Chapter 2

Introduction to Linear Polynomials

Many everyday situations grow or shrink by the same amount every step: a taxi fare, a savings account, a water tank emptying. This chapter gives you the simple algebra that describes all of them, and shows you how that algebra turns into a straight line on a graph.

What terms, variables, coefficients and constants are, using real examples
What a polynomial is, and how to find its degree
What makes a polynomial "linear", and why linear patterns go up or down by equal steps
How to turn word problems into equations and solve them
Linear growth and linear decay, with tables of values
How to find a rule like y = ax + b from two facts
How to draw y = ax + b as a straight line, and what a and b do to that line

Class 9, MathematicsOpen the original PDF

2.1A quick refresher: algebraic expressions

In earlier classes you met algebraic expressions: combinations of numbers, letters and operation signs like +, −, ×. The letters stand for numbers we do not know yet, or numbers that can change. Let us look at three real situations to see how such expressions are built.

Example 1: Raju's pens and pencils

Raju goes to a shop that sells sealed boxes. Each red box has 4 pens. Each blue box has 5 pencils. Raju buys x red boxes and y blue boxes. The shopkeeper also gives him 3 extra pens free. How many pens and pencils does he get altogether?

Book Fig. 2.1: the red boxes (4 pens each) and blue boxes (5 pencils each) in the shop.

x red boxes with 4 pens each give 4 × x = 4x pens.
y blue boxes with 5 pencils each give 5 × y = 5y pencils.
Plus 3 free pens.
Total: 4x + 5y + 3.

This expression has some parts that have names. Learn them now, because the whole chapter uses them:

Word	Meaning	In 4x + 5y + 3
Term	One piece of the expression, separated from the others by + or − signs	4x, 5y and 3
Variable	A letter that stands for a number that can change (earlier books called these "letter-numbers")	x and y
Coefficient	The number multiplying a variable	4 (of x) and 5 (of y)
Constant	A term with no variable, so its value never changes	3

Think of a shopping bill

The variables are like quantities you have not decided yet (how many boxes). The coefficients are like the price tags (how much each box gives you). The constant is like a fixed free gift that comes no matter what you buy.

Example 2: fencing and planting a garden

A rectangular garden is l metres long and w metres wide.

A wire fence goes along both lengths, costing ₹100 per metre.
A wooden fence goes along both widths, costing ₹80 per metre.
Seeds are sown over the whole garden, costing ₹50 per square metre.

The garden. Orange: the two long sides (wire fence). Green: the two short sides (wooden fence). The inside is sown with seeds.

Wire fence: there are two lengths, so 2l metres. Cost = 2l × 100 = ₹200l.
Wooden fence: two widths, so 2w metres. Cost = 2w × 80 = ₹160w.
Seeds: these cover the area, which is l × w square metres. Cost = 50 × l × w = ₹50lw.
Total cost = ₹(200l + 160w + 50lw).

Think and reflect: name the terms, variables and coefficients of 200l + 160w + 50lw. How is it different from Example 1?Tap to reveal

Terms: 200l, 160w and 50lw. Variables: l and w. Coefficients: 200 (of l), 160 (of w) and 50 (of lw).

Differences from Example 1: here one term, 50lw, has two variables multiplied together. In Example 1 every term had at most one variable. Also, this expression has no constant term, while Example 1 had the constant 3.

Example 3: bending a wire into rectangles

A wire 20 cm long is bent into a rectangle. You could make a 7 cm by 3 cm rectangle, or a 5.5 cm by 4.5 cm rectangle, and many more. Can we write one expression for the area of all of them?

The wire becomes the perimeter. A rectangle's perimeter is 2 × (length + width) = 20, so length + width = 10. If the length is x cm, the width must be what is left over: (10 − x) cm.

Area = length × width = x(10 − x) = 10x − x².

Length x (cm)	Width 10 − x (cm)	Area 10x − x² (cm²)
2	8	16
3	7	21
4.5	5.5	24.75
5	5	25 (a square)
7	3	21

Think and reflect: name the terms, variables and coefficients of 10x − x². How does it compare with Example 1?Tap to reveal

Terms: 10x and −x². Variable: only x. Coefficients: 10 (of x) and −1 (of x²). The −1 is hidden: −x² means −1 × x².

Compared with Example 1: both are algebraic expressions. But Example 1 uses two variables (x and y), while this uses only one (x). Also, this one has x squared (x × x), while every variable in Example 1 appears only to the first power.

2.1Polynomials and their degree

Examples 1 and 2 used two variables. Example 3 used just one. From now on, this chapter only uses expressions with one variable, like 4x, x² + 1, 2y − 5 or 3z + 7.

Notice that a variable can appear with different powers. A power tells you how many times the variable is multiplied by itself: x² means x × x, and y³ means y × y × y.

Polynomial (in one variable)

An algebraic expression made of one variable, its whole-number powers, coefficients and a constant, joined by + and −. Examples: x² + 5x + 1, and 5y³ + y² + 2y − 1. Because it has only one variable, it is also called a univariate polynomial ("uni" means one, "variate" means variable).

Degree

The highest power of the variable in the polynomial. In x² + 5x + 1 the highest power is 2, so the degree is 2. In 5y³ + y² − 8 the highest power is 3, so the degree is 3.

Polynomials get special names based on their degree:

Degree	Name	Example
0	Constant polynomial	8
1	Linear polynomial	3z + 7
2	Quadratic polynomial	x² + 5x + 1
3	Cubic polynomial	5y³ + y² + 2y − 1

Why is a plain number like 8 "degree 0"?

Any number raised to the power 0 equals 1. So x⁰ = 1, and 8 can be written as 8 × x⁰ = 8x⁰. The power of x there is 0, so the degree is 0.

Let us read one polynomial carefully: 5y³ + y² + 2y − 1.

The coefficient of y³ is 5.
The coefficient of y² is 1 (y² means 1 × y², the 1 is just not written).
The coefficient of y is 2.
The constant term is −1 (the minus sign belongs to it).

Three traps

1. The sign belongs to the coefficient. In x⁴ − 3x³, the coefficient of x³ is −3, not 3.

2. An invisible 1. In y² or x, the coefficient is 1. In −x², it is −1.

3. A missing term has coefficient 0. In 4z³ + 5z² − 11 there is no plain z term, so the coefficient of z is 0.

Exercise Set 2.1, solved

1. Find the degree of (i) 2x² − 5x + 3 (ii) y³ + 2y − 1 (iii) −9 (iv) 4z − 3Tap to reveal

(i) Highest power of x is 2, so degree 2.
(ii) Highest power of y is 3, so degree 3.
(iii) −9 is a constant (−9x⁰), so degree 0.
(iv) 4z is z to the power 1, so degree 1.

2. Write polynomials of degree 1, 2 and 3.Tap to reveal

Many answers are correct. For example: degree 1: 3x + 2; degree 2: x² − 4x + 1; degree 3: 2x³ + x − 5. Just make sure the highest power is the one you want.

3. What are the coefficients of x² and x³ in x⁴ − 3x³ + 6x² − 2x + 7?Tap to reveal

Coefficient of x² is 6. Coefficient of x³ is −3 (remember, the minus sign comes with it).

4. What is the coefficient of z in 4z³ + 5z² − 11?Tap to reveal

There is no plain z term, so the coefficient of z is 0. You could write the polynomial as 4z³ + 5z² + 0z − 11.

5. What is the constant term of 9x³ + 5x² − 8x − 10?Tap to reveal

The term with no x is −10.

2.2Linear polynomials: equal steps, equal jumps

Remember, a linear polynomial has degree 1. It looks like (number) × x + (number), for example 4x, 2x + 3, or 200 + 50m. The rest of this chapter is all about these.

Example 4: the perimeter of a square

A square with side x has perimeter 4x (four equal sides). 4x is a linear polynomial.

Side (cm)	1	1.5	2	2.5	3
Perimeter 4x (cm)	4	6	8	10	12

Answering the book's question: every time the side grows by 0.5 cm, the perimeter grows by 2 cm, every single time.

Example 5: a chess club

A chess club charges a joining fee of ₹200, plus ₹50 for every match you play.

Matches played	1	2	3	4	5	m
Amount paid (₹)	250	300	350	400	450	200 + 50m

For m matches, the cost is ₹(200 + 50m). That is a linear polynomial in m. Each extra match adds exactly ₹50.

Think and reflect: a player paid ₹750. How many matches did he play?Tap to reveal

Set the cost equal to 750: 200 + 50m = 750.
Take away the joining fee from both sides: 50m = 550.
Divide both sides by 50: m = 11.

He played 11 matches.

The signature of a linear polynomial

When the input goes up in equal steps, the output goes up (or down) in equal jumps. The perimeter always jumps by 2. The chess fee always jumps by 50. A list of numbers that changes by the same amount each time is called a linear pattern.

Compare this with the wire rectangles from Example 3, where the area was 10x − x² (not linear). As x goes 1, 2, 3, 4, 5, the area goes 9, 16, 21, 24, 25. The jumps are 7, 5, 3, 1: not equal. That is how you can tell a quadratic apart from a linear pattern.

From a linear polynomial to a linear equation

When you set a linear polynomial equal to a number, you get a linear equation, which you can solve to find the unknown.

Example 6: two numbers

The sum of two numbers is 64. One is 10 more than the other. Find them.

Name the unknown. Let the smaller number be x. Then the larger is x + 10.
Write the equation. Their sum is 64: x + (x + 10) = 64, which is 2x + 10 = 64.
Solve. Subtract 10 from both sides: 2x = 54. Divide by 2: x = 27.
Answer. The numbers are 27 and 27 + 10 = 37.
Check. 27 + 37 = 64. Correct.

Solving is like a balance scale

An equation is a balanced scale: the left side weighs the same as the right side. Whatever you do to one side (add, subtract, multiply, divide), do the same to the other side, and it stays balanced. Keep doing this until x is alone on one side.

A polynomial is an input-output machine

Take the linear polynomial 2x + 3. Put in any number for x, and you get exactly one number out:

Put in x = 4: 2 × 4 + 3 = 8 + 3 = 11.
Put in x = −6: 2 × (−6) + 3 = −12 + 3 = −9.

Putting a number in place of the variable is called substituting. A process like this, where every input gives exactly one output, is called a function. We say "2x + 3 is a function of x". You will learn much more about functions in later classes.

Think and reflect: the wire rectangle's area 10x − x² is also a function of x. What is the area when x = 6 cm?Tap to reveal

Yes, it is an input-output process: put in a length, get out an area. For x = 6: 10 × 6 − 6² = 60 − 36 = 24 cm². That is the 6 cm by 4 cm rectangle.

Note: 2x + 3 is a linear function, while 10x − x² is a quadratic function.

Exercise Set 2.2, solved

A 4-step recipe for word problems

1. Choose a letter for the unknown. 2. Write every other quantity using that letter. 3. Turn the sentence that gives a total or a condition into an equation. 4. Solve, then check your answer in the original words.

1. Find 5x − 3 when (i) x = 0 (ii) x = −1 (iii) x = 2Tap to reveal

(i) 5 × 0 − 3 = 0 − 3 = −3
(ii) 5 × (−1) − 3 = −5 − 3 = −8
(iii) 5 × 2 − 3 = 10 − 3 = 7

2. Find 7s² − 4s + 6 when (i) s = 0 (ii) s = −3 (iii) s = 4Tap to reveal

(i) 7 × 0 − 0 + 6 = 6
(ii) 7 × (−3)² − 4 × (−3) + 6 = 7 × 9 + 12 + 6 = 63 + 12 + 6 = 81. Careful: (−3)² = 9 (positive), and −4 × (−3) = +12.
(iii) 7 × 16 − 16 + 6 = 112 − 16 + 6 = 102

3. Salil's mother is three times Salil's age. In 5 years their ages will add up to 70. Find their ages now.Tap to reveal

Let Salil's age now be x. His mother's age now is 3x.
In 5 years: Salil is x + 5, mother is 3x + 5.
Their sum then is 70: (x + 5) + (3x + 5) = 70, so 4x + 10 = 70.
4x = 60, so x = 15.

Salil is 15 years old and his mother is 45.

Check: in 5 years they will be 20 and 50, and 20 + 50 = 70.

4. Two positive integers differ by 63 and are in the ratio 2 : 5. Find them.Tap to reveal

A ratio of 2 : 5 means the numbers are 2 "parts" and 5 "parts". Call one part k, so the numbers are 2k and 5k.
Their difference is 5k − 2k = 3k = 63, so k = 21.

The integers are 2 × 21 = 42 and 5 × 21 = 105.

Check: 105 − 42 = 63, and 42 : 105 simplifies to 2 : 5 (divide both by 21).

5. Ruby has 3 times as many two-rupee coins as five-rupee coins. Altogether she has ₹88. How many coins of each?Tap to reveal

Let the number of five-rupee coins be x. Then she has 3x two-rupee coins.
Money from five-rupee coins: 5x. From two-rupee coins: 2 × 3x = 6x.
Total: 5x + 6x = 11x = 88, so x = 8.

8 five-rupee coins and 24 two-rupee coins.

Check: 8 × 5 = 40 and 24 × 2 = 48, and 40 + 48 = 88.

6. A 300 foot fence is cut into two pieces. The longer is 4 times the shorter. How long is each?Tap to reveal

Let the shorter piece be x feet. The longer is 4x feet.
Together: x + 4x = 5x = 300, so x = 60.

The pieces are 60 feet and 240 feet.

7. A rectangle's length is three more than twice its width. Its perimeter is 24 cm. Find its dimensions.Tap to reveal

Let the width be w cm. "Three more than twice the width" means length = 2w + 3.
Perimeter = 2 × (length + width) = 2 × (2w + 3 + w) = 2 × (3w + 3) = 6w + 6.
6w + 6 = 24, so 6w = 18, so w = 3.
Length = 2 × 3 + 3 = 9.

The rectangle is 9 cm long and 3 cm wide.

Check: 2 × (9 + 3) = 24.

2.3Exploring linear patterns

Look at this growing pattern of square tiles. Each new stage has two more tiles than the one before.

A growing pattern of square tiles (book Fig. 2.4): 1, 3, 5, 7 tiles. Our drawing arranges them as an L; the book's picture may arrange them differently, but the counts are the same.

Stage	1	2	3	4	5	6	7
Tiles	1	3	5	7	9	11	13

The next three stages after Stage 4 have 9, 11 and 13 tiles. But how could we find, say, Stage 100 without counting 100 stages? We need a rule.

How to find the rule, step by step

Find the jump. The tiles go up by 2 each time. So the rule will have 2n in it (n is the stage number).
Test 2n. For stages 1, 2, 3 it gives 2, 4, 6. But the real counts are 1, 3, 5.
Fix the difference. Each real count is exactly 1 less than 2n. So the rule is 2n − 1.
Check. Stage 5: 2 × 5 − 1 = 9. Correct.

Shortcut for any linear pattern

If the pattern jumps by d each time and starts at a, then the rule is d × n + (a − d). Here d = 2 and a = 1, so the rule is 2n + (1 − 2) = 2n − 1.

2n − 1 has degree 1, so it is a linear polynomial. That fits: the counts jump by the same amount, 2, every time.

Think and reflect: how many tiles in Stage 15 and Stage 26? Which stage has 21 tiles? 47 tiles?Tap to reveal

Stage 15: 2 × 15 − 1 = 29 tiles.
Stage 26: 2 × 26 − 1 = 51 tiles.
21 tiles: 2n − 1 = 21, so 2n = 22, so n = 11.
47 tiles: 2n − 1 = 47, so 2n = 48, so n = 24.

Example 7: Bela's pocket money

Bela has ₹100 and spends ₹5 every day. After how many days will she have ₹40 left?

Day	0	1	2	3	4	n
Money left (₹)	100	95	90	85	80	100 − 5n

Each day takes away 5, so after n days she has 100 − 5n. To have ₹40 left: 100 − 5n = 40, so 5n = 60, so n = 12 days.

Think and reflect: how much is left on day 15? When is all the money gone?Tap to reveal

Day 15: 100 − 5 × 15 = 100 − 75 = ₹25.

All gone when 100 − 5n = 0, so 5n = 100, so n = 20 days.

Example 8: an auto-rickshaw fare

The fare starts at ₹25 and stays ₹25 for the first 2 km. After that it rises by ₹15 for every extra km.

Km travelled	1	2	3	4	5	6
Fare (₹)	25	25	40	55	70	85

For 10 km: the first 2 km cost ₹25, and there are 10 − 2 = 8 extra km. Fare = 25 + 15 × 8 = ₹145.

For any n km (with n at least 2), the extra km are n − 2, so:

Fare = 25 + 15(n − 2) = 25 + 15n − 30 = 15n − 5

Why "n − 2" and not "n"?

The ₹15 charge only starts after the first 2 km, which are already covered by the ₹25. If you wrote 25 + 15n, you would be charging the first 2 km twice.

Think and reflect: for how many km is the fare ₹130?Tap to reveal

15n − 5 = 130, so 15n = 135, so n = 9 km. Check: 25 + 15 × 7 = 25 + 105 = 130.

Linear pattern

A list of numbers where the difference between any two neighbours is always the same. In every example above, the n-th number is given by a linear expression in n. You will study these more in the chapter on sequences and progressions.

Exercise Set 2.3, solved

1. A student has ₹500 in her account and gets ₹150 every month. How much does she have month by month? Find an expression for month n.Tap to reveal

End of month 1: 500 + 150 = ₹650. Month 2: ₹800. Month 3: ₹950. Month 4: ₹1100. Each month adds ₹150.

Amount at the end of month n = 500 + 150n

2. A rally starts with 120 members, and 9 drop out each hour. How many remain after 1, 2, 3 hours? Find an expression for hour n.Tap to reveal

After 1 hour: 111. After 2 hours: 102. After 3 hours: 93.

Members after n hours = 120 − 9n

3. A rectangle is 13 cm long. Find its area when the breadth is 12, 10 and 8 cm. What is the linear pattern?Tap to reveal

13 × 12 = 156 cm². 13 × 10 = 130 cm². 13 × 8 = 104 cm².

Area = 13b, where b is the breadth

Each time the breadth drops by 2 cm, the area drops by 13 × 2 = 26 cm², always the same jump.

4. A box has length 7 cm and breadth 11 cm. Find its volume for heights 5, 9 and 13 cm. What is the linear pattern?Tap to reveal

The base is 7 × 11 = 77 cm². Volume = base × height.

77 × 5 = 385 cm³. 77 × 9 = 693 cm³. 77 × 13 = 1001 cm³.

Volume = 77h, where h is the height

Each extra 4 cm of height adds 77 × 4 = 308 cm³.

5. Sarita reads 20 pages a day of a 500-page book. How many pages are left after 15 days? Write the linear pattern.Tap to reveal

After 15 days she has read 20 × 15 = 300 pages, so 500 − 300 = 200 pages are left.

Pages left after n days = 500 − 20n

2.4Linear growth and linear decay

Linear expressions are perfect for describing things that grow or shrink steadily.

Example 9: the cost of a journey (growth)

The cost of a journey is C(d) = 100 + 60d, where d is the distance in km and C is the cost in rupees.

C(d) is not "C times d"

The brackets here are function notation. Read C(d) as "C of d", meaning "the cost when the distance is d". So C(3) means "the cost for 3 km" = 100 + 60 × 3 = ₹280. Nothing is being multiplied by C.

Distance d (km)	0	1	2	3	4	5
Cost C (₹)	100	160	220	280	340	400

Every extra km adds the same ₹60. That steady increase is called linear growth.

Think and reflect: what does 15 km cost? For how many km is the cost ₹700?Tap to reveal

C(15) = 100 + 60 × 15 = 100 + 900 = ₹1000.

100 + 60d = 700, so 60d = 600, so d = 10 km.

Example 10: water in a tank (decay)

At the start of summer, the water in a tank is 3 m high. After t months, the height is h(t) = 3 − 0.5t metres.

Month t	0	1	2	3	4
Height h (m)	3	2.5	2	1.5	1

Every month takes away the same 0.5 m. That steady decrease is called linear decay.

Think and reflect: how high is the water after 5 months?Tap to reveal

h(5) = 3 − 0.5 × 5 = 3 − 2.5 = 0.5 m. (One more month and the tank would be empty, since h(6) = 0.)

Linear growth

A quantity increases by the same amount over equal time steps. Example: +₹60 per km.

Linear decay

A quantity decreases by the same amount over equal time steps. Example: −0.5 m per month.

Exercise Set 2.4, solved

1. A plant is 1.75 feet tall and grows 0.5 feet a month. (i) Height after 7 months? (ii) Table for 0 to 10 months. (iii) Expression, and why it is linear growth.Tap to reveal

(i) 1.75 + 0.5 × 7 = 1.75 + 3.5 = 5.25 feet.

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

(iii) h = 1.75 + 0.5t. It is linear growth because the height goes up by the same 0.5 feet every month.

2. A phone costs ₹10,000 and loses ₹800 of value each year. (i) Value after 3 years? (ii) Table for 0 to 8 years. (iii) Expression, and why it is linear decay.Tap to reveal

(i) 10000 − 800 × 3 = 10000 − 2400 = ₹7,600.

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

(iii) v = 10000 − 800t. It is linear decay because the value drops by the same ₹800 every year.

3. A village has 750 people, and 50 people move in each year. (i) Population after 6 years? (ii) Table for 0 to 10 years. (iii) Expression, and why it is linear growth.Tap to reveal

(i) 750 + 50 × 6 = 750 + 300 = 1050 people.

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

(iii) P = 750 + 50t. Linear growth: the same 50 people are added every year.

4. A ₹600 recharge loses ₹15 each day. (i) Write b(x) for the balance after x days. (ii) When does it run out? (iii) Table for days 1 to 10.Tap to reveal

(i) b(x) = 600 − 15x. It is linear decay: the balance drops by the same ₹15 each day.

(ii) 600 − 15x = 0, so 15x = 600, so x = 40 days.

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

2.5Linear relationships: y = ax + b

A linear relationship connects two variables x and y by a rule of the form y = ax + b, where a and b are fixed numbers. For the tile pattern, if x is the stage and y is the number of tiles, the relationship is y = 2x − 1 (so a = 2 and b = −1).

Often we do not know a and b. We only know a couple of facts, and we have to work a and b out.

Example 11: a phone bill

A company charges a fixed monthly fee plus a fixed cost per GB of data. Using 10 GB gave a bill of ₹350. Using 20 GB gave ₹550. The bill is y = ax + b, where x is the GB used. Find a and b.

Turn each fact into an equation. x = 10, y = 350 gives 350 = 10a + b. And x = 20, y = 550 gives 550 = 20a + b.
Get b on its own from the first one. b = 350 − 10a.
Put that into the second one. 550 = 20a + (350 − 10a) = 10a + 350.
Solve. 10a = 200, so a = 20.
Find b. b = 350 − 10 × 20 = 350 − 200 = 150.
Write the rule. y = 20x + 150.

This method, where you rearrange one equation and put it into the other, is called substitution.

A quicker way to see a

Going from 10 GB to 20 GB (10 more GB) raised the bill from 350 to 550 (₹200 more). So each GB costs 200 ÷ 10 = ₹20. That is a. Then the fixed part is 350 − 20 × 10 = ₹150. That is b.

Think and reflect: what do 20 and 150 mean in y = 20x + 150?Tap to reveal

20 is the cost of each GB of data (₹20 per GB). 150 is the fixed monthly fee you pay even if you use no data at all (put x = 0 and you get y = 150).

Exercise Set 2.5, solved

1. A learning platform: 10 modules cost ₹400, 14 modules cost ₹500. Find a and b in y = ax + b.Tap to reveal

4 more modules cost ₹100 more, so a = 100 ÷ 4 = 25 (₹25 per module).

Then 400 = 25 × 10 + b, so b = 400 − 250 = 150 (fixed monthly fee).

y = 25x + 150

2. A gym: 10 hours of badminton cost ₹800, 15 hours cost ₹1100. Find a and b.Tap to reveal

5 more hours cost ₹300 more, so a = 300 ÷ 5 = 60 (₹60 per hour).

Then 800 = 60 × 10 + b, so b = 800 − 600 = 200.

y = 60x + 200

3. Celsius and Fahrenheit: °C = a °F + b. Ice melts at 0 °C = 32 °F, water boils at 100 °C = 212 °F. Find a and b.Tap to reveal

Two facts: 0 = 32a + b and 100 = 212a + b.
Subtract the first from the second: 100 − 0 = (212 − 32)a, so 100 = 180a.
a = 100 ÷ 180 = 5/9.
From the first: b = −32a = −32 × 5/9 = −160/9.

°C = (5/9) °F − 160/9, which is the same as °C = (5/9)(°F − 32)

Check with boiling water: (5/9)(212 − 32) = (5/9) × 180 = 100. Correct.

2.6Seeing linear relationships as straight lines

Here is the most beautiful fact in this chapter: if you plot every pair (x, y) that fits y = ax + b on a coordinate grid, the points always form a perfectly straight line. That is why these are called "linear".

How to draw y = 2x + 1

A straight line is fixed once you know two points on it (only one straight line passes through two given points). So:

Pick an easy x, like x = 0. Then y = 2 × 0 + 1 = 1. Point A (0, 1).
Pick another x, like x = 3. Then y = 2 × 3 + 1 = 7. Point B (3, 7).
Plot A and B, join them with a ruler, and extend the line in both directions.

Redrawn Fig. 2.5. Two points are enough to draw the line. The green point (1, 3) is another point that sits on it.

Think and reflect: complete the table for y = 2x + 1 with x = 1, 2, 5, 7, 9, 12, 20.Tap to reveal

x	1	2	5	7	9	12	20
y	3	5	11	15	19	25	41

Every one of these points lies on the same line.

Is a point on the line? Substitute and see

A point lies on a line only if its coordinates make the equation true. Is (7, 15) on y = 2x + 1? Put x = 7: 2 × 7 + 1 = 15, which equals y. Yes. Is (4, 10) on it? 2 × 4 + 1 = 9, not 10. No.

Examples 12 and 13: guessing the rule from points

Plot (−1, −3), (0, 0), (1, 3), (3, 9), (4, 12). They lie on one straight line. Look at each pair: the y-coordinate is always 3 times the x-coordinate. So the line is y = 3x.

Redrawn Fig. 2.6. Every point has y = 3 × x.

Now plot (−3, 6), (−2, 4), (0, 0), (1, −2), (2, −4), (3, −6). This time the y-coordinate is always −2 times the x-coordinate. The line is y = −2x, and it slopes downward.

Redrawn Fig. 2.7. Every point has y = −2 × x, so the line falls as you move right.

What does a do? (Examples 14 and 15)

Draw y = ½x, y = x and y = 2x on the same grid. To plot y = ½x, the book suggests the points (0, 0) and (4, 2): check, ½ × 4 = 2.

Redrawn Figs. 2.8 and 2.9. All three lines pass through the origin. The bigger the number in front of x, the steeper the line.

What we notice:

Lines of the form y = ax always pass through the origin (0, 0), because when x = 0, y = a × 0 = 0.
y = x is exactly in the middle: it is equally tilted towards both axes.
When a is bigger than 1, the line is steeper than y = x.
When a is between 0 and 1, the line is less steep than y = x.

Slope

The number a in y = ax (or y = ax + b) is called the slope of the line. It tells you how much y changes when x goes up by 1. A slope of 2 means "up 2 for every 1 step right". You will learn more about slope in a later chapter.

Now try negative values of a: y = −⅓x, y = −x and y = −3x.

Redrawn Figs. 2.10 and 2.11. With a negative slope, the lines fall from left to right. Again, the bigger the size of the number, the steeper the line.

Positive slope rises, negative slope falls

Reading from left to right: a positive slope makes the line go up (like climbing a hill). A negative slope makes it go down. That links straight back to section 2.4: linear growth gives a line with positive slope, and linear decay gives a line with negative slope.

The slope is also the "jump" in a linear pattern. The tile pattern 1, 3, 5, 7 jumps by 2, and its line y = 2x − 1 has slope 2. Same number, two ways of seeing it.

Think and reflect: how do the graphs of y = 3x + 1 and y = −3x + 1 differ?Tap to reveal

Both cross the y-axis at the same point, (0, 1), and both are equally steep. But y = 3x + 1 rises from left to right (slope +3), while y = −3x + 1 falls (slope −3). Each is the mirror image of the other in the y-axis.

What does b do? (Example 16)

Draw y = 2x − 1, y = 2x + 1 and y = 2x + 5. They all have the same a = 2, but different b.

Redrawn Figs. 2.12 and 2.13. Same slope, so the lines are parallel. Changing b just slides the line up or down.

The three lines never meet: they are parallel. Changing b slides the whole line up or down without tilting it.

Where does the line cross the y-axis?

Draw y = x + 3, y = 2x + 5 and y = 3x − 2, and watch where each one cuts the y-axis.

Redrawn Fig. 2.14. Each line cuts the y-axis at the point (0, b).

The pattern: y = ax + b always cuts the y-axis at (0, b). That makes sense: on the y-axis x = 0, and then y = a × 0 + b = b.

y-intercept

The number b in y = ax + b. It tells you where the line crosses the y-axis. The y-intercept of y = x + 3 is 3 (it crosses 3 units above the origin). The y-intercept of y = 3x − 2 is −2 (it crosses 2 units below the origin).

The big summary of y = ax + b

y = ax + b

a = slope (how steep, and whether it rises or falls). b = y-intercept (where it crosses the y-axis).

Change a with b fixed: the line tilts around the point (0, b). Change b with a fixed: the line slides up or down and stays parallel. Lines with equal slopes and different intercepts are parallel.

Exercise Set 2.6, solved

(i) y = 4x, y = 2x, y = x. What do a and b do?Tap to reveal

Here b = 0 for all three, so all pass through the origin. The slopes are 4, 2 and 1, all positive, so all rise. The larger a is, the steeper the line: y = 4x is the steepest.

(ii) y = −6x, y = −3x, y = −xTap to reveal

All pass through the origin (b = 0). All slopes are negative, so all fall from left to right. The bigger the size of a (6 is bigger than 3, which is bigger than 1), the steeper the fall.

(iii) y = 5x, y = −5xTap to reveal

Both pass through the origin and are equally steep. One rises and the other falls: they are mirror images of each other in the y-axis (and also in the x-axis).

(iv) y = 3x − 1, y = 3x, y = 3x + 1Tap to reveal

Same slope a = 3, so the three lines are parallel. They differ only in b, so they cross the y-axis at −1, 0 and 1: each is slid up by 1 from the previous one.

(v) y = −2x − 3, y = −2x, y = 2x + 3Tap to reveal

y = −2x − 3 and y = −2x have the same slope (−2), so they are parallel, crossing the y-axis at −3 and 0. The third line, y = 2x + 3, has slope +2, so it rises instead and is not parallel to them. In fact, y = 2x + 3 is the mirror image of y = −2x − 3 in the x-axis.

ExEnd-of-chapter exercises, solved

Try each one first, then open it. Starred (*) questions are harder.

1. Write a degree 3 polynomial in x where the coefficient of x² is −7.Tap to reveal

For example: x³ − 7x² + 2x + 1. Any answer works if the highest power is 3 and the x² term is −7x².

2. Find (i) 5x² − 3x + 7 at x = 1 (ii) 4t³ − t² + 6 at t = a.Tap to reveal

(i) 5 × 1 − 3 × 1 + 7 = 5 − 3 + 7 = 9.

(ii) Replace every t with a: 4a³ − a² + 6. The answer is still an expression, because a is a letter.

3. Multiply a number by 5/2 and add 2/3. The result is −7/12. Find the number.Tap to reveal

Let the number be x. Then (5/2)x + 2/3 = −7/12.
Subtract 2/3 from both sides. Make the denominators the same: 2/3 = 8/12. So (5/2)x = −7/12 − 8/12 = −15/12 = −5/4.
Divide by 5/2, which means multiply by 2/5: x = −5/4 × 2/5 = −10/20 = −1/2.

The number is −1/2.

Check: (5/2) × (−1/2) = −5/4 = −15/12, and −15/12 + 8/12 = −7/12.

4. One positive number is 5 times another. Add 21 to both, and one new number becomes twice the other. Find the numbers.Tap to reveal

Let the numbers be x and 5x.
After adding 21: x + 21 and 5x + 21. The bigger one is twice the smaller: 5x + 21 = 2(x + 21).
5x + 21 = 2x + 42, so 3x = 21, so x = 7.

The numbers are 7 and 35.

Check: 7 + 21 = 28 and 35 + 21 = 56, and 56 = 2 × 28.

5. You have ₹800 and save ₹250 a month. How much after (i) 6 months (ii) 2 years? Write the linear pattern.Tap to reveal

After n months: 800 + 250n.

(i) 800 + 250 × 6 = 800 + 1500 = ₹2,300.

(ii) 2 years = 24 months: 800 + 250 × 24 = 800 + 6000 = ₹6,800.

*6. The digits of a two-digit number differ by 3. Adding the number to its reverse gives 143. Find both numbers.Tap to reveal

A two-digit number with tens digit a and units digit b is worth 10a + b. (For example, 85 = 10 × 8 + 5.)
Its reverse is 10b + a. The sum is 10a + b + 10b + a = 11a + 11b = 11(a + b).
11(a + b) = 143, so a + b = 13.
The digits differ by 3. Two numbers that add to 13 and differ by 3 are 8 and 5 (8 + 5 = 13, 8 − 5 = 3).

The numbers are 85 and 58.

Check: 85 + 58 = 143.

*7. Draw and find slope and y-intercept: (i) y = −3x + 4 (ii) 2y = 4x + 7 (iii) 5y = 6x − 10 (iv) 3y = 6x − 11. Are any parallel?Tap to reveal

First, get each one into the form y = ax + b by dividing both sides by the number in front of y.

Line	As y = ax + b	Slope a	y-intercept b	Cuts y-axis at
(i)	y = −3x + 4	−3	4	(0, 4)
(ii)	y = 2x + 7/2	2	7/2 = 3.5	(0, 3.5)
(iii)	y = (6/5)x − 2	6/5	−2	(0, −2)
(iv)	y = 2x − 11/3	2	−11/3 ≈ −3.67	(0, −11/3)

Parallel: lines (ii) and (iv), because both have slope 2.

*8. Kelvin x and Fahrenheit y are linked by y = (9/5)(x − 273) + 32. (i) Convert 313 K. (ii) Convert 158 °F to Kelvin.Tap to reveal

(i) y = (9/5)(313 − 273) + 32 = (9/5) × 40 + 32 = 72 + 32 = 104 °F.

(ii) 158 = (9/5)(x − 273) + 32. Subtract 32: 126 = (9/5)(x − 273). Multiply by 5/9: x − 273 = 70. So x = 343 K.

*9. Work = force × distance. Write this as a linear equation with force 3, draw it, and find the work when the distance is 2.Tap to reveal

With force 3: w = 3d. This is a line through the origin with slope 3.

For d = 2: w = 3 × 2 = 6 units. On the graph, go up from d = 2 to the line and across to the w-axis: it reads 6.

*10. The graph of p(x) passes through (1, 5) and (3, 11). Find p(x) and where it cuts the axes.Tap to reveal

p(x) = ax + b. From (1, 5): a + b = 5. From (3, 11): 3a + b = 11.
Subtract: 2a = 6, so a = 3. Then b = 5 − 3 = 2.
p(x) = 3x + 2.
y-axis (x = 0): p(0) = 2, so (0, 2).
x-axis (y = 0): 3x + 2 = 0, so x = −2/3, giving (−2/3, 0).

*11. p(x) = ax + b and q(x) = cx + d with p(0) = 5, p(x) − q(x) cutting the x-axis at (3, 0), and p(x) + q(x) = 6x + 4. Find p and q.Tap to reveal

p(0) = a × 0 + b = b, so b = 5.
p(x) + q(x) = (a + c)x + (b + d) = 6x + 4. So a + c = 6 and b + d = 4, which gives d = −1.
p(x) − q(x) = (a − c)x + (b − d) = (a − c)x + 6. It cuts the x-axis at x = 3, so it equals 0 there: 3(a − c) + 6 = 0, so a − c = −2.
Now a + c = 6 and a − c = −2. Adding them: 2a = 4, so a = 2, and c = 4.

p(x) = 2x + 5 and q(x) = 4x − 1

Check: p + q = 6x + 4. Correct. p − q = −2x + 6, which is 0 at x = 3. Correct.

*12. A matchstick hexagon pattern grows by one hexagon each stage, sharing a side. Find the rule and answer the questions.Tap to reveal

Each new hexagon shares one side with the previous one, so it needs only 5 new matchsticks.

(i) The first hexagon needs 6 sticks. Each new hexagon borrows one side, so it adds only 5. Stage 4 needs 6 + 5 + 5 + 5 = 21, and Stage 5 needs 26.

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

(iii) Rule: the jump is 5 and the first term is 6, so the rule is 5n + (6 − 5) = 5n + 1.

(iv) Stage 15: 5 × 15 + 1 = 76 matchsticks.

(v) For 200: 5n + 1 = 200, so 5n = 199, so n = 39.8. A stage number must be a whole number, so no, 200 matchsticks cannot form a stage. (Stage 39 uses 196 and Stage 40 uses 201.)

*13. p(x) passes through (2, 3) and (6, 11). q(x) passes through (4, −1) and is parallel to p(x). Find both and where they meet the x-axis.Tap to reveal

p(x) = ax + b: 2a + b = 3 and 6a + b = 11. Subtract: 4a = 8, so a = 2, then b = 3 − 4 = −1. p(x) = 2x − 1.
Parallel means the same slope, so q(x) = 2x + d. Through (4, −1): −1 = 8 + d, so d = −9. q(x) = 2x − 9.
p meets the x-axis where 2x − 1 = 0: x = 1/2, the point (1/2, 0).
q meets the x-axis where 2x − 9 = 0: x = 9/2, the point (9/2, 0), which is (4.5, 0).

*14. What do all functions f(x) = ax + a, with a greater than 0, have in common?Tap to reveal

Factor it: f(x) = a(x + 1). At x = −1 this is a × 0 = 0, whatever a is. So every one of these lines passes through the point (−1, 0).

Also, since a is positive, they all rise from left to right, and each one's y-intercept equals its slope (both are a), so they all cross the y-axis above the origin.

RecapThe whole chapter on one screen

An algebraic expression combines numbers, variables and operations. In 2x² + 5xy − 3y², the terms are 2x², 5xy and −3y², and the coefficients are 2, 5 and −3.
A univariate polynomial uses one variable. Its degree is the highest power: x² + 5x + 3 has degree 2.
Degree 0: constant. Degree 1: linear. Degree 2: quadratic. Degree 3: cubic.
A linear pattern changes by the same amount each step. Its n-th term is a linear expression in n.
Linear growth: up by a fixed amount each step. Linear decay: down by a fixed amount each step.
A linear relationship is y = ax + b, and its graph is a straight line.
a is the slope (steepness and direction). b is the y-intercept (where it crosses the y-axis, at (0, b)). If b = 0, the line passes through the origin.
Growth gives a positive slope; decay gives a negative slope.
Lines with the same slope and different b are parallel.

WordsWords to know

Term: One piece of an expression, like 4x or −3.
Variable: A letter standing for a number that can change.
Coefficient: The number multiplying the variable in a term.
Constant: A term with no variable; its value never changes.
Polynomial: An expression of a variable, its whole-number powers, coefficients and a constant.
Degree: The highest power of the variable in a polynomial.
Linear polynomial: A polynomial of degree 1, like 2x + 3.
Substitute: To put a number in place of a variable.
Function: A rule that gives exactly one output for each input.
Linear equation: A linear polynomial set equal to a number, like 2x + 10 = 64.
Linear pattern: A list of numbers that changes by the same amount each step.
Slope: The a in y = ax + b: how much y changes when x goes up by 1.
y-intercept: The b in y = ax + b: where the line crosses the y-axis.
Parallel lines: Lines that never meet; they have the same slope.

CheckQuick self-check

What is the degree of 7 − 2x + x³?Answer

3. The order the terms are written in does not matter; the highest power is 3.

Is 3, 7, 11, 15 a linear pattern? What is its rule?Answer

Yes, it jumps by 4 each time. Rule: 4n + (3 − 4) = 4n − 1.

Where does y = −4x + 7 cross the y-axis? Does it rise or fall?Answer

It crosses at (0, 7). The slope is −4, so it falls from left to right.

Are y = 5x − 2 and y = 5x + 9 parallel?Answer

Yes. Both have slope 5; only their y-intercepts differ.

A taxi charges ₹40 plus ₹12 per km. Write the cost for d km, and find the cost of 5 km.Answer

Cost = 40 + 12d. For 5 km: 40 + 60 = ₹100.

Does the point (2, 9) lie on y = 3x + 3?Answer

3 × 2 + 3 = 9, which matches y. Yes.