Class 9 Mathematics Chapter 2: Introduction to Linear Polynomials

Counting what you cannot see

A shopkeeper sells sealed boxes. Every red box holds 4 pens; every blue box holds 5 pencils.

You buy some red boxes and some blue boxes, and the shopkeeper throws in 3 loose pens. How many items do you have?

Without opening anything, you can still answer — provided you give the answer in terms of the numbers you do not yet know. Call the red boxes $x$ and the blue boxes $y$:

$$4x + 5y + 3$$

That is an algebraic expression: a piece of arithmetic with letters standing in for numbers you have not been told.

The names of the pieces

Every part of that expression has a name, and questions use these names constantly.

• The pieces added together — $4x$, $5y$ and $3$ — are the terms.
• The letters $x$ and $y$ are the variables.
• The numbers multiplying the variables, $4$ and $5$, are the coefficients.
• The lone number $3$, with no variable attached, is the constant term.

A term carries its own sign. In $2y - 5$ the terms are $2y$ and $-5$, not $2y$ and $5$.

One variable at a time

The expression above uses two variables. This whole chapter sticks to expressions in one variable only — things like $4x$, $2y - 5$, $3z + 7$ or $5y^3 + y^2 + 2y - 1$.

An algebraic expression built from one variable and its whole-number powers is called a polynomial in that variable. (The full name is a univariate polynomial, "univariate" meaning "one variable"; when the context is clear we just say polynomial.)

Degree: the highest power

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

• In $x^2 + 5x + 1$ the highest power is $2$, so the degree is $2$.
• In $5y^3 + y^2 - 8$ the highest power is $3$, so the degree is $3$.

Degree is the single most useful thing to know about a polynomial, because it decides what the polynomial is called and how its graph behaves.

The family names

The last row surprises people. Why is the plain number $8$ a polynomial of degree $0$? Because $8$ can be written as $8x^0$, and $x^0 = 1$. The variable is there with a power of zero, so the highest power present is $0$.

This chapter is about the degree-1 row. Linear polynomials are where we spend the rest of it.

Reading a polynomial carefully

Two traps hide in the vocabulary.

Invisible coefficients. In $5y^3 + y^2 + 2y - 1$, what is the coefficient of $y^2$? It is $1$, not $0$. Nobody writes $1y^2$, but the $1$ is there.

Missing terms. In $x^4 - 3x^3 + 6x^2 - 2x + 7$, what is the coefficient of $x^3$? It is $-3$ — take the sign with it. And in $4z^3 + 5z^2 - 11$, what is the coefficient of $z$? There is no $z$ term at all, so the coefficient is $0$.

A polynomial as a machine

Here is the most useful way to picture one. A polynomial is a machine: you feed a number in, and one number comes out.

Take $2x + 3$. Feed in $4$:

$$2(4) + 3 = 8 + 3 = 11$$

Feed in $-6$:

$$2(-6) + 3 = -12 + 3 = -9$$

Each input produces exactly one output. Written this way, a polynomial is a function of its variable, and $2x + 3$ is a linear function. (An expression like $10x - x^2$ is a machine too, but a quadratic one.)

Evaluating: put the number in brackets

Find the value of $5x - 3$ when $x = 0$, $x = -1$ and $x = 2$.

• $x = 0$: $;5(0) - 3 = -3$
• $x = -1$: $;5(-1) - 3 = -5 - 3 = -8$
• $x = 2$: $;5(2) - 3 = 10 - 3 = 7$

Always write the substituted number inside brackets. With $7s^2 - 4s + 6$ at $s = -3$, writing $7 \times -3^2$ invites you to square only the 3. Brackets make it safe:

$$7(-3)^2 - 4(-3) + 6 = 7(9) + 12 + 6 = 63 + 18 = 81$$

Note $-4 \times (-3) = +12$. A negative input makes signs flip in every odd-powered term, and that is where marks go.

From polynomial to equation

Set a linear polynomial equal to a number and you get a linear equation — something with an answer rather than a value that depends on the input.

That single step is what turns a story into mathematics.

A worked word problem

The sum of two numbers is 64, and one is 10 more than the other. Find them.

Name the smaller one $x$. Then the larger is $x + 10$, and their sum gives the equation:

$$x + (x + 10) = 64$$ $$2x + 10 = 64$$ $$2x = 54$$ $$x = 27$$

So the numbers are $27$ and $37$. Check both conditions: $27 + 37 = 64$, and $37 - 27 = 10$. Both hold.

Another one, with a ratio

Ruby has three times as many two-rupee coins as five-rupee coins, and ₹88 in total. How many of each?

Let the number of five-rupee coins be $f$. Then the two-rupee coins number $3f$, and the money they are worth is:

$$2(3f) + 5f = 88$$ $$6f + 5f = 88$$ $$11f = 88 \quad\Rightarrow\quad f = 8$$

So $8$ five-rupee coins and $24$ two-rupee coins. Check: $48 + 40 = 88$. Correct.

Notice the shape of the method: name the smaller quantity, write everything else in terms of it, then set up one equation.

Choosing the letter well

A 300-foot fence is cut into two pieces, the longer four times the shorter.

Let the shorter piece be $s$. Then the longer is $4s$, and $s + 4s = 300$, so $5s = 300$ and $s = 60$. The pieces are $60$ ft and $240$ ft.

Naming the longer piece first would have given $L + \tfrac{L}{4} = 300$ — correct, but with a fraction to carry. Name the smallest quantity and the arithmetic stays whole.

Slips that cost marks

• Forgetting the sign of a term: the terms of $2y - 5$ are $2y$ and $-5$.
• Calling the coefficient of $y^2$ in $y^2 + 3$ zero. It is $1$.
• Saying a missing term has coefficient $1$. If there is no $z$ term, the coefficient of $z$ is $0$.
• Substituting without brackets, so $-3$ squared comes out as $-9$.
• Claiming that a plain number has no degree. A non-zero constant has degree $0$.

What to write in the exam

For a degree question, name the highest power and stop. For a value question, show the substitution with brackets before you simplify. For a word problem, state what your letter stands for in words before the first equation, and check your answer against every condition in the question — a pair of numbers that fits the sum but not the difference earns nothing.

A pattern that grows one step at a time

Square tiles are laid out in stages. Stage 1 uses 1 tile. Stage 2 uses 3. Stage 3 uses 5, and stage 4 uses 7.

$$1,; 3,; 5,; 7,; \ldots$$

Two questions follow immediately, and the second is the interesting one.

How many tiles at stage 5? Easy — keep going: 9.

How many tiles at stage 100? You are not going to count that far. You need a rule.

Finding the rule

Look at the stage number beside the tile count:

At each stage the number of tiles is one less than twice the stage number. Check it: stage 2 gives $2(2) - 1 = 3$; stage 5 gives $2(5) - 1 = 9$. Both right.

So the rule for stage $n$ is:

$$2n - 1$$

That expression has degree 1 — it is a linear polynomial. Now stage 100 is one substitution away: $2(100) - 1 = 199$ tiles.

Test a rule before you trust it

A rule spotted from three or four terms is a guess until it is checked. Substitute a stage you already know and see whether the right number comes back.

For $2n - 1$ at stage 6: $2(6) - 1 = 11$, and the table says 11. At stage 7: $2(7) - 1 = 13$, and the table agrees again.

Two successful checks on values you did not use to build the rule are enough to proceed with confidence. Skipping this step is how a rule that fits the first two terms and nothing else survives into the final answer.

What makes this pattern linear

Look at the gaps between consecutive tile counts:

$$1 \xrightarrow{+2} 3 \xrightarrow{+2} 5 \xrightarrow{+2} 7 \xrightarrow{+2} 9$$

The gap is always 2. Never 2 then 3, never doubling — the same amount added every time.

A linear pattern is a sequence in which the difference between consecutive terms is constant.

And that constant difference is exactly the coefficient of $n$ in the rule. In $2n - 1$ the coefficient is 2, and the pattern goes up in 2s. The two facts are the same fact.

Reading far ahead, and backwards

Once you have the rule, both directions are open.

Forwards. Stage 15 holds $2(15) - 1 = 29$ tiles; stage 26 holds $2(26) - 1 = 51$.

Backwards. Which stage holds 21 tiles? Set the rule equal to 21:

$$2n - 1 = 21 \quad\Rightarrow\quad 2n = 22 \quad\Rightarrow\quad n = 11$$

Stage 11. And for 47 tiles, $2n = 48$ gives stage 24.

A backwards question can have no answer. Can a stage hold exactly 200 tiles? $2n - 1 = 200$ gives $2n = 201$, so $n = 100.5$ — not a whole number, so no stage qualifies. Every value of $2n - 1$ is odd, and 200 is even.

Linear growth

The same shape describes money, distance, population — anything that rises by a fixed amount each step.

Linear growth means a quantity climbs by the same fixed step every time the count advances by one.

A chess club charges 200 rupees to join, then 50 rupees per match.

Each extra match adds exactly 50, so after $m$ matches the total is:

$$200 + 50m$$

The 200 is where the pattern starts; the 50 is the constant difference.

A member paid 750 rupees. How many matches? Set $200 + 50m = 750$, so $50m = 550$ and $m = 11$ matches.

Linear decay

Run the same idea downhill.

Linear decay means a quantity drops by the same fixed step every time the count advances by one.

Bela has 100 rupees and spends 5 rupees a day.

After $n$ days she has:

$$100 - 5n$$

On day 12 that leaves $100 - 60 = 40$ rupees; on day 15, $100 - 75 = 25$ rupees.

When is the money gone? $100 - 5n = 0$ gives $n = 20$ days.

The only difference is a sign

Growth and decay are not two separate topics. Both have the form

$$(\text{starting value}) + (\text{constant step}) \times n$$

and the step is positive for growth, negative for decay. That is the whole distinction.

A water tank 3 m deep losing 0.5 m a month follows $h = 3 - 0.5t$: after 5 months $3 - 2.5 = 0.5$ m remains, and the tank is dry at $t = 6$. Same shape, negative step.

A pattern that starts late

Real situations are not always tidy from the first step.

An auto-rickshaw charges 25 rupees for the first 2 km, then 15 rupees per kilometre after that.

For a 10 km ride, the first 2 km cost 25 and the remaining 8 km cost $15 \times 8 = 120$, so the fare is 145 rupees.

For $n$ kilometres with $n \geq 2$:

$$25 + 15(n - 2) = 25 + 15n - 30 = 15n - 5$$

When is the fare 130 rupees? $15n - 5 = 130$ gives $15n = 135$, so $n = 9$ km.

Notice the caution: the rule $15n - 5$ is only valid from 2 km onwards. At $n = 1$ it would give 10 rupees, but the real fare is 25. A rule taken from a table applies only over the range the table describes.

Getting the rule from a table

When you are handed a table rather than a story, work in this order:

1. Check the differences. If they are not constant, the pattern is not linear and none of this applies.
2. The constant difference is the coefficient of $n$.
3. Find the starting value by working back to $n = 0$, or by substituting any one row and solving.

Try it. A rally starts with 120 members and loses 9 each hour. The difference is $-9$, the value at hour 0 is 120, so after $n$ hours the number remaining is $120 - 9n$.

Slips that cost marks

• Assuming a pattern is linear without checking that the differences are constant.
• Losing the minus sign in a decay rule, so $100 - 5n$ becomes $100 + 5n$.
• Mixing up the starting value with the first term. In $200 + 50m$ the starting value is 200, but the first payment is 250.
• Giving a fractional answer to a "which stage" question. Stage numbers are whole numbers, so a fraction means no such stage exists.
• Using a rule outside the range it was built for, like the fare rule below 2 km.

What to write in the exam

State the rule in terms of the letter the question uses, and say what that letter counts. Show the constant difference — it is usually worth a mark by itself. For a backwards question, write the equation, solve it, and then say whether the answer makes sense: a stage number must be a whole number, and a number of days cannot be negative.

Two quantities moving together

A student checks her internet bill twice. In a month when she used 10 GB she paid 350 rupees; in a month when she used 20 GB she paid 550 rupees.

She wants a rule — something that turns any amount of data into the bill, so she can budget before the month ends rather than after.

Two readings, and from them a whole rule. Here is how.

The form of a linear relationship

When two quantities are connected so that a fixed change in one produces a fixed change in the other, the connection can be written as:

$$y = ax + b$$

This is a linear relationship. The letters $a$ and $b$ are numbers fixed by the situation, and $x$ and $y$ are the two quantities.

For the bill, $x$ is the data used in GB and $y$ is the amount paid in rupees. Finding the rule means finding $a$ and $b$.

Finding a and b from two readings

Substitute each reading into $y = ax + b$ and you get two equations.

From $x = 10$, $y = 350$: $;350 = 10a + b$ From $x = 20$, $y = 550$: $;550 = 20a + b$

Rearrange the first to $b = 350 - 10a$, and put that into the second:

$$550 = 20a + (350 - 10a)$$ $$550 = 10a + 350$$ $$10a = 200 \quad\Rightarrow\quad a = 20$$

Then $b = 350 - 10(20) = 350 - 200 = 150$. So the rule is:

$$y = 20x + 150$$

Check it against both readings before going further: $20(10) + 150 = 350$ ✓ and $20(20) + 150 = 550$ ✓.

What the two numbers mean

They are not just algebra. In this situation:

• $a = 20$ is the cost per GB — each extra gigabyte adds 20 rupees.
• $b = 150$ is the fixed monthly fee, paid even at zero GB.

Now the rule earns its keep: a month using 15 GB will cost $20(15) + 150 = 450$ rupees.

Whenever you find $a$ and $b$ from a real situation, ask what each one is. It is often worth a mark, and it catches an answer that came out backwards.

Drawing the line: two points are enough

A linear relationship drawn on the coordinate plane is a straight line, and two points fix a straight line completely. So to plot one, find any two points and join them.

Take $y = 2x + 1$. Choose easy values of $x$:

• $x = 0$ gives $y = 1$, so $A(0, 1)$ is on the line.
• $x = 3$ gives $y = 7$, so $B(3, 7)$ is on the line.

Plot $A$ and $B$, join them with a ruler, and extend in both directions. That line is the whole relationship.

Use a third point as a check. $x = 1$ gives $y = 3$, and $(1, 3)$ should land exactly on the ruled line. If it misses, one of your three calculations is wrong.

Does a point lie on the line?

There is no need to plot anything. Substitute the coordinates and see whether the equation holds.

Is $(7, 15)$ on $y = 2x + 1$? Substituting: $2(7) + 1 = 15$. Yes, it matches, so the point lies on the line.

Is $(5, 12)$ on it? $2(5) + 1 = 11$, not 12. No, it does not.

A point lies on a line exactly when its coordinates satisfy the line's equation.

Lines through the origin

When $b = 0$ the relationship is simply $y = ax$, and something is guaranteed: putting $x = 0$ gives $y = 0$.

Every line of the form $y = ax$ passes through the origin. $y = \tfrac{1}{2}x$, $y = x$, $y = 2x$, $y = -3x$ — all of them go through $(0, 0)$, and they differ only in how steeply they rise or fall.

The slope: what a does

The number $a$ is called the slope of the line, and it measures steepness.

Compare $y = \tfrac{1}{2}x$, $y = x$ and $y = 2x$ on one set of axes:

• $y = x$ rises one unit for every unit across.
• $y = 2x$ rises two units for every unit across — steeper.
• $y = \tfrac{1}{2}x$ rises half a unit for every unit across — shallower.

So among lines with $a > 0$: the larger $a$ is, the steeper the line. And the slope is the same constant difference met in the last topic — the amount $y$ changes for a step of one in $x$.

A negative slope goes downhill

If $a$ is negative the line falls as you move right. $y = -x$, $y = -3x$ and $y = -\tfrac{1}{3}x$ all slope downwards, the steepest being $y = -3x$.

This joins the two halves of the chapter:

Linear growth is a line with positive slope. Linear decay is a line with negative slope.

A pattern rising by 50 each step and a line of slope 50 are the same object described twice.

The y-intercept: what b does

Put $x = 0$ into $y = ax + b$ and you get $y = b$. So the line passes through $(0, b)$ — it cuts the y-axis at height $b$.

That value is called the y-intercept.

• $y = 2x + 5$ cuts the y-axis at $(0, 5)$.
• $y = x + 3$ cuts it at $(0, 3)$.
• $y = 3x - 2$ cuts it at $(0, -2)$, two units below the origin.

Reading a y-intercept off an equation takes no working at all: it is the number on its own.

Same slope, different intercepts: parallel lines

Now fix $a$ and change $b$. Compare $y = 2x - 1$, $y = 2x + 1$ and $y = 2x + 5$.

All three climb at exactly the same rate, because all three have slope 2. They cut the y-axis at $-1$, $1$ and $5$, so they sit at different heights — but they never meet.

Lines with equal slopes and different y-intercepts are parallel.

Changing $b$ slides a line up or down without tilting it. Changing $a$ tilts it without moving where it crosses the y-axis.

Slips that cost marks

• Plotting only two points and never checking with a third.
• Reading the y-intercept from an equation not yet in the form $y = ax + b$ — in $2y = 4x + 7$ you must divide by 2 first, giving $y = 2x + 3.5$ and an intercept of 3.5, not 7.
• Saying two lines are parallel because they look it, instead of comparing slopes.
• Forgetting that a negative intercept puts the crossing point below the origin.
• Finding $a$ and $b$ and never checking them against the readings you started from.

What to write in the exam

Put the equation into the form $y = ax + b$ before reading anything off it. State the slope and the y-intercept separately, and give the crossing point as a coordinate pair $(0, b)$. When plotting, show the table of the points you used, label the axes, and mark the third checking point. And when $a$ and $b$ come from a real situation, say in words what each of them means.