Chapter 2 exercise answers: Introduction to Linear Polynomials

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(i) 2x² − 5x + 3

1. List the powers of x that appear: 2, 1 and 0.
2. The highest power is 2.
3. So the degree is 2.

Answer2

(ii) y³ + 2y − 1

1. List the powers of y that appear: 3, 1 and 0.
2. The highest power is 3.
3. So the degree is 3.

Answer3

(iii) −9

1. −9 is a constant term, with no letter written next to it.
2. Write it as −9x⁰, since x⁰ = 1 for any x.
3. The only power of x here is 0, so the degree is 0.

Answer0

(iv) 4z − 3

1. List the powers of z that appear: 1 and 0.
2. The highest power is 1.
3. So the degree is 1.

Answer1

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Open NCERT p. 18Checked by computerAnswers can differ: one example

1. A polynomial of degree 1 has highest power 1: for example, x + 1.
2. A polynomial of degree 2 has highest power 2: for example, x² − 3.
3. A polynomial of degree 3 has highest power 3: for example, x³ + 2x − 1.

Answerx + 1 (degree 1), x² − 3 (degree 2), x³ + 2x − 1 (degree 3). Many other examples also work, as long as the highest power matches.

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1. Write down the number in front of each power: 1 for x⁴, −3 for x³, 6 for x², −2 for x, and 7 on its own.
2. The coefficient of x² is 6.
3. The coefficient of x³ is −3 (the minus sign belongs to the coefficient).

AnswerCoefficient of x² is 6; coefficient of x³ is −3.

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1. Write the polynomial with every power showing: 4z³ + 5z² + 0z − 11.
2. There is no term with just z on its own, so its coefficient is 0.
3. The coefficient of z is 0.

Answer0

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1. The constant term is the part with no x attached to it — the number on its own.
2. That number is −10.
3. So the constant term is −10.

Answer−10

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(i) x = 0

1. Put x = 0 into 5x − 3.
2. 5 × 0 − 3 = 0 − 3.
3. = −3.

Answer-3

(ii) x = −1

1. Put x = −1 into 5x − 3.
2. 5 × (−1) − 3 = −5 − 3.
3. = −8.

Answer-8

(iii) x = 2

1. Put x = 2 into 5x − 3.
2. 5 × 2 − 3 = 10 − 3.
3. = 7.

Answer7

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(i) s = 0

1. Put s = 0 into 7s² − 4s + 6.
2. 7 × 0 − 4 × 0 + 6 = 0 − 0 + 6.
3. = 6.

Answer6

(ii) s = −3

1. Put s = −3 into 7s² − 4s + 6.
2. (−3)² = 9, so 7 × 9 = 63.
3. −4 × (−3) = 12.
4. 63 + 12 + 6 = 81.

Answer81

(iii) s = 4

1. Put s = 4 into 7s² − 4s + 6.
2. 4² = 16, so 7 × 16 = 112.
3. −4 × 4 = −16.
4. 112 − 16 + 6 = 102.

Answer102

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1. Let Salil's present age be x years.
2. His mother's present age is 3x years (three times Salil's).
3. After 5 years, Salil will be x + 5 and his mother will be 3x + 5.
4. Their ages add up to 70: (x + 5) + (3x + 5) = 70.
5. 4x + 10 = 70, so 4x = 60, so x = 15.
6. Salil is 15 years old, and his mother is 3 × 15 = 45 years old.

AnswerSalil is 15 years old; his mother is 45 years old.

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1. Let the two integers be 2k and 5k, since their ratio is 2:5.
2. Their difference is 63: 5k − 2k = 63.
3. 3k = 63, so k = 21.
4. Smaller integer = 2 × 21 = 42.
5. Larger integer = 5 × 21 = 105.

AnswerThe two integers are 42 and 105.

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1. Let the number of five-rupee coins be x.
2. Then the number of two-rupee coins is 3x (three times as many).
3. Total value: 5 × x + 2 × 3x = 88.
4. 5x + 6x = 88, so 11x = 88, so x = 8.
5. Five-rupee coins = 8, and two-rupee coins = 3 × 8 = 24.

Answer24 two-rupee coins and 8 five-rupee coins.

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1. Let the shorter piece be x feet.
2. The longer piece is 4x feet (four times as long).
3. Together they make the whole fence: x + 4x = 300.
4. 5x = 300, so x = 60.
5. Shorter piece = 60 feet, and longer piece = 4 × 60 = 240 feet.
6. Check: 60 + 240 = 300 feet, and 240 is four times 60.

AnswerThe shorter piece is 60 feet and the longer piece is 240 feet.

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1. Let the width be w cm.
2. The length is 2w + 3 cm (three more than twice the width).
3. Perimeter = 2 × (length + width) = 24.
4. 2 × ((2w + 3) + w) = 24, so 2(3w + 3) = 24.
5. 6w + 6 = 24, so 6w = 18, so w = 3.
6. Width = 3 cm, and length = 2 × 3 + 3 = 9 cm.

AnswerWidth = 3 cm, length = 9 cm.

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Open NCERT p. 23Checked by computerReads two ways: both answers shown

1. The question can be read in two ways and the book does not say which it means, so both are worked out. The first takes the words exactly as printed.
2. Reading 1: ₹500 is what she has now, and ₹150 is added at the end of every month, the first month included.
3. End of month 1: 500 + 150 = ₹650. End of month 2: 650 + 150 = ₹800. End of month 3: ₹950. End of month 4: ₹1100.
4. After n months, ₹150 has been added n times, so at the end of the n^th month she has 500 + 150n.
5. Reading 2: ₹500 is what she has in the first month, and ₹150 is added from the second month onwards (the phrase 'from the second month onwards' can be taken this way).
6. Month 1: ₹500. Month 2: 500 + 150 = ₹650. Month 3: ₹800. Month 4: ₹950.
7. In the n^th month, ₹150 has been added (n − 1) times, so she has 500 + 150(n − 1) = 150n + 350.
8. Both are linear expressions: the amount goes up by the same ₹150 every month, so 150 multiplies n in each.

AnswerRead as ₹150 added every month, the first included: ₹800, ₹950, ₹1100, … at the end of months 2, 3, 4, …, and the amount in the n^th month is 500 + 150n. Read as ₹500 in the first month and ₹150 added from the second month: ₹650, ₹800, ₹950, … in months 2, 3, 4, …, and the amount in the n^th month is 150n + 350.

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1. Each hour, 9 members leave, so after n hours the group has lost 9n members from the original 120: 120 − 9n.
2. After 1 hour: 120 − 9 = 111.
3. After 2 hours: 120 − 18 = 102.
4. After 3 hours: 120 − 27 = 93.
5. So the expression for the nth hour is 120 − 9n.

Answer111, 102 and 93 members remain after 1, 2 and 3 hours; in general, 120 − 9n members remain after n hours.

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(i) 12 cm

1. Area of a rectangle = length × breadth.
2. Area = 13 × 12.
3. = 156 cm².

Answer156 cm²

(ii) 10 cm

1. Area = 13 × 10.
2. = 130 cm².

Answer130 cm²

(iii) 8 cm

1. Area = 13 × 8.
2. = 104 cm².

Answer104 cm²

general

1. The length is fixed at 13 cm and the breadth is b cm.
2. Area = length × breadth = 13 × b.
3. So the area in terms of b is 13b cm², a linear pattern in b.

Answer13b

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(i) 5 cm

1. Volume of a box = length × breadth × height.
2. Base area = 7 × 11 = 77 cm².
3. Volume = 77 × 5.
4. = 385 cm³.

Answer385 cm³

(ii) 9 cm

1. Volume = 77 × 9.
2. = 693 cm³.

Answer693 cm³

(iii) 13 cm

1. Volume = 77 × 13.
2. = 1001 cm³.

Answer1001 cm³

general

1. The base of the box (length × breadth) is fixed at 7 × 11 = 77 cm².
2. Volume = base area × height = 77 × h.
3. So the volume in terms of h is 77h cm³, a linear pattern in h.

Answer77h

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1. Pages read in d days = 20d.
2. Pages left = 500 − 20d.
3. After 15 days: 500 − 20 × 15 = 500 − 300 = 200.

Answer200 pages are left after 15 days; in general, 500 − 20d pages are left after d days.

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(i) Find the height after 7 months

1. Height after t months: h = 1.75 + 0.5t.
2. After 7 months: h = 1.75 + 0.5 × 7 = 1.75 + 3.5.
3. = 5.25 feet.

Answer5.25 feet

(ii) table of values for t varying from 0 to 10 months

1. Use h = 1.75 + 0.5t for t = 0, 1, 2, ..., 10.

2. t (months)	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

AnswerHeights run 1.75, 2.25, 2.75, 3.25, 3.75, 4.25, 4.75, 5.25, 5.75, 6.25, 6.75 feet.

(iii) expression that relates h and t, and explain why

1. The starting height (at t = 0) is 1.75 feet, and it grows by 0.5 feet every month.
2. So h = 1.75 + 0.5t.
3. Each month adds the same fixed amount, 0.5 feet — an equal gain over every equal interval — which is exactly what linear growth means.

Answerh = 1.75 + 0.5t; it grows by the same 0.5 feet every month, so it is linear growth.

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(i) Find the value of the phone after 3 years

1. Value after t years: v = 10000 − 800t.
2. After 3 years: v = 10000 − 800 × 3 = 10000 − 2400.
3. = ₹7600.

Answer₹7600

(ii) table of values for t varying from 0 to 8 years

1. Use v = 10000 − 800t for t = 0, 1, 2, ..., 8.

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

AnswerValues run ₹10000, 9200, 8400, 7600, 6800, 6000, 5200, 4400, 3600.

(iii) expression that relates v and t, and explain why

1. The starting value (at t = 0) is ₹10000, and it falls by ₹800 every year.
2. So v = 10000 − 800t.
3. Each year subtracts the same fixed amount, ₹800 — an equal loss over every equal interval — which is exactly what linear decay means.

Answerv = 10000 − 800t; it loses the same ₹800 every year, so it is linear decay.

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(i) Find the population of the village after 6 years

1. Population after t years: P = 750 + 50t.
2. After 6 years: P = 750 + 50 × 6 = 750 + 300.
3. = 1050.

Answer1050

(ii) table of values for t varying from 0 to 10 years

1. Use P = 750 + 50t for t = 0, 1, 2, ..., 10.

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

AnswerPopulations run 750, 800, 850, 900, 950, 1000, 1050, 1100, 1150, 1200, 1250.

(iii) expression that relates P and t, and explain why

1. The starting population (at t = 0) is 750, and it gains 50 people every year.
2. So P = 750 + 50t.
3. Each year adds the same fixed amount, 50 people — an equal gain over every equal interval — which is exactly what linear growth means.

AnswerP = 750 + 50t; it gains the same 50 people every year, so it is linear growth.

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(i) equation that models the remaining balance b(x) … for x days

1. Starting balance is ₹600, and it falls by ₹15 for every day used.
2. So b(x) = 600 − 15x.
3. Each day subtracts the same fixed amount, ₹15 — an equal loss over every equal interval — which is exactly what linear decay means.

Answerb(x) = 600 − 15x; it loses the same ₹15 every day, so it is linear decay.

(ii) After how many days will the balance run out?

1. The balance runs out when b(x) = 0.
2. 600 − 15x = 0, so 15x = 600.
3. x = 40.

Answer40 days

(iii) table of values for x varying from 1 to 10 days

1. Use b(x) = 600 − 15x for x = 1, 2, ..., 10.

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

AnswerBalances run ₹585, 570, 555, 540, 525, 510, 495, 480, 465, 450.

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1. Let the monthly bill be y = ax + b, where x is the number of modules used.
2. At x = 10, y = 400, so 10a + b = 400.
3. At x = 14, y = 500, so 14a + b = 500.
4. Subtract the first equation from the second: 4a = 100, so a = 25.
5. Put a = 25 back into 10a + b = 400: b = 400 − 250 = 150.

Answera = 25, b = 150

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1. Let the monthly bill be y = ax + b, where x is the number of hours used.
2. At x = 10, y = 800, so 10a + b = 800.
3. At x = 15, y = 1100, so 15a + b = 1100.
4. Subtract the first equation from the second: 5a = 300, so a = 60.
5. Put a = 60 back into 10a + b = 800: b = 800 − 600 = 200.

Answera = 60, b = 200

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1. The relation is °C = a×°F + b.
2. Ice melts at °C = 0, °F = 32: 0 = 32a + b.
3. Water boils at °C = 100, °F = 212: 100 = 212a + b.
4. Subtract the first equation from the second: 180a = 100, so a = 5/9.
5. Put a = 5/9 into 0 = 32a + b: b = −32 × 5/9 = −160/9.

Answera = 5/9, b = −160/9

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Open NCERT p. 36One way to think about it

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

1. At x = 0 all three lines give y = 0. At x = 1: y = 4x gives 4, y = 2x gives 2, y = x gives 1.
2. Plot (0, 0) and (1, 4) for y = 4x; (0, 0) and (1, 2) for y = 2x; (0, 0) and (1, 1) for y = x.
3. Join each pair of points with a ruler and extend the line both ways.

In shortAll three lines pass through the origin (0, 0), since b = 0 for each. A bigger value of a makes the line steeper: y = 4x is the steepest, y = x is the gentlest, and y = 2x is in between.

(ii) y = −6x, y = −3x, y = −x

1. At x = 0 all three give y = 0. At x = 1: y = −6x gives −6, y = −3x gives −3, y = −x gives −1.
2. Plot (0, 0) and (1, −6) for y = −6x; (0, 0) and (1, −3) for y = −3x; (0, 0) and (1, −1) for y = −x.
3. Join each pair of points and extend both ways.

In shortAll three lines again pass through the origin, since b = 0. All three fall, because a is negative in each case. y = −6x is the steepest fall, and y = −x is the gentlest.

(iii) y = 5x, y = −5x

1. At x = 0 both give y = 0. At x = 1: y = 5x gives 5, and y = −5x gives −5.
2. Plot (0, 0) and (1, 5) for y = 5x; (0, 0) and (1, −5) for y = −5x.
3. Join each pair of points and extend both ways.

In shortBoth lines pass through the origin and are equally steep, since |a| = 5 for both. But y = 5x rises while y = −5x falls — they are mirror images of each other across the x-axis.

(iv) y = 3x − 1, y = 3x, …

1. All three have a = 3. At x = 0: y = 3x − 1 gives −1, y = 3x gives 0, y = 3x + 1 gives 1. At x = 1: they give 2, 3 and 4.
2. Plot (0, −1), (1, 2) for y = 3x − 1; (0, 0), (1, 3) for y = 3x; (0, 1), (1, 4) for y = 3x + 1.
3. Join each pair of points and extend both ways.

In shortAll three lines are equally steep (same a = 3), so they are parallel to one another and never meet. Changing b only slides the line up or down — it does not tilt it.

(v) y = −2x − 3, y = −2x, …

1. y = −2x − 3: at x = 0, y = −3; at x = 1, y = −5.
2. y = −2x: at x = 0, y = 0; at x = 1, y = −2.
3. y = 2x + 3: at x = 0, y = 3; at x = 1, y = 5.
4. Plot each pair of points and join with a ruler, extending both ways.

In shorty = −2x − 3 and y = −2x have the same a = −2, so they are parallel to each other. y = 2x + 3 has a = 2 (opposite sign), so it is not parallel to the other two — it rises while they fall, and it crosses both of them.

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Open NCERT p. 36Checked by computerAnswers can differ: one example

1. A degree-3 polynomial in x must have an x³ term, with the power 3 actually present.
2. The coefficient of the x² term must be −7, so include the term −7x².
3. The coefficients of the x³ term, the x term and the constant term can be any numbers — pick simple ones: 1 for x³, 2 for x, and −5 for the constant.

Answerx³ − 7x² + 2x − 5 (one valid example; the x³ coefficient just needs to be non-zero, and the x and constant terms can be any numbers).

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(i) 5x² − 3x + 7 if x = 1

1. Substitute x = 1 into 5x² − 3x + 7.
2. 5(1)² = 5, and −3(1) = −3.
3. 5 − 3 + 7 = 9.

Answer9

(ii) 4t³ − t² + 6 if t = a

1. Substitute t = a into 4t³ − t² + 6.
2. This gives 4a³ − a² + 6 directly — substituting a letter works exactly the same way as substituting a number.

Answer4a³ − a² + 6

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1. Let the number be n.
2. Multiplying by 5/2 and adding 2/3 gives (5/2)n + 2/3.
3. This is equal to −7/12: (5/2)n + 2/3 = −7/12.
4. Subtract 2/3 from both sides: (5/2)n = −7/12 − 8/12 = −15/12 = −5/4.
5. Divide both sides by 5/2 (multiply by 2/5): n = −5/4 × 2/5 = −1/2.

Answer−1/2

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1. Let the smaller number be x, so the positive number (5 times it) is 5x.
2. Adding 21 to both gives x + 21 and 5x + 21.
3. 'One new number is twice the other', and 5x + 21 is the bigger one, so 5x + 21 = 2(x + 21).
4. Solve: 5x + 21 = 2x + 42, so 3x = 21, giving x = 7.
5. So the numbers are 7 and 5 × 7 = 35. Check: 7 + 21 = 28, 35 + 21 = 56, and 56 = 2 × 28. ✓

AnswerThe numbers are 7 and 35.

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(i) 6 months

1. The amount after n months follows the linear pattern: amount = 800 + 250n, since you start with ₹800 and save ₹250 every month.
2. For n = 6: 800 + 250 × 6 = 800 + 1500 = 2300.

Answer₹2,300 (using the pattern amount = 800 + 250n)

(ii) 2 years

1. 2 years = 24 months, so use n = 24 in the same pattern, amount = 800 + 250n.
2. 800 + 250 × 24 = 800 + 6000 = 6800.

Answer₹6,800 (using the pattern amount = 800 + 250n)

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1. Let the tens digit be x and the units digit be y, so the number is 10x + y.
2. The digits differ by 3, so x − y = 3 (taking x as the bigger digit).
3. Interchanging the digits gives the number 10y + x.
4. Adding the original and interchanged numbers: (10x + y) + (10y + x) = 11x + 11y = 143, so x + y = 13.
5. Solve x − y = 3 and x + y = 13 together: adding the two equations gives 2x = 16, so x = 8, and then y = 5.
6. The two-digit number is 85 (tens digit 8, units digit 5); reading the same pair of digits the other way round gives 58, which fits the question equally well.

AnswerThe two numbers are 58 and 85.

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(i) y = −3x + 4

1. The equation is already in the form y = ax + b: y = −3x + 4.
2. Slope a = −3. Constant term b = 4, so the line cuts the y-axis at (0, 4).

AnswerSlope = −3, y-intercept = 4, cuts the y-axis at (0, 4).

(ii) 2y = 4x + 7

1. Divide throughout by 2 to write it as y = ax + b: y = 2x + 7/2.
2. Slope a = 2. Constant term b = 7/2, so it cuts the y-axis at (0, 7/2).

AnswerSlope = 2, y-intercept = 7/2, cuts the y-axis at (0, 7/2).

(iii) 5y = 6x − 10

1. Divide throughout by 5: y = (6/5)x − 2.
2. Slope a = 6/5. Constant term b = −2, so it cuts the y-axis at (0, −2).

AnswerSlope = 6/5, y-intercept = −2, cuts the y-axis at (0, −2).

(iv) 3y = 6x − 11

1. Divide throughout by 3: y = 2x − 11/3.
2. Slope a = 2. Constant term b = −11/3, so it cuts the y-axis at (0, −11/3).
3. Compare all four slopes: −3, 2, 6/5 and 2. Lines (ii) and (iv) both have slope 2, so they are parallel to each other; no other pair shares a slope.

AnswerSlope = 2, y-intercept = −11/3, cuts the y-axis at (0, −11/3). Lines (ii) and (iv) are parallel to each other (both have slope 2); no other pair is parallel.

Watch this explained “The name for a”, 2:35 into What a and b do to the line: slope, y-intercept, and parallel families · हिंदी में देखें

Open NCERT p. 37Checked by computer

(i) the temperature of the liquid is 313 K

1. Substitute x = 313 into y = 9/5 (x − 273) + 32.
2. 313 − 273 = 40.
3. 9/5 × 40 = 72.
4. 72 + 32 = 104.

Answer104 °F

(ii) If the temperature is 158 °F

1. Substitute y = 158 into the equation: 158 = 9/5 (x − 273) + 32.
2. Subtract 32 from both sides: 126 = 9/5 (x − 273).
3. Multiply both sides by 5/9: x − 273 = 70.
4. Add 273 to both sides: x = 343.

Answer343 K

Watch this explained “Forwards, and backwards”, 2:10 into Linear growth and linear decay · हिंदी में देखें

Open NCERT p. 37Checked by computer

1. Work = constant force × distance, so w = F × d.
2. Taking the constant force as 3 units, this becomes w = 3d — a linear equation in the two variables w and d.
3. To plot it, use two points: at d = 0, w = 0; at d = 2, w = 6.
4. Substitute d = 2 into w = 3d: w = 3 × 2 = 6, matching the plotted point.

Answerw = 3d. When d = 2 units, w = 6 units.

Watch this explained “Two inputs, chosen not found”, 0:36 into Why two points are enough to draw the line · हिंदी में देखें

Open NCERT p. 38Checked by computer

(i) Find the polynomial p(x)

1. Let p(x) = ax + b.
2. p(1) = 5 gives a + b = 5.
3. p(3) = 11 gives 3a + b = 11.
4. Subtract: 2a = 6, so a = 3.
5. Then b = 5 − 3 = 2.

Answerp(x) = 3x + 2

(ii) Find the coordinates where the graph of p(x) cuts the axes

1. The graph cuts the x-axis where p(x) = 0: 3x + 2 = 0, so x = −2/3.
2. The graph cuts the y-axis where x = 0: p(0) = 2.

AnswerCuts the x-axis at (−2/3, 0) and the y-axis at (0, 2).

(iii)

1. Plot the two given points (1, 5) and (3, 11), and join them with a ruler, extending the line both ways.
2. Check that the same line also passes through (−2/3, 0) and (0, 2) — both should lie exactly on it.

AnswerThe line through (1, 5) and (3, 11) also passes through (−2/3, 0) and (0, 2), confirming the answers to (i) and (ii).

Watch this explained “Two equations, solved”, 2:55 into Recovering y = ax + b from two observations · हिंदी में देखें

Open NCERT p. 38Checked by computer

1. p(0) = 5 means the constant term of p(x) is b = 5, since p(0) = a×0 + b = b.
2. p(x) − q(x) = (a − c)x + (b − d); this cuts the x-axis at (3, 0), so 3(a − c) + (b − d) = 0.
3. p(x) + q(x) = 6x + 4 for all x, so matching coefficients: a + c = 6 and b + d = 4.
4. From b + d = 4 and b = 5: d = 4 − 5 = −1.
5. Substitute b = 5 and d = −1 into 3(a − c) + (b − d) = 0: 3(a − c) + 6 = 0, so a − c = −2.
6. Solve a + c = 6 and a − c = −2 together: adding gives 2a = 4, so a = 2, and then c = 4.

Answerp(x) = 2x + 5 and q(x) = 4x − 1.

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Open NCERT p. 38Checked by computer

(i) Draw the next two stages of the pattern

1. Stage 1 has 1 hexagon, using all 6 of its sides: 6 matchsticks.
2. Each new hexagon shares one side with the hexagon before it, so it only needs 5 new matchsticks (6 sides minus the 1 already shared).
3. Stage 4 = stage 3's count + 5 = 16 + 5 = 21. Stage 5 = 21 + 5 = 26.

AnswerStage 4 needs 21 matchsticks; Stage 5 needs 26 matchsticks.

(ii) Complete the following table

1. Stage 1: 6. Stage 2: 6 + 5 = 11. Stage 3: 11 + 5 = 16. Stage 4: 21. Stage 5: 26 (each stage adds 5 to the one before).

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

Answer1 → 6, 2 → 11, 3 → 16, 4 → 21, 5 → 26, and n → 5n + 1.

(iii) Find a rule to determine the number of matchsticks

1. From the drawing: stage n has 1 starting hexagon (6 sticks) plus (n − 1) more hexagons added, each contributing 5 new sticks.
2. Total = 6 + 5(n − 1) = 6 + 5n − 5 = 5n + 1.

AnswerNumber of matchsticks at stage n = 5n + 1.

(iv) How many matchsticks will be required for the 15th stage

1. Substitute n = 15 into 5n + 1: 5 × 15 + 1 = 75 + 1 = 76.

Answer76 matchsticks.

(v) Can 200 matchsticks form a stage in this pattern

1. Set 5n + 1 = 200 and solve for n: 5n = 199, so n = 199/5 = 39.8.
2. A stage number must be a whole number, and 39.8 is not one.

AnswerNo — 200 matchsticks cannot form any stage, since 5n + 1 = 200 gives n = 39.8, which is not a whole number.

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Open NCERT p. 39Checked by computer

1. p(x) = ax + b passes through (2, 3) and (6, 11): a = (11 − 3)/(6 − 2) = 8/4 = 2.
2. Then b = 3 − 2×2 = −1, so p(x) = 2x − 1.
3. q(x) is parallel to p(x), so it has the same slope: c = a = 2.
4. q(x) passes through (4, −1): d = −1 − 2×4 = −9, so q(x) = 2x − 9.
5. p(x) meets the x-axis where 2x − 1 = 0, i.e. x = 1/2, giving (1/2, 0).
6. q(x) meets the x-axis where 2x − 9 = 0, i.e. x = 9/2, giving (9/2, 0).

Answerp(x) = 2x − 1 and q(x) = 2x − 9. p(x) meets the x-axis at (1/2, 0) and q(x) meets the x-axis at (9/2, 0).

Watch this explained “Same a, different b”, 8:29 into What a and b do to the line: slope, y-intercept, and parallel families · हिंदी में देखें

Open NCERT p. 39One way to think about it

1. Try a few values of a: for a = 1, f(x) = x + 1; for a = 2, f(x) = 2x + 2; for a = 3, f(x) = 3x + 3.
2. Find where each crosses the x-axis by setting f(x) = 0: x + 1 = 0 gives x = −1; 2x + 2 = 0 gives x = −1; 3x + 3 = 0 gives x = −1.
3. In general, f(x) = ax + a = a(x + 1), which is zero exactly when x + 1 = 0, i.e. x = −1 — and this does not depend on a at all (as long as a is not zero).
4. Also, since a > 0 in every case, every one of these lines has a positive slope, so every one of them rises.

In shortThey all pass through the same point on the x-axis, (−1, 0), because f(x) = ax + a = a(x + 1) is zero exactly when x = −1, whatever positive value a takes. They are also all rising lines, since a > 0 in every case.

Watch this explained “Why nothing added means the origin”, 1:11 into What a and b do to the line: slope, y-intercept, and parallel families · हिंदी में देखें