Key Concepts & Formulae at a Glance
- Algebraic expression: combines numbers, variables and operations, e.g. \(2x^2+5xy-3y^2\).
- Univariate polynomial: an algebraic expression in one variable. Its degree is the highest power of that variable.
- Linear polynomial: a polynomial of degree 1, e.g. \(2x+3\).
- Linear growth — quantity increases by a fixed amount over equal intervals; linear decay — quantity decreases by a fixed amount over equal intervals.
- Linear pattern: a sequence where the difference between consecutive terms is constant.
- Parallel lines share the same slope a but have different y-intercepts b.
Think and Reflect
TR(After Example 2: garden fencing, 200l + 160w + 50lw)
1. Can you identify the terms, variables and coefficients of this algebraic expression?
2. How is it different from the algebraic expression in Example 1?
1. Algebraic expression: \(200l + 160w + 50lw\)
The terms are 200l, 160w and 50lw.
The variables used are l and w.
Coefficient of l = 200;
coefficient of w = 160;
coefficient of lw = 50.
There is no constant term.
2. Example 1 (4x + 5y + 3) has a constant term, and every term involves only one variable. Example 2 has no constant term, and its term 50lw is a product of both variables — a degree-2 term — unlike Example 1, where every term has degree 1 or 0.
TR(After Example 3: wire bent into a rectangle, 10x – x²)
1. Can you identify the terms, variables and coefficients of this algebraic expression?
2. Can you point out any similarity or difference between the algebraic expressions obtained in Examples 1 and 3?
1. Algebraic expression: \(10x - x^2\)
The terms are 10x and –x².
Only one variable is used: x.
Coefficient of x = 10; coefficient of x² = –1.
2. Example 3 involves only one variable (x), while Example 1 involves two (x and y). Example 3 also contains a degree-2 term (x²), while every term in Example 1 has degree 1 or 0.
Exercise Set 2.1
1Find the degrees of the following polynomials:
(i) 2x² – 5x + 3
(ii) y³ + 2y – 1
(iii) –9
(iv) 4z – 3
The degree of a polynomial is the highest power of the variable that appears in it.
(i) In \(2x^2-5x+3\), the powers of x present are 2, 1 and 0. The highest power is 2, so the degree is 2 (a quadratic polynomial).
(ii) In \(y^3+2y-1\), the powers of y present are 3, 1 and 0. The highest power is 3, so the degree is 3 (a cubic polynomial).
(iii) \(-9\) can be written as \(-9x^0\), so the only power present is 0. The degree is 0 (a constant polynomial).
(iv) In \(4z-3\), the powers of z present are 1 and 0. The highest power is 1, so the degree is 1 (a linear polynomial).
(ii) Degree 3
(iii) Degree 0
(iv) Degree 1
2Write polynomials of degrees 1, 2 and 3.
We need one polynomial each where the highest power of the variable is 1, 2 and 3 respectively. Any correct example is acceptable.
Degree 1 (linear): \(2x + 3\) — highest power of x here is 1.
Degree 2 (quadratic): \(x^2 + 2x + 1\) — highest power of x here is 2.
Degree 3 (cubic): \(x^3 + x^2 + x + 1\) — highest power of x here is 3.
Degree 2: x² + 2x + 1
Degree 3: x³ + x² + x + 1
3What are the coefficients of x² and x³ in the polynomial x⁴ – 3x³ + 6x² – 2x + 7?
Given: \(x^4 - 3x^3 + 6x^2 - 2x + 7\)
The term containing \(x^3\) is \(-3x^3\), so its coefficient is –3.
The term containing \(x^2\) is \(6x^2\), so its coefficient is 6.
Coefficient of x³ = –3
4What is the coefficient of z in the polynomial 4z³ + 5z² – 11?
Given: \(4z^3 + 5z^2 - 11\)
Scanning the polynomial, the powers of z present are 3, 2 and 0 — there is no term with z raised to the power 1.
Since there is no \(z^1\) term, its coefficient must be 0.
5What is the constant term of the polynomial 9x³ + 5x² – 8x – 10?
Given: \(9x^3 + 5x^2 - 8x - 10\)
The constant term is the term with no variable attached to it — the term that remains if x = 0. Here it is –10.
Think and Reflect
TR(After Example 4: perimeter 4x) 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?
Perimeter of a square of side x = \(4x\)
x = 1: \(4(1)=4\) cm
x = 1.5: \(4(1.5)=6\) cm
x = 2: \(4(2)=8\) cm
x = 2.5: \(4(2.5)=10\) cm
x = 3: \(4(3)=12\) cm
| Side (cm) | 1 | 1.5 | 2 | 2.5 | 3 |
|---|---|---|---|---|---|
| Perimeter (cm) | 4 | 6 | 8 | 10 | 12 |
Each time the side increases by 0.5 cm, the perimeter increases by \(6-4=2\), \(8-6=2\), \(10-8=2\), \(12-10=2\) — a constant 2 cm every time.
TR(After Example 5: chess club, 200 + 50m) If a player paid ₹750, how many matches did he play?
Total cost for m matches = \(200+50m\); total paid = ₹750
\(200+50m = 750\)
\(50m = 750-200 = 550\)
\(m = \dfrac{550}{50} = 11\)
TR(Input-output discussion) 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?
Here x, the length of the rectangle in cm, is the input fed into the machine. The machine computes the corresponding area using the rule \(10x - x^2\), and the area is the output.
At x = 6: \(10(6) - 6^2 = 60 - 36 = 24\)
Exercise Set 2.2
1Find the value of the linear polynomial 5x – 3 if:
(i) x = 0
(ii) x = –1
(iii) x = 2
To evaluate a polynomial at a given value, substitute that value in place of the variable and simplify.
(i) At x = 0: \(5(0) - 3 = 0 - 3 = -3\)
(ii) At x = –1: \(5(-1) - 3 = -5 - 3 = -8\)
(iii) At x = 2: \(5(2) - 3 = 10 - 3 = 7\)
(ii) –8
(iii) 7
2Find the value of the quadratic polynomial 7s² – 4s + 6 if:
(i) s = 0
(ii) s = –3
(iii) s = 4
(i) At s = 0: \(7(0)^2 - 4(0) + 6 = 0 - 0 + 6 = 6\)
(ii) At s = –3: \(7(-3)^2 - 4(-3) + 6 = 7(9) + 12 + 6 = 63 + 12 + 6 = 81\)
(iii) At s = 4: \(7(4)^2 - 4(4) + 6 = 7(16) - 16 + 6 = 112 - 16 + 6 = 102\)
(ii) 81
(iii) 102
3The 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.
Let Salil's present age = x years. Then his mother's present age = 3x years.
After 5 years, Salil's age = \(x+5\) and his mother's age = \(3x+5\).
Since their ages will add up to 70:
\((x+5) + (3x+5) = 70\)
\(4x + 10 = 70\)
\(4x = 60\)
\(x = 15\)
So Salil's present age = 15 years, and his mother's present age = \(3(15) = 45\) years.
Check: after 5 years, Salil = 20, mother = 50, and \(20+50=70\), which matches.
4The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.
Since the ratio of the integers is 2:5, let the integers be 2k and 5k for some positive number k.
Their difference is 63:
\(5k - 2k = 63\)
\(3k = 63\)
\(k = 21\)
So the integers are \(2k = 2(21) = 42\) and \(5k = 5(21) = 105\).
Check: \(105 - 42 = 63\), and \(42:105 = 2:5\), which matches.
5Ruby 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?
Let the number of ₹5 coins = x. Then the number of ₹2 coins = 3x.
Total value of all coins is ₹88:
\(5x + 2(3x) = 88\)
\(5x + 6x = 88\)
\(11x = 88\)
\(x = 8\)
So she has 8 coins of ₹5, and \(3(8)=24\) coins of ₹2.
Check: \(5(8) + 2(24) = 40 + 48 = 88\), which matches.
6A 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?
Let the shorter piece = x feet. Then the longer piece = 4x feet.
Together they make up the full fence:
\(x + 4x = 300\)
\(5x = 300\)
\(x = 60\)
So the shorter piece = 60 feet, and the longer piece = \(4(60) = 240\) feet.
Check: \(60+240=300\), which matches.
7If 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?
Let the width = w cm. Then the length = \(2w+3\) cm.
Using the perimeter formula, Perimeter = \(2(\text{length}+\text{width})\):
\(2(w + 2w+3) = 24\)
\(2(3w+3) = 24\)
\(6w+6 = 24\)
\(6w = 18\)
\(w = 3\)
So the width = 3 cm, and the length = \(2(3)+3 = 9\) cm.
Check: Perimeter = \(2(9+3) = 24\), which matches.
Think and Reflect
TRPredict 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.
Fig. 2.4 (redrawn from the textbook): A growing pattern of square tiles
Stages 1 to 4 give 1, 3, 5, 7 tiles — each stage adds 2 more tiles than the previous one.
Stage 5 = 7 + 2 = 9. Stage 6 = 9 + 2 = 11. Stage 7 = 11 + 2 = 13.
TRUsing 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?
Stage 15: \(2(15)-1 = 30-1 = 29\) tiles
Stage 26: \(2(26)-1 = 52-1 = 51\) tiles
For 21 tiles: \(2n-1=21 \Rightarrow 2n = 22 \Rightarrow n = 11\)
For 47 tiles: \(2n-1=47 \Rightarrow 2n = 48 \Rightarrow n = 24\)
Stage 26 → 51 tiles
21 tiles occurs at Stage 11
47 tiles occurs at Stage 24
TR(Example 7: Bela's pocket money, 100 – 5n) What amount will be left on the 15th day? How many days will it take for the entire amount to be spent?
Amount left on day 15: \(100 - 5(15) = 100 - 75 = 25\)
The amount becomes 0 when: \(100 - 5n = 0 \Rightarrow 5n = 100 \Rightarrow n = 20\)
TR(Auto-rikshaw fare, 15n – 5) For how many km will the fare be ₹130?
\(15n - 5 = 130\)
\(15n = 135\)
\(n = 9\)
Exercise Set 2.3
1A 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 nth month.
Starting amount = ₹500. After n months, she has received \(150n\) rupees in pocket money.
Balance at the end of every month: month 2 → ₹800, month 3 → ₹950, month 4 → ₹1100, and so on.
For the nth month: \(500 + 150n\)
| Month, n | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Amount (₹) | 650 | 800 | 950 | 1100 |
2A 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 nth hour.
Starting members = 120. After n hours, \(9n\) members have dropped out.
Members remaining after n hours \(= 120 - 9n\)
| Hour, n | 1 | 2 | 3 |
|---|---|---|---|
| Members | 111 | 102 | 93 |
3Suppose 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.
Area = length × breadth = \(13 \times b\), where b is the breadth.
(i) At b = 12: \(13 \times 12 = 156\) cm²
(ii) At b = 10: \(13 \times 10 = 130\) cm²
(iii) At b = 8: \(13 \times 8 = 104\) cm²
(i) 156 cm²
(ii) 130 cm²
(iii) 104 cm²
4Suppose 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.
Volume = length × breadth × height = \(7 \times 11 \times h = 77h\)
(i) At h = 5: \(77 \times 5 = 385\) cm³
(ii) At h = 9: \(77 \times 9 = 693\) cm³
(iii) At h = 13: \(77 \times 13 = 1001\) cm³
(i) 385 cm³
(ii) 693 cm³
(iii) 1001 cm³
5Sarita 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.
Pages left after n days \(= 500 - 20n\)
At n = 15: \(500 - 20(15) = 500 - 300 = 200\)
Think and Reflect
TR(Example 9: journey cost, C = 100 + 60d) What is the cost for travelling 15 km? For how many kilometres will the cost of the journey be ₹700?
At d = 15: \(C = 100 + 60(15) = 100 + 900 = 1000\)
Solving \(100 + 60d = 700\): \(60d = 600 \Rightarrow d = 10\)
TR(Example 10: water tank, h = 3 – 0.5t) What will be the height of the water at the end of 5 months?
At t = 5: \(h(5) = 3 - 0.5(5) = 3 - 2.5 = 0.5\)
Exercise Set 2.4
1Suppose 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.
(i) At t = 7: \(h = 1.75 + 0.5(7) = 1.75 + 3.5 = 5.25\) feet
(ii) Table of values:
| t | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| h | 1.75 | 2.25 | 2.75 | 3.25 | 3.75 | 4.25 | 4.75 | 5.25 | 5.75 | 6.25 | 6.75 |
h(t) = 1.75 + 0.5t — a straight line with positive slope 0.5; the point (7, 5.25) is the height after 7 months.
(iii) \(h(t) = 1.75 + 0.5t\). This represents linear growth because h increases by a fixed amount, 0.5 feet, for every one-month increase in t — a constant rate of increase, i.e. a positive slope.
(iii) h(t) = 1.75 + 0.5t — linear growth (positive slope of 0.5)
2A 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.
(i) At t = 3: \(v = 10000 - 800(3) = 10000 - 2400 = 7600\)
(ii) Table of values:
| t | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| v (₹) | 10000 | 9200 | 8400 | 7600 | 6800 | 6000 | 5200 | 4400 | 3600 |
v(t) = 10000 − 800t — a straight line with negative slope −800; the point (3, 7600) is the value after 3 years.
(iii) \(v(t) = 10000 - 800t\). This represents linear decay because v decreases by a fixed amount, ₹800, every year — a constant rate of decrease, i.e. a negative slope.
(iii) v(t) = 10000 – 800t — linear decay (negative slope of –800)
3The 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.
(i) At t = 6: \(P = 750 + 50(6) = 750 + 300 = 1050\)
(ii) Table of values:
| t | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| P | 750 | 800 | 850 | 900 | 950 | 1000 | 1050 | 1100 | 1150 | 1200 | 1250 |
P(t) = 750 + 50t — a straight line with positive slope 50; the point (6, 1050) is the population after 6 years.
(iii) \(P(t) = 750 + 50t\). This is linear growth because P increases by the fixed amount 50 every year — a positive slope of 50.
(iii) P(t) = 750 + 50t — linear growth (positive slope of 50)
4A 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.
(i) \(b(x) = 600 - 15x\). This is linear decay because the balance decreases by the fixed amount ₹15 every day — a negative slope of –15.
(ii) Solving \(600 - 15x = 0\): \(15x = 600 \Rightarrow x = 40\)
(iii) Table of values:
| x | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| b(x) | 585 | 570 | 555 | 540 | 525 | 510 | 495 | 480 | 465 | 450 |
b(x) = 600 − 15x — a straight line with negative slope −15; it reaches 0 at x = 40 days.
Think and Reflect
TR(Example 11: telecom bill, y = 20x + 150) Can you guess what the numbers 20 and 150 in the equation y = 20x + 150 represent?
In \(y = 20x + 150\), x is the amount of data used (GB) and y is the total bill (₹).
20 is the slope — the extra cost charged for every additional GB of data used.
150 is the y-intercept — the fixed monthly fee the student pays even if she uses 0 GB of data.
Exercise Set 2.5
1A 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.
When x = 10, y = 400: \(400 = 10a+b\) ...
(i)
When x = 14, y = 500: \(500 = 14a+b\) ...
(ii)
Subtracting (i) from (ii): \(4a = 100 \Rightarrow a = 25\)
Substituting a = 25 into (i): \(400 = 10(25)+b = 250+b \Rightarrow b = 150\)
2A 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.
When x = 10, y = 800: \(800 = 10a+b\) ...
(i)
When x = 15, y = 1100: \(1100 = 15a+b\) ...
(ii)
Subtracting (i) from (ii): \(5a = 300 \Rightarrow a = 60\)
Substituting a = 60 into (i): \(800 = 10(60)+b = 600+b \Rightarrow b = 200\)
3Consider 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.)
When °F = 32, °C = 0: \(0 = 32a+b\) ...
(i)
When °F = 212, °C = 100: \(100 = 212a+b\) ...
(ii)
Subtracting (i) from (ii): \(180a = 100 \Rightarrow a = \dfrac{100}{180} = \dfrac{5}{9}\)
Substituting \(a = \dfrac{5}{9}\) into (i): \(0 = 32\left(\dfrac{5}{9}\right)+b \Rightarrow b = -\dfrac{160}{9}\)
Think and Reflect
TR(Line y = 2x + 1) Identify other points on the line by completing the following table (x = 1, 2, 5, 7, 9, 12, 20).
Substituting each x-value into \(y = 2x + 1\):
x = 1: \(2(1)+1=3\)
x = 2: \(2(2)+1=5\)
x = 5: \(2(5)+1=11\)
x = 7: \(2(7)+1=15\)
x = 9: \(2(9)+1=19\)
x = 12: \(2(12)+1=25\)
x = 20: \(2(20)+1=41\)
| x | 1 | 2 | 5 | 7 | 9 | 12 | 20 |
|---|---|---|---|---|---|---|---|
| y | 3 | 5 | 11 | 15 | 19 | 25 | 41 |
The straight line y = 2x + 1, with the points A(0, 1), B(3, 7), (1, 3) and (7, 15) marked.
TR(Fig. 2.9: lines y = ax, a > 0) Does this help you to conclude anything about the linear equation y = ax, a > 0 as a varies? What happens when a > 1 and when a < 1?
Every line of the form y = ax passes through the origin (0, 0), since substituting x = 0 gives y = 0 regardless of a.
When a > 1, the line is steeper than y = x. When 0 < a < 1, the line is flatter (less steep) than y = x. When a = 1, the line coincides exactly with y = x.
y = x/2, y = x and y = 2x — all pass through the origin; the larger a is, the steeper the line.
TR(Fig. 2.11: lines y = –ax, a > 0) Does this help you to conclude anything about the linear equation y = –ax, a > 0, as a varies? What will happen when a > 1 and when a < 1?
These lines also pass through the origin, but slope downward since the coefficient of x is negative.
When a > 1, the line falls more steeply than y = –x. When 0 < a < 1, the line falls more gently than y = –x.
y = −x/3, y = −x and y = −3x — all pass through the origin and slope downward; the larger a is, the steeper the fall.
TRDifferentiate between the graphs of the equations y = 3x + 1, and y = –3x + 1.
Both lines have the same y-intercept, since at x = 0 both give y = 1 — they cross the y-axis at the same point, (0, 1).
y = 3x + 1 has slope +3 (rises left to right). y = –3x + 1 has slope –3, equal in magnitude but opposite in sign (falls left to right).
y = 3x + 1 rises and y = −3x + 1 falls; both cut the y-axis at (0, 1).
TR(Fig. 2.13: y = 2x – 1, y = 2x + 1, y = 2x + 5) Does this help you to conclude anything about the linear equation y = ax + b when a is fixed but b varies?
All three lines have the same slope, a = 2, so they are equally steep. Only b, the y-intercept, changes: –1, 1, and 5.
y = 2x − 1, y = 2x + 1 and y = 2x + 5 — parallel lines with slope 2 and y-intercepts −1, 1 and 5.
Exercise Set 2.6
1(i)y = 4x, y = 2x, y = x
All three lines pass through (0, 0) since b = 0 in each case. A second point for each:
y = 4x → (1, 4)
y = 2x → (1, 2)
y = x → (1, 1)
y = 4x, y = 2x and y = x — all pass through the origin; y = 4x is the steepest.
1(ii)y = –6x, y = –3x, y = –x
All three lines pass through (0, 0). A second point for each:
y = –6x → (1, –6)
y = –3x → (1, –3)
y = –x → (1, –1)
y = −6x, y = −3x and y = −x — all pass through the origin and slope downward; y = −6x is the steepest.
1(iii)y = 5x, y = –5x
y = 5x passes through (0, 0) and (1, 5).
y = –5x passes through (0, 0) and (1, –5).
y = 5x and y = −5x — same steepness, opposite directions; each is the mirror image of the other in the x-axis.
1(iv)y = 3x – 1, y = 3x, y = 3x + 1
y = 3x – 1 passes through (0, –1) and (1, 2).
y = 3x passes through (0, 0) and (1, 3).
y = 3x + 1 passes through (0, 1) and (1, 4).
y = 3x − 1, y = 3x and y = 3x + 1 — parallel lines with slope 3 and y-intercepts −1, 0 and 1.
1(v)y = –2x – 3, y = –2x, y = 2x + 3
y = –2x – 3 passes through (0, –3) and (1, –5).
y = –2x passes through (0, 0) and (1, –2).
y = 2x + 3 passes through (0, 3) and (1, 5).
y = −2x − 3 and y = −2x are parallel (slope −2); y = 2x + 3 has slope 2, so it is not parallel to them.
End-of-Chapter Exercises
1Write a polynomial of degree 3 in the variable x, in which the coefficient of the x² term is –7.
We need the highest power of x to be 3 (a cubic polynomial), and the term with x² must have coefficient –7.
\(x^3 - 7x^2 + 2x + 5\) satisfies both conditions: highest power is 3, and the coefficient of x² is –7.
2Find 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
(i) At x = 1: \(5(1)^2 - 3(1) + 7 = 5 - 3 + 7 = 9\)
(ii) Replacing t with a in the expression directly gives \(4a^3 - a^2 + 6\).
(ii) 4a³ – a² + 6
3If we multiply a number by \(\frac{5}{2}\) and add \(\frac{2}{3}\) to the product, we get \(-\frac{7}{12}\). Find the number.
Let the number be x.
\(\dfrac{5}{2}x + \dfrac{2}{3} = -\dfrac{7}{12}\)
\(\dfrac{5}{2}x = -\dfrac{7}{12} - \dfrac{2}{3}\)
Taking LCM 12: \(-\dfrac{7}{12} - \dfrac{8}{12} = -\dfrac{15}{12} = -\dfrac{5}{4}\)
\(\dfrac{5}{2}x = -\dfrac{5}{4}\)
\(x = -\dfrac{5}{4} \times \dfrac{2}{5} = -\dfrac{10}{20} = -\dfrac{1}{2}\)
4A 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?
Let the smaller number = x. Then the larger number = 5x.
Adding 21 to both: new smaller number = x + 21; new larger number = 5x + 21.
Since 5x + 21 was already the larger quantity, it becomes twice the other new number:
\(5x + 21 = 2(x+21)\)
\(5x+21 = 2x+42\)
\(3x = 21\)
\(x = 7\)
So the smaller number = 7, and the larger number = \(5(7) = 35\).
Check: new numbers are 28 and 56, and indeed \(56 = 2 \times 28\).
5If 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.
Amount after n months \(= 800 + 250n\)
(i) At n = 6: \(800 + 250(6) = 800 + 1500 = 2300\)
(ii) 2 years = 24 months. At n = 24: \(800 + 250(24) = 800 + 6000 = 6800\)
(i) ₹2,300
(ii) ₹6,800
6*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.
Let the tens digit = x and the units digit = y. Original number \(= 10x+y\). Interchanged number \(= 10y+x\).
Sum of the two numbers is 143:
\((10x+y)+(10y+x) = 143\)
\(11x+11y = 143\)
\(11(x+y) = 143\)
\(x+y = 13\) ...
(i)
The digits differ by 3:
\(x-y = 3\) ...
(ii)
Adding (i) and (ii): \(2x = 16 \Rightarrow x=8\)
From (i): \(y = 13-8 = 5\)
Original number = \(10(8)+5 = 85\); interchanged number = \(10(5)+8 = 58\)
Check: \(85+58=143\), and \(8-5=3\), both matching.
7*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?
Rewriting each equation in the form y = ax + b:
(i) \(y=-3x+4\) is already in this form; slope = –3, y-intercept = 4
(ii) \(2y=4x+7 \Rightarrow y=2x+3.5\); slope = 2, y-intercept = 3.5
(iii) \(5y=6x-10 \Rightarrow y=1.2x-2\); slope = 1.2, y-intercept = –2
(iv) \(3y=6x-11 \Rightarrow y=2x-\dfrac{11}{3}\); slope = 2, y-intercept = \(-\dfrac{11}{3} \approx -3.67\)
Points where each line cuts the y-axis:
(i) (0, 4)
(ii) (0, 3.5)
(iii) (0, –2)
(iv) \((0, -\dfrac{11}{3})\)
The four lines of Q7 with their y-intercepts marked; lines (ii) and (iv) have the same slope 2, so they are parallel.
Comparing the slopes: lines (ii) and (iv) both have slope 2, so these two lines are parallel to each other.
8*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 \(y = \frac{9}{5}(x - 273) + 32\).
(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.
(i) At x = 313: \(y = \dfrac{9}{5}(313-273)+32 = \dfrac{9}{5}(40)+32 = 72+32 = 104\)
(ii) At y = 158: \(158 = \dfrac{9}{5}(x-273)+32\)
\(126 = \dfrac{9}{5}(x-273)\)
\(x-273 = 126 \times \dfrac{5}{9} = 70\)
\(x = 343\)
(ii) 343 K
9*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.
Work \(w = \) force \(\times\) distance \(d\). With force = 3 units: \(w = 3d\)
This passes through the origin (0, 0) with slope 3; a second point is (2, 6).
w = 3d — a straight line through the origin; at d = 2 the work done is w = 6.
At d = 2: \(w = 3(2) = 6\)
10*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.
(i) Let p(x) = ax + b.
At (1, 5): \(5=a+b\) ... (1)
At (3, 11): \(11=3a+b\) ... (2)
Subtracting (1) from (2): \(2a=6 \Rightarrow a=3\)
Substituting into (1): \(5=3+b \Rightarrow b=2\)
So \(p(x) = 3x + 2\)
(ii) y-intercept: at x = 0, \(p(0)=2\), so the line cuts the y-axis at (0, 2).
x-intercept: \(0=3x+2 \Rightarrow x=-\dfrac{2}{3}\), so the line cuts the x-axis at \((-\dfrac{2}{3}, 0)\).
(iii) Graph:
p(x) = 3x + 2, passing through (1, 5) and (3, 11); it cuts the axes at (0, 2) and (−2/3, 0).
The graph confirms the line passes through (1, 5) and (3, 11), and crosses the axes at (0, 2) and \((-\dfrac{2}{3}, 0)\) as calculated.
11*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).
From p(0) = 5: \(p(0)=a(0)+b=b\), so \(b=5\)
From \(p(x)+q(x) = 6x+4\): \((ax+b)+(cx+d) = (a+c)x+(b+d)\)
Comparing with 6x + 4: \(a+c=6\) and \(b+d=4\)
Since b = 5: \(5+d=4 \Rightarrow d=-1\)
From p(x) – q(x) cutting the x-axis at (3, 0): \(p(x)-q(x) = (a-c)x+(b-d)\), which equals 0 at x = 3:
\(3(a-c)+(5-(-1))=0\)
\(3(a-c)+6=0 \Rightarrow a-c=-2\)
Solving \(a+c=6\) and \(a-c=-2\) together: adding gives \(2a=4 \Rightarrow a=2\), then \(c=6-2=4\)
So \(p(x) = 2x + 5\) and \(q(x) = 4x - 1\)
Check: \(p(x)-q(x) = -2x+6\), which is 0 at x = 3, and \(p(x)+q(x)=6x+4\), both matching.
12*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 nth 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.
The first three stages of the pattern (redrawn from the textbook).
Stage 1 is a single hexagon, using all 6 of its sides = 6 matchsticks. Every new hexagon added shares exactly 1 side (1 matchstick) with the hexagon before it, so it only needs 5 new matchsticks.
(i) Next two stages:
Stage 4 — 4 hexagons, 21 matchsticks.
Stage 5 — 5 hexagons, 26 matchsticks.
Stage 4 = 3(5) + 6 = 21 matchsticks. Stage 5 = 4(5) + 6 = 26 matchsticks.
(ii) Table of values:
| Stage, n | 1 | 2 | 3 | 4 | 5 | … | n |
|---|---|---|---|---|---|---|---|
| Matchsticks | 6 | 11 | 16 | 21 | 26 | … | 5n+1 |
(iii) The first hexagon needs 6 matchsticks, and each of the remaining (n – 1) hexagons adds 5 more:
Matchsticks \(= 6 + 5(n-1) = 6+5n-5 = 5n+1\)
(iv) At n = 15: \(5(15)+1 = 75+1 = 76\)
(v) Solving \(5n+1=200\): \(5n=199 \Rightarrow n=39.8\). Since n must be a whole number, this is not possible.
Stage 4 = 21, Stage 5 = 26
15th stage = 76 matchsticks
200 matchsticks cannot form any stage since n = 39.8 is not a whole number
13*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.
Let p(x) = ax + b.
At (2, 3): \(3=2a+b\) ... (1)
At (6, 11): \(11=6a+b\) ... (2)
Subtracting (1) from (2): \(4a=8 \Rightarrow a=2\)
Substituting into (1): \(3=2(2)+b=4+b \Rightarrow b=-1\)
So \(p(x) = 2x - 1\)
Since q(x) is parallel to p(x), their slopes are equal, so c = a = 2. Let q(x) = 2x + d.
q passes through (4, –1): \(-1=2(4)+d=8+d \Rightarrow d=-9\)
So \(q(x) = 2x - 9\)
x-intercept of p(x): \(2x-1=0 \Rightarrow x=\dfrac{1}{2}\), giving \((\dfrac{1}{2}, 0)\)
x-intercept of q(x): \(2x-9=0 \Rightarrow x=\dfrac{9}{2}\), giving \((\dfrac{9}{2}, 0)\)
p(x) = 2x − 1 and q(x) = 2x − 9 — parallel lines (slope 2); they cut the x-axis at (1/2, 0) and (9/2, 0).
14*What do all linear functions of the form f(x) = ax + a, a > 0, have in common?
\(f(x) = ax+a = a(x+1)\)
At x = –1: \(f(-1) = a(-1+1) = a(0) = 0\), true for every value of a.
Extra Practice Questions
Seven extra questions in the style of the textbook's own exercises, for independent practice once you've gone through the solved questions above. The two with a graph are shown open with their full working, since the diagram is part of reading the question — attempt each one on paper first, then tap to check.
1Classify each of the following as a linear, quadratic, or constant polynomial, and state its degree:
(i) \(7-2x\)
(ii) \(5x^2+x\)
(iii) \(12\)
The degree is the highest power of the variable present; that determines the type.
(i) In \(7-2x\), the highest power of x is 1 — this is a linear polynomial, degree 1.
(ii) In \(5x^2+x\), the highest power of x is 2 — this is a quadratic polynomial, degree 2.
(iii) \(12\) can be written as \(12x^0\), so the only power present is 0 — this is a constant polynomial, degree 0.
(ii) Quadratic, degree 2
(iii) Constant, degree 0.
2If \(f(x)=-4x+9\), find \(f(0)\), \(f(2)\), and \(f(-3)\).
Substitute each value of x into \(f(x)=-4x+9\).
\(f(0)=-4(0)+9=0+9=9\)
\(f(2)=-4(2)+9=-8+9=1\)
\(f(-3)=-4(-3)+9=12+9=21\)
3A water tank holds 800 litres and drains at 25 litres per minute. Write a linear expression for the amount of water left after m minutes. How much is left after 12 minutes?
The tank starts at 800 litres and loses 25 litres for every minute that passes.
Amount left after m minutes: \(w(m)=800-25m\)
After 12 minutes: \(w(12)=800-25(12)=800-300=500\)
4A candle is 20 cm tall and burns down 1.5 cm every hour.
(i) Write an expression for its height after h hours.
(ii) Find its height after 6 hours.
(iii) After how many hours will it be 5 cm tall?
H(h) = 20 − 1.5h — the height falls steadily; it is 11 cm at h = 6 and 5 cm at h = 10 hours.
The candle starts at 20 cm and loses 1.5 cm for every hour that passes.
(i) Height after h hours: \(H(h)=20-1.5h\)
(ii) After 6 hours: \(H(6)=20-1.5(6)=20-9=11\) cm
(iii) Set \(H(h)=5\): \(20-1.5h=5 \Rightarrow 1.5h=15 \Rightarrow h=10\) hours
(ii) 11 cm after 6 hours
(iii) 10 hours until it reaches 5 cm.
5A tutor charges \(y=ax+b\), where x is the number of sessions and y is the total cost in ₹. 4 sessions cost ₹1400; 7 sessions cost ₹2000. Find a and b, then find the cost for 10 sessions.
Substitute both data points into \(y=ax+b\) to get two equations.
From (4, 1400): \(4a+b=1400\)
From (7, 2000): \(7a+b=2000\)
Subtracting the first equation from the second: \(3a=600 \Rightarrow a=200\)
Substituting back: \(4(200)+b=1400 \Rightarrow 800+b=1400 \Rightarrow b=600\)
So \(y=200x+600\). For 10 sessions: \(y=200(10)+600=2000+600=2600\)
6Find the zero of the linear polynomial \(p(x)=3x-15\). If p(x) models a company's monthly profit (in ₹ thousands) after x months of operation, what does this zero represent?
The zero of a polynomial is the value of x that makes it equal to 0.
Set \(p(x)=0\): \(3x-15=0 \Rightarrow 3x=15 \Rightarrow x=5\)
In context, x = 5 is the month in which the company's profit is exactly ₹0 — the break-even point. Before month 5 the company is running at a loss (p(x) is negative), and after month 5 it turns profitable (p(x) is positive).
7Determine which of the following lines are parallel, using their slopes: \(y=5x-2\), \(y=-5x+2\), \(y=5x+8\).
y = 5x − 2 and y = 5x + 8 are parallel (slope 5); y = −5x + 2 has slope −5, so it is not parallel to them.
Two lines \(y=m_1x+c_1\) and \(y=m_2x+c_2\) are parallel exactly when their slopes are equal: \(m_1=m_2\).
\(y=5x-2\) has slope 5.
\(y=-5x+2\) has slope −5.
\(y=5x+8\) has slope 5.
The first and third lines both have slope 5, so they are parallel. The second line has a different slope (−5), so it is not parallel to the other two.