Introduction to Linear Polynomials | CBSE Class 9 Maths Notes

This Mathematics note covers algebraic expressions, terms and coefficients, polynomial degree, evaluation, linear equations, growing patterns, linear growth and decay, relationships between quantities, straight-line graphs, slope, intercepts and parallel lines.

How do algebraic expressions describe quantities?

An algebraic expression combines numbers, variables and mathematical operations. A variable is a letter used to represent a quantity. Writing an expression lets us describe a calculation before we know the particular value of every quantity involved.

Suppose Raju buys red boxes containing four pens each and blue boxes containing five pencils each. Let $x$ be the number of red boxes and $y$ the number of blue boxes. He also receives three extra pens.

1. The pens in the red boxes number $4x$.
2. The pencils in the blue boxes number $5y$.
3. Adding the extra pens gives the total number of items: $$4x+5y+3.$$

The terms are $4x$, $5y$ and $3$. The numbers multiplying the variables are their coefficients: four for $x$ and five for $y$. The constant term is three because it does not depend on either box count.

How can the same variables occur in different terms?

Let $l$ and $w$ denote a rectangular garden's length and width in metres. Wire fencing costs ₹100 per metre along the two lengths, wooden fencing costs ₹80 per metre along the two widths, and seeds cost ₹50 per square metre.

1. The cost along the two lengths, in rupees, is $2l\times100=200l$.
2. The cost along the two widths, in rupees, is $2w\times80=160w$.
3. The seed cost, in rupees, is $50\times l\times w=50lw$.
4. The combined cost, in rupees, is $200l+160w+50lw$.

The term $50lw$ contains both variables multiplied together. Counting terms and counting variables are different tasks: this expression has three terms but two variables. Its coefficient of $lw$ is fifty.

Note: A number of terms does not tell us the number of variables. Identify the letters first, then identify the separate terms and their numerical coefficients.

What makes a polynomial linear, quadratic or cubic?

A univariate polynomial is a polynomial in one variable. The same variable may occur with different powers in several terms. The degree is the highest power of the variable appearing with a non-zero coefficient.

For example, let $z$ be a variable in $3z+7$. Its highest power is one, so it is a linear polynomial. The constant term does not raise its degree. A polynomial can be linear even when it has both a variable term and a constant.

Here $x$ and $y$ are variables in their respective expressions. The classification depends on the highest power, not the largest coefficient. In the cubic expression, the coefficients of the cubic, square and first-power terms are five, one and two respectively.

How do signs and absent terms affect coefficients?

A coefficient includes its sign. When the numerical multiplier is omitted, a positive variable term has coefficient $1$, while a negative variable term has coefficient $-1$. An absent power has coefficient zero: in $4z^3+5z^2-11$, the coefficient of the first-power term $z$ is zero.

Worked example 1. Find the coefficients of $x^2$ and $x^3$ in $x^4-3x^3+6x^2-2x+7$, where $x$ is the variable.

1. Locate the square term: it is $6x^2$.
2. Read its numerical multiplier: the coefficient of $x^2$ is $6$.
3. Locate the cubic term: it is $-3x^3$.
4. Keep the negative sign: the coefficient of $x^3$ is $-3$.

Answer: The coefficients are $6$ and $-3$, respectively.

The degree of this polynomial is four because of its first term. Asking for a particular coefficient is different from asking for the degree; inspect the requested power even when another term has a higher power.

How do we evaluate a polynomial as an input-output rule?

Evaluation means replacing the variable by a given number and carrying out the operations. A polynomial can be viewed as an input-output process: the input is a value of the variable, and the output is the resulting value of the expression.

For the expression $2x+3$, where $x$ denotes the input, the rule is to double the input and then add three. The expression stays the same while different inputs produce different outputs. This is an example of a function of the variable.

Worked example 2. Evaluate $2x+3$ for the inputs $x=4$ and $x=-6$.

1. For the first input, substitute four: $2x+3=2(4)+3$.
2. Multiply before adding: $2(4)+3=8+3=11$.
3. For the second input, use brackets around the negative number: $2x+3=2(-6)+3$.
4. Complete the calculation: $2(-6)+3=-12+3=-9$.

Answer: The outputs are $11$ and $-9$, respectively.

Can a geometric rule be a quadratic function?

A wire of length 20 cm is bent into a rectangle. Let $x$ be its length in centimetres, $w$ its width in centimetres, and $A$ its area in square centimetres. The fixed perimeter determines the width once the length is chosen.

1. Write the perimeter condition: $2x+2w=20$.
2. Divide by two and rearrange: $x+w=10$, so $w=10-x$.
3. Multiply length by width: $A=x(10-x)=10x-x^2$.
4. For the given length of six centimetres, calculate $A=10(6)-6^2=60-36=24$.

The area is 24 square centimetres at that input. The presence of the square term makes this a quadratic function, even though it uses only one variable. Having one variable does not by itself make a polynomial linear.

Note: When substituting a negative input into a square term, square the whole input. Brackets preserve the intended operation and help distinguish multiplication by a negative number from subtraction.

How does a linear expression become an equation to solve?

A linear expression gives a value for a chosen input. Equating a linear polynomial to a constant produces a linear equation. Solving the equation finds the input that gives the specified output.

The expression and the equation play different roles. In a word problem, first decide what the variable represents. Then use the relationships in the question to express other quantities in terms of that variable before writing the equality.

Worked example 3. Two numbers have sum 64, and one is ten more than the other. Find both numbers.

1. Let $x$ be the smaller number. The larger number is $x+10$.
2. Use the given sum: $x+(x+10)=64$.
3. Collect the variable terms: $2x+10=64$.
4. Subtract ten from both sides: $2x=54$.
5. Divide both sides by two: $x=27$.
6. Find the larger number: $x+10=27+10=37$.
7. Check both conditions: $27+37=64$ and $37-27=10$.

Answer: The two numbers are $27$ and $37$.

How can a fixed charge be separated from a variable charge?

A chess club charges a joining fee of ₹200 and ₹50 for each match. Let $m$ be the number of matches. The total payment in rupees is $200+50m$. The joining fee stays fixed, while the match charge changes with the input.

Worked example 4. With a joining fee of ₹200 and a charge of ₹50 per match, how many matches correspond to a total payment of ₹750?

1. Write the payment equation: $200+50m=750$.
2. Remove the joining fee: $50m=750-200=550$.
3. Divide by the charge per match: $m=550\div50=11$.
4. Check the total: $200+50(11)=200+550=750$.

Answer: The player played $11$ matches.

The final check should use the original conditions. Checking only the simplified equation could leave a modelling mistake unnoticed. Here, the answer is a whole number, as required for counting matches.

How can a growing pattern be represented by a linear rule?

A linear pattern has a constant difference between consecutive terms. For the tile pattern with counts $1,3,5,7,9,11,13$, each new stage adds two tiles. The stage number and the number of tiles are related, but they are not the same quantity.

Four stages are shown. The first contains one square. The following stages form two adjacent columns with the right column one square taller, containing three, five and seven squares in total.

See Fig. 2.4 in your NCERT textbook

Result: The tile count is one less than twice the stage number

Let $n$ denote the stage number and $T$ the number of tiles at that stage. Start with one tile, then add two tiles for each stage after the first.

1. The number of additions by stage $n$ is $n-1$.
2. The tiles added after the first stage number $2(n-1)$.
3. The total is $T=1+2(n-1)$.
4. Expand and simplify: $T=1+2n-2=2n-1$.

The rule is linear because the stage variable has degree one. It also reproduces the first stage, which is a useful check on the constant term. Using twice the stage number without subtracting one would shift every count.

Worked example 5. For the tile pattern $T=2n-1$, find the count at stage 15 and the stage containing 21 tiles.

1. Substitute the requested stage: $T=2(15)-1$.
2. Calculate the count: $T=30-1=29$.
3. For the inverse question, set the tile count to 21: $2n-1=21$.
4. Add one: $2n=22$.
5. Divide by two: $n=11$.
6. Check the inverse answer: $2(11)-1=22-1=21$.

Answer: Stage $15$ contains $29$ tiles, and stage $11$ contains $21$ tiles.

A pattern rule therefore supports two kinds of question. Substitution finds the count at a known stage; solving an equation finds the stage for a known count. Read carefully which quantity the question gives.

What does linear growth mean in practical situations?

Linear growth occurs when a quantity increases by a fixed amount over equal intervals. The important condition is equal intervals: a constant increase per kilometre or per month links the changing quantity to a linear expression.

Let $d$ be distance travelled in kilometres and $C(d)$ the journey cost in rupees. The notation $C(d)$ means the cost corresponding to the distance input. Consider the rule $C(d)=100+60d$.

The constant term is the starting cost. The coefficient of distance tells us how much the cost increases for each extra kilometre. Keeping these roles separate helps when constructing a rule from a verbal description.

Worked example 6. For $C(d)=100+60d$, find the cost of travelling 15 km and the distance corresponding to ₹700.

1. For fifteen kilometres, substitute $d=15$: $C(15)=100+60(15)$.
2. Calculate the variable charge: $60(15)=900$.
3. Add the starting charge: $C(15)=100+900=1000$.
4. For the given cost, write $100+60d=700$.
5. Subtract the starting charge: $60d=600$.
6. Divide by sixty: $d=10$.
7. Check: $C(10)=100+600=700$.

Answer: The cost is ₹1000, and the requested distance is 10 km.

Why must the interval stay the same?

The fixed increase here is ₹60 for each kilometre. A comparison across unequal distance intervals would involve unequal changes in cost. To identify a linear pattern from a table, compare outputs for equal changes in input.

The same principle applies to a plant growing by a fixed height each month or a village gaining a fixed number of residents each year. The starting amount and the increase per interval together determine the linear rule.

How do linear decay and restricted fare rules work?

Linear decay occurs when a quantity decreases by a fixed amount over equal intervals. Bela begins with ₹100 and spends ₹5 each day. Let $n$ be the number of days elapsed. Her remaining money in rupees is $100-5n$.

Worked example 7. Bela has ₹100 and spends ₹5 daily. After how many days will ₹40 remain?

1. Equate the remaining-money expression to the required amount: $100-5n=40$.
2. Subtract one hundred: $-5n=-60$.
3. Divide by negative five: $n=12$.
4. Check the spending: $5(12)=60$.
5. Check the remainder: $100-60=40$.

Answer: ₹40 remains after $12$ days.

A water-height model gives another example. Let $t$ be elapsed months and $h(t)$ the water height in metres. The rule $h(t)=3-0.5t$ starts at three metres and decreases by half a metre each month.

Why should a formula's stated range be kept?

An auto-rickshaw fare is ₹25 for the initial two kilometres and ₹15 per kilometre thereafter. The extra charge applies to the distance beyond the initial two kilometres. It does not apply to the entire journey distance.

Worked example 8. Find the fare for 10 km when the initial 2 km cost ₹25 and each further kilometre costs ₹15.

1. Let $n$ be the total distance in kilometres. The additional distance is $n-2$.
2. Let $F$ be the fare in rupees. For $n\geq2$, write $F=25+15(n-2)$.
3. Simplify: $F=25+15n-30=15n-5$.
4. For ten kilometres, calculate the extra distance: $10-2=8$.
5. Calculate the fare: $F=25+15(8)=25+120=145$.

Answer: The fare is ₹145.

Note: The simplified fare rule $F=15n-5$ is stated for $n\geq2$. For the initial two kilometres, use the fixed fare of ₹25. A correct simplification does not remove the original condition.

How can two observations determine a linear relationship?

Let $x$ represent an input quantity and $y$ its corresponding output. A linear relationship can be written as $y=ax+b$, where $a$ is the fixed coefficient of the input and $b$ is the constant term.

In a fixed-fee problem, the coefficient represents the charge for each unit used, and the constant represents the fixed charge. If these are unknown, two observations provide equations that can be solved together.

Worked example 9. An internet bill consists of a fixed monthly fee plus a charge per GB. Using 10 GB costs ₹350; using 20 GB costs ₹550. Find the billing rule.

1. Let $x$ be data used in GB and $y$ the bill in rupees. Let $a$ be the charge per GB and $b$ the fixed fee. Write $y=ax+b$.
2. Substitute the first observation: $350=10a+b$.
3. Rearrange for the fixed fee: $b=350-10a$.
4. Substitute the second observation and this expression: $550=20a+(350-10a)$.
5. Collect terms: $550=10a+350$, so $10a=200$.
6. Divide by ten: $a=20$.
7. Find the fixed fee: $b=350-10(20)=350-200=150$.
8. Insert both values into the rule: $y=20x+150$.
9. Check both bills: $20(10)+150=350$ and $20(20)+150=550$.

Answer: The rule is $y=20x+150$, with ₹20 per GB and a fixed fee of ₹150.

What do the two constants tell us?

The increase between the observations comes from additional usage because the fixed fee is unchanged. After finding the unit charge, substitute it into either original observation to obtain the fixed fee.

A common mistake is to divide the whole bill by the usage and treat the result as the variable rate. That mixes the fixed and variable parts. The two-equation method preserves their different roles and allows both original bills to be checked.

How do we plot and check a straight-line graph?

A point on a coordinate plane is described by an ordered pair $(x,y)$. Here $x$ is its horizontal coordinate and $y$ its vertical coordinate. The order matters because exchanging the entries generally changes the point.

To graph a relationship such as $y=2x+1$, choose input values and calculate their corresponding outputs. Two distinct points determine the straight line. Plot them, join them using a ruler and extend the line in both directions.

Worked example 10. Find two points for the graph of $y=2x+1$ using $x=0$ and $x=3$, and check whether $(7,15)$ lies on the line.

1. At the first input, calculate $y=2(0)+1=1$, giving the point $(0,1)$.
2. At the second input, calculate $y=2(3)+1=6+1=7$, giving the point $(3,7)$.
3. Plot these two points and draw the straight line through them.
4. For the proposed point, substitute its horizontal coordinate: $2(7)+1=14+1=15$.
5. The calculated value equals the point's vertical coordinate, so the point satisfies the equation.

Answer: Use $(0,1)$ and $(3,7)$; the point $(7,15)$ also lies on the line.

A rising line passes through the labelled points $A(0,1)$ and $B(3,7)$. The horizontal and vertical axes are labelled, and the line crosses the vertical axis at the first point.

See Fig. 2.5 in your NCERT textbook

Result: Coordinates on the line satisfy its equation

A graph and an equation describe the same relationship in different ways. Substitution provides an exact check of a plotted point: its coordinates must make the equality true. This is especially useful when a drawing is too small to read precisely.

The input-output view now becomes visual. Each input and its output form one ordered pair, and the collection of pairs for a linear relationship lies on a straight line.

How does slope describe direction and steepness?

In $y=ax+b$, the coefficient $a$ is the slope and $b$ is the vertical intercept. Slope describes how the output changes with the input. A positive slope represents linear growth; a negative slope represents linear decay.

Property: Lines with no constant term pass through the origin

The origin is the point $(0,0)$. For a line written as $y=ax$, substituting a zero horizontal coordinate also gives a zero vertical coordinate. The following numbered steps explain this shared point.

1. Start with the relation $y=ax$.
2. Set the horizontal coordinate to zero: $x=0$.
3. Calculate the vertical coordinate: $y=a(0)=0$.
4. Therefore the graph contains the point $(0,0)$.

Three rising straight lines share the origin. They are labelled $y=\frac12x$, $y=x$ and $y=2x$. The line with coefficient two is steepest; the line with coefficient one-half is least steep.

See Fig. 2.9 in your NCERT textbook

For positive coefficients, a value greater than one gives a line steeper than $y=x$, while a value between zero and one gives a less steep line. The line $y=x$ is equally inclined to the axes when their unit scales are equal.

Three falling lines pass through the origin, labelled $y=-\frac13x$, $y=-x$ and $y=-3x$. The line with coefficient negative three falls most steeply, while the negative one-third line falls least steeply.

See Fig. 2.11 in your NCERT textbook

How is slope connected to a sequence?

For the tile relationship $y=2x-1$, the horizontal coordinate represents the stage number and the vertical coordinate represents the tile count. Increasing the stage by one adds two tiles. Its slope is two, matching that constant difference.

The sign and size of the slope answer different questions. The sign distinguishes rising from falling as we move right; the magnitude describes steepness when graphs are compared on the same axes.

How do intercepts explain shifted and parallel lines?

The vertical intercept tells us where a line meets the vertical axis. For $y=ax+b$, it is the signed value $b$, and the intersection point is $(0,b)$. An intercept value and the coordinates of an intersection point are related but different answers.

Property: The constant term gives the vertical intercept

1. At the vertical axis, the horizontal coordinate is zero: $x=0$.
2. Substitute into the linear relationship: $y=a(0)+b$.
3. Simplify: $y=b$.
4. Write the complete intersection point: $(0,b)$.

The lines $y=2x+5$, $y=x+3$ and $y=3x-2$ meet the vertical axis at points labelled $A$, $B$ and $C$. Their coordinates are $(0,5)$, $(0,3)$ and $(0,-2)$, respectively.

See Fig. 2.14 in your NCERT textbook

A negative intercept is below the origin. In $y=3x-2$, the intercept is negative two, meaning an intersection two units along the negative vertical direction. It is clearer to distinguish this signed coordinate from the positive distance of two units.

Property: Equal slopes and different intercepts give parallel lines

Changing the constant while keeping the slope fixed shifts a line without changing its direction. Conversely, changing the slope while keeping the constant fixed changes the steepness or direction while preserving the vertical-axis intersection.

The three lines labelled $y=2x-1$, $y=2x+1$ and $y=2x+5$ rise in the same direction. Their vertical-axis intersections differ, and the lines are parallel.

See Fig. 2.13 in your NCERT textbook

Before reading a slope or intercept, rearrange an equation so that the coefficient of the output variable is one. Reading coefficients directly from an unrearranged equation can give an incorrect answer.

1. Consider the given equation $2y=4x+7$, where $x$ and $y$ are coordinates.
2. Divide every term by two: $y=2x+\frac72$.
3. Read the slope as $2$ and the vertical intercept as $\frac72$.
4. State the vertical-axis intersection as $(0,\frac72)$.

Glossary

• Algebraic expression — A combination of numbers, variables and operations used to represent a quantity or calculation.
• Variable — A letter representing a quantity whose value can change between different inputs.
• Coefficient — The numerical multiplier of a variable term, including any negative sign belonging to it.
• Constant term — A term whose value does not depend on the variable in the expression.
• Univariate polynomial — A polynomial involving one variable, which may appear in several terms with different powers.
• Degree — The highest power of the variable appearing with a non-zero coefficient in a polynomial.
• Linear polynomial — A polynomial of degree one, possibly including a constant term as well.
• Linear pattern — A sequence in which the difference between each pair of consecutive terms is constant.
• Linear growth — An increase in a quantity by a fixed amount over equal input intervals.
• Linear decay — A decrease in a quantity by a fixed amount over equal input intervals.
• Ordered pair — A pair giving a point's horizontal coordinate first and its vertical coordinate second.
• Slope — The coefficient of the input in a linear relationship, describing its direction and steepness.
• Vertical intercept — The signed vertical coordinate of the point where a line meets the vertical axis.
• Parallel lines — Distinct straight lines having equal slopes and different vertical intercepts in the given linear form.

Common errors and misconceptions

• Misconception: The largest coefficient gives the degree. Correct: The degree comes from the highest power of the variable with a non-zero coefficient.
• Misconception: A term's coefficient is always positive. Correct: Include its sign; the coefficient of $x^3$ in $-3x^3$ is $-3$.
• Misconception: Every polynomial in one variable is linear. Correct: The expression $10x-x^2$ is quadratic because its highest power is two.
• Misconception: The ₹15 per kilometre auto-rickshaw charge applies from the beginning. Correct: The initial two kilometres cost ₹25; the additional charge applies only to the remaining distance.
• Misconception: Any two numbers can be used as coordinates on a given line. Correct: The ordered pair must satisfy the line's equation.
• Misconception: A negative vertical intercept means a negative distance. Correct: Its sign identifies the negative vertical direction; the distance from the origin is positive.
• Misconception: The slope in $2y=4x+7$ is four. Correct: Rearrange to $y=2x+\frac72$; the slope is two.
• Misconception: Equal slopes force equal vertical intercepts. Correct: Equal slopes with different intercepts describe distinct parallel lines.

Exam-style questions with model answers

Q1. Find the degree of $4z-3$, where $z$ is the variable, and the coefficient of $z$ in $4z^3+5z^2-11$. [2 marks]

1. The highest power in $4z-3$ is one, so its degree is $1$.
2. The second polynomial has no first-power term. Its coefficient of $z$ is therefore $0$.

Q2. Evaluate $5x-3$, where $x$ is the input variable, for $x=0$, $x=-1$ and $x=2$. [3 marks]

1. Replace the variable with each input separately, keeping the expression unchanged. For zero, the multiplication term vanishes: $5(0)-3=0-3=-3$.
2. For negative one, put the input in brackets before multiplying: $5(-1)-3=-5-3=-8$.
3. For two, multiply first and then subtract: $5(2)-3=10-3=7$. Thus the three outputs, in the same order as the inputs, are $-3$, $-8$ and $7$.

Q3. Salil's mother is three times his present age. After five years, their ages will total 70 years. Find both present ages and check the result. [4 marks]

1. Let $x$ denote Salil's present age in years. His mother's present age is $3x$ years. Five years later, their ages will be $x+5$ and $3x+5$.
2. Use the future total: $(x+5)+(3x+5)=70$.
3. Collect terms and subtract ten: $4x+10=70$, hence $4x=60$.
4. Divide by four: $x=15$. His mother is $3(15)=45$ years old.
5. Check the future ages: $15+5=20$ and $45+5=50$, giving $20+50=70$. Their present ages are 15 and 45 years.

Q4. A cylindrical tank's water height is modelled by $h(t)=3-0.5t$, where $t$ is elapsed time in months and $h(t)$ is height in metres. Find the height after five months and explain why the rule represents linear decay. [3 marks]

1. Substitute five for the elapsed-month input: $h(5)=3-0.5(5)$. The total decrease over those five months is $0.5(5)=2.5$ metres.
2. Subtract this decrease from the initial height: $h(5)=3-2.5=0.5$. The water height after five months is therefore half a metre.
3. The rule represents linear decay because the height decreases by the same amount, half a metre, in each equal interval of one month.

Q5. A learning platform charges a fixed monthly fee and a constant charge per module. A bill for 10 modules is ₹400, and a bill for 14 modules is ₹500. Find the linear billing rule, identify both charges and check both bills. [5 marks]

1. Let $x$ be the number of modules and $y$ the bill in rupees. Let $a$ be the charge per module and $b$ the fixed monthly fee. The required relationship is $y=ax+b$.
2. The first observation gives $400=10a+b$, so the fixed fee can be expressed as $b=400-10a$.
3. The second observation gives $500=14a+b$. Substitute the expression for the fee: $500=14a+400-10a=4a+400$.
4. Subtract four hundred and divide by four: $4a=100$, hence $a=25$.
5. Substitute back: $b=400-10(25)=150$. The rule is $y=25x+150$, with ₹25 per module and a ₹150 fixed fee.
6. Check the two original bills independently: $25(10)+150=400$ and $25(14)+150=500$. Both given observations are satisfied by the same rule.

Q6. For the lines $2y=4x+7$ and $3y=6x-11$, where $x$ and $y$ are coordinates, find each slope, vertical intercept and vertical-axis intersection. Decide whether the lines are parallel. [4 marks]

1. Put the first line into the form with output coefficient one: $y=2x+\frac72$. Its slope is $2$, its vertical intercept is $\frac72$, and its vertical-axis intersection is $(0,\frac72)$.
2. Divide the second equation throughout by three: $y=2x-\frac{11}{3}$. Its slope is $2$, its vertical intercept is $-\frac{11}{3}$, and its intersection is $(0,-\frac{11}{3})$.
3. The lines have equal slopes and different vertical intercepts. They are therefore distinct parallel lines.

Q7. A linear polynomial $p(x)$, where $x$ is the input and $p(x)$ the output, has a graph through $(1,5)$ and $(3,11)$. Find the polynomial and its intersections with both axes. Give an algebraic check. [5 marks]

1. Write $p(x)=ax+b$, where $a$ is its slope and $b$ its constant term. The given points give $a+b=5$ and $3a+b=11$.
2. Subtract the first equality from the second to remove the constant: $2a=6$. Therefore $a=3$.
3. Substitute into the first equality: $3+b=5$, giving $b=2$. Hence $p(x)=3x+2$.
4. At the vertical axis, set the input to zero: $p(0)=2$. The intersection is $(0,2)$.
5. At the horizontal axis, the output is zero: $3x+2=0$. Thus $3x=-2$ and $x=-\frac23$, giving $(-\frac23,0)$.
6. Check the given points: $3(1)+2=5$ and $3(3)+2=11$. Check the horizontal intersection: $3(-\frac23)+2=0$. These calculations confirm the rule and its axis intersections.

Key takeaways

• Identify variables, terms and coefficients separately; the number of terms does not determine the number of variables.
• A polynomial's degree is its highest variable power with a non-zero coefficient; degree one means linear.
• Evaluation substitutes a known input, while solving an equation finds an input that produces a specified output.
• Linear patterns have constant differences between consecutive terms, allowing stage counts to be represented by linear rules.
• Linear growth and decay involve fixed changes over equal intervals, with the sign distinguishing increase from decrease.
• Two observations can determine a fixed charge and a unit charge when the relationship is known to be linear.
• Points on a straight-line graph satisfy its equation; substitution checks an ordered pair exactly.
• Equal slopes and different vertical intercepts produce parallel lines; the constant term gives the vertical-axis intersection.

Test yourself

What is the degree of the non-zero constant polynomial $-9$?

Its degree is zero because it is a non-zero constant polynomial.

What is the constant term in $9x^3+5x^2-8x-10$, where $x$ is the variable?

The constant term is $-10$; retain the negative sign attached to the term.

In the rule $T=2n-1$, where $T$ is tile count and $n$ is stage number, what does the coefficient two describe?

It describes the increase of two tiles for each increase of one stage.

Why does $y=3x$, where $x$ and $y$ are coordinates, pass through the origin?

A zero input gives a zero output, so the ordered pair $(0,0)$ satisfies the equation.

What distinguishes linear growth from linear decay when comparing equal input intervals?

Growth adds a fixed amount, whereas decay subtracts a fixed amount over those equal intervals.

Where does $y=3x-2$, where $x$ and $y$ are coordinates, meet the vertical axis?

It meets the vertical axis at $(0,-2)$, two units below the origin.

Why are $y=2x-1$ and $y=2x+5$, with coordinate variables $x$ and $y$, parallel?

Both lines have slope two, but their vertical intercepts are different.

Why should the fare rule $F=15n-5$, where $F$ is rupee fare and $n$ distance in kilometres, retain the condition $n\geq2$?

The first two kilometres have a fixed fare; the linear expression incorporates additional distance after that initial portion.