $\dfrac{199}{5}$

⇒ n = 39.8

Hence, 200 matchsticks cannot form a stage in this pattern, since 39.8 is not a whole number.

Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:
(i) The graph of p(x) passes through the points (2, 3) and (6, 11).
(ii) The graph of q(x) passes through the point (4, –1).
(iii) The graph of q(x) is parallel to the graph of p(x).

Find the polynomials p(x) and q(x). Also, find the coordinates of the point where these lines meet the x-axis.

Answer

Given:

p(x) = ax + b and q(x) = cx + d

Using condition (i): p(x) passes through (2, 3) and (6, 11)

When x = 2, p(x) = 3:

3 = 2a + b ...(I)

When x = 6, p(x) = 11:

11 = 6a + b ...(II)

Subtracting (I) from (II):

(6a + b) - (2a + b) = 11 - 3

⇒ 4a = 8

⇒ a = 2

Substituting a = 2 in (I):

3 = 2(2) + b

⇒ 3 = 4 + b

⇒ b = -1

∴ p(x) = 2x - 1.

Using condition (iii): q(x) is parallel to p(x)

Parallel lines have the same slope. So, c = a = 2.

Using condition (ii): q(x) passes through (4, -1)

When x = 4, q(x) = -1:

-1 = c(4) + d

⇒ -1 = 2(4) + d

⇒ -1 = 8 + d

⇒ d = -9

∴ q(x) = 2x - 9.

Finding the coordinates of the points where the lines meet the x-axis:

For p(x) = 2x - 1, when y = 0:

2x - 1 = 0

⇒ 2x = 1

⇒ x = $\dfrac{1}{2}$

So, p(x) meets the x-axis at $\left(\dfrac{1}{2}, 0\right)$.

For q(x) = 2x - 9, when y = 0:

2x - 9 = 0

⇒ 2x = 9

⇒ x = $\dfrac{9}{2}$

So, q(x) meets the x-axis at $\left(\dfrac{9}{2}, 0\right)$.

Hence, p(x) = 2x - 1 and q(x) = 2x - 9. The graph of p(x) meets the x-axis at $\left(\dfrac{1}{2}, 0\right)$ and the graph of q(x) meets the x-axis at $\left(\dfrac{9}{2}, 0\right)$.

What do all linear functions of the form f(x) = ax + a, a > 0, have in common?

Answer

The given linear function is f(x) = ax + a, where a > 0.

This can be rewritten as f(x) = a(x + 1).

To find a point common to all such functions, we look for a value of x that makes f(x) the same for any value of a.

When x = -1:

f(-1) = a(-1) + a = -a + a = 0

So, for every value of a > 0, the line f(x) = ax + a passes through the point (-1, 0).

Also, the slope of each line is 'a' (positive, since a > 0), so all these lines rise from bottom-left to top-right and represent linear growth.

The y-intercept of each line is 'a', which is the same as the slope.

Hence, all linear functions of the form f(x) = ax + a, where a > 0, pass through the common point (-1, 0) on the x-axis. They all have positive slopes (so they all represent linear growth), and the slope of each line is equal to its y-intercept.