Introduction to Linear Polynomials — Notes

Terms, Variables, and Coefficients

Algebraic expressions are formed using variables (symbols that can take any value, often represented by letters like $x,y,l,w$) and constants (fixed numerical values). These expressions are made up of terms, which are parts separated by addition or subtraction.

In an algebraic expression, the numerical factor multiplying a variable is called its coefficient. A term that does not contain any variable is called a constant term.

For example, in the expression $4x+5y+3$:

• The terms are $4x$, $5y$, and $3$.
• The variables are $x$ and $y$.
• The coefficient of $x$ is $4$.
• The coefficient of $y$ is $5$.
• $3$ is the constant term.

Example 1: Total Items

Given: Raju buys $x$ red boxes with 4 pens each and $y$ blue boxes with 5 pencils each. He gets 3 extra pens free.

To find: An algebraic expression for the total number of pens and pencils.

Solution:

• Number of pens from red boxes = Number of red boxes × Pens per red box = $x\times 4=4x$.
• Number of pencils from blue boxes = Number of blue boxes × Pencils per blue box = $y\times 5=5y$.
• Extra pens received = $3$.
• Total quantity = Pens from red boxes + Pencils from blue boxes + Extra pens
• Total quantity = $4x+5y+3$.

Answer: The algebraic expression for the total quantity is $4x+5y+3$.

Example 2: Garden Cost

Given: A rectangular garden of length $l$ metres and width $w$ metres. Wire fencing along length costs 100 per metre. Wooden fencing along width costs 80 per metre. Seeds cost `50 per square metre.

To find: An algebraic expression for the total cost.

Solution:

• Cost of wire fencing along the garden length = (Perimeter along length) × (Cost per metre) = $(2l)\times 100=200l$.
• Cost of wooden fencing along the garden width = (Perimeter along width) × (Cost per metre) = $(2w)\times 80=160w$.
• Cost of sowing seeds = (Area of garden) × (Cost per square metre) = $(l\times w)\times 50=50lw$.
• Total cost = Cost of wire fencing + Cost of wooden fencing + Cost of seeds.
• Total cost = $200l+160w+50lw$.

Answer: The algebraic expression for the total cost is $200l+160w+50lw$.

Example 3: Rectangle Area

Given: A wire of length 20 cm is bent to form rectangles. The length of the rectangle is $x$ cm.

To find: An expression for the area of such rectangles.

Solution:

• The perimeter of the rectangle is 20 cm.
• Perimeter = $2\times (\text{length}+\text{width})$.
• $20=2\times (x+\text{width})$.
• $10=x+\text{width}$.
• Width = $10-x$ cm.
• Area of the rectangle = length × width = $x\times (10-x)$.
• Area = $10x-x^2$.

Answer: The expression for the area is $10x-x^2$.

Remember

Algebraic expressions can involve one or more variables. The terms, coefficients, and variables are identified based on their position and relation to the operations of addition, subtraction, and multiplication.

Exam-smart

Questions might present a real-life scenario (like shopping or construction) and ask you to form an algebraic expression representing a quantity (like total cost, total items, or area). You need to identify the variables, constants, and how they combine using mathematical operations.

Degree of a Polynomial

An algebraic expression involving only one variable and where the powers of the variable are non-negative integers is called a polynomial in one variable.

• Polynomials involving one variable are also called univariate polynomials.
• The highest power of the variable in a polynomial is called its degree.

Based on the degree, polynomials are classified:

Example 1: Degrees of Polynomials

Given: Polynomials (i) $2x^2\text{–}5x+3$ (ii) $y^3+2y\text{–}1$ (iii) $-9$ (iv) $4z\text{–}3$

To find: The degree of each polynomial.

Solution:

• (i) In $2x^2\text{–}5x+3$, the powers of $x$ are 2, 1, and 0. The highest power is 2.
• (ii) In $y^3+2y\text{–}1$, the powers of $y$ are 3, 1, and 0. The highest power is 3.
• (iii) $-9$ is a constant polynomial, which can be written as $-9x^0$. The power of $x$ is 0.
• (iv) In $4z\text{–}3$, the powers of $z$ are 1 and 0. The highest power is 1.

Answer:

• (i) Degree is 2.
• (ii) Degree is 3.
• (iii) Degree is 0.
• (iv) Degree is 1.

Example 2: Writing Polynomials by Degree

Given: Requirement for polynomials of specific degrees.

To find: Examples of polynomials with degrees 1, 2, and 3.

Solution:

• Degree 1 (Linear Polynomial): We need the highest power to be 1. Example: $5x+2$.
• Degree 2 (Quadratic Polynomial): We need the highest power to be 2. Example: $3x^2-7x+1$.
• Degree 3 (Cubic Polynomial): We need the highest power to be 3. Example: $2y^3-4y^2+y-9$.

Answer:

• Degree 1: $5x+2$
• Degree 2: $3x^2-7x+1$
• Degree 3: $2y^3-4y^2+y-9$

Example 3: Coefficients

Given: Polynomials (i) $x^4\text{–}3x^3+6x^2\text{–}2x+7$ (ii) $4z^3+5z^2\text{–}11$ (iii) $9x^3+5x^2\text{–}8x\text{–}10$

To find: Specific coefficients and the constant term.

Solution:

• (i) In $x^4\text{–}3x^3+6x^2\text{–}2x+7$:
• The coefficient of $x^2$ is $6$.
• The coefficient of $x^3$ is $-3$.
• (ii) In $4z^3+5z^2\text{–}11$:
• The coefficient of $z$ is $0$ (since the term $0z$ is not explicitly written).
• (iii) In $9x^3+5x^2\text{–}8x\text{–}10$:
• The constant term is $-10$.

Answer:

• (i) Coefficient of $x^2$ is $6$, coefficient of $x^3$ is $-3$.
• (ii) Coefficient of $z$ is $0$.
• (iii) Constant term is $-10$.

Remember

For a polynomial like $a{x}^n+b{x}^{n-1}+\mathrm{.}\mathrm{.}\mathrm{.}+k$, the highest power $n$ is the degree. When a power of a variable is missing, its coefficient is zero.

Exam-smart

You might be given an algebraic expression and asked to identify its degree, or the coefficient of a specific term. Pay close attention to the highest power present and the number multiplying each variable term. For polynomials with missing terms, remember their coefficients are zero.

Linear Polynomials

A linear polynomial is a polynomial of degree 1. This means the highest power of the variable in the polynomial is 1.

General form: $ax+b$, where $a$ and $b$ are constants, and $a\mathrm{\ne }0$.

• $x$ is the variable.
• $a$ is the coefficient of the variable $x$.
• $b$ is the constant term.

Linear polynomials describe relationships where a change in one quantity results in a proportional change in another quantity, with a possible constant offset.

Linear Patterns

Linear patterns are sequences of numbers where the difference between consecutive terms is constant. This constant difference is often referred to as the common difference. These patterns arise from linear polynomials.

Example 1: Perimeter of a Square

Consider a square with side length $x$. Its perimeter is given by $P=4x$. If the side length increases by a fixed amount, say $0.5$ cm, the perimeter also increases by a fixed amount ($4\times 0.5=2$ cm).

• Side = 1 cm, Perimeter = $4\times 1=4$ cm
• Side = 1.5 cm, Perimeter = $4\times 1.5=6$ cm (Increase of 2 cm)
• Side = 2 cm, Perimeter = $4\times 2=8$ cm (Increase of 2 cm)

The pattern of perimeters (4, 6, 8, ...) has a constant difference of 2.

Example 2: Chess Club Fees

A chess club charges a joining fee of 200 and 50 per match. The total cost $C$ for playing $m$ matches is $C=200+50m$.

Here, for every additional match played, the cost increases by `50. This is a linear pattern.

Example 1 (from textbook)

Given: The perimeter of a square of side $x$ is $4x$. To find: The perimeter of squares with sides 1 cm, 1.5 cm, 2 cm, 2.5 cm, and 3 cm. Solution: The formula for the perimeter of a square with side $x$ is $P=4x$.

• For side $x=1$ cm, $P=4\times 1=4$ cm.
• For side $x=1.5$ cm, $P=4\times 1.5=6$ cm.
• For side $x=2$ cm, $P=4\times 2=8$ cm.
• For side $x=2.5$ cm, $P=4\times 2.5=10$ cm.
• For side $x=3$ cm, $P=4\times 3=12$ cm. If the sides increase by 0.5 cm, the perimeters increase by $4\times 0.5=2$ cm each time. Answer: The perimeters are 4 cm, 6 cm, 8 cm, 10 cm, and 12 cm.

Example 2 (from textbook)

Given: A chess club charges a joining fee of 200 plus 50 for every match. The number of matches is $m$. To find: The total amount a player will have to pay. Solution: The joining fee is a constant amount (200). The cost per match is 50. If $m$ is the number of matches played, the total cost $C$ is given by:

This is a linear polynomial in $m$. Answer: The total cost is `(200 + 50m).

Example 3 (from textbook)

Given: The sum of two numbers is 64. One number is 10 more than the other. To find: The two numbers. Solution: Let the smaller number be $x$. Then the larger number is $x+10$. The sum of the two numbers is 64.

Combine like terms:

This is a linear equation formed by a linear polynomial. Subtract 10 from both sides:

Divide by 2:

The smaller number is $x=27$. The larger number is $x+10=27+10=37$. Answer: The two numbers are 27 and 37.

Example 4 (from textbook)

Given: The linear polynomial is $5x-3$. To find: The value of the polynomial for $x=0,x=-1,x=2$. Solution: We substitute the given values of $x$ into the polynomial $5x-3$.

• For $x=0$:

• For $x=-1$:

• For $x=2$:

Answer: The values are -3, -8, and 7 respectively.

Remember

A linear polynomial of degree 1 can be written as $ax+b$, where $a\mathrm{\ne }0$. The graph of a linear polynomial is always a straight line.

Exam-smart

In case/source/data questions, you might be given a table of values or a word problem describing a situation with a constant rate of change. You'll need to identify the variable, the constant term, and the coefficient to form a linear polynomial (or equation) and then use it to find unknown values. For example, a question might describe ticket prices where there's a fixed booking charge and a per-ticket cost.

Try this: Find the value of the linear polynomial $3y+7$ when $y=-2$. Answer: $1$.

Exploring Linear Patterns

Linear patterns describe situations where a quantity changes by a constant amount for each unit increase in another quantity. This constant change is often referred to as a common difference.

Identifying Linear Patterns from Sequences

A sequence of numbers represents a linear pattern if the difference between any two consecutive terms is constant. This constant difference is key to understanding and predicting future terms in the sequence.

Example 1: Growing Pattern of Square Tiles

Given: A pattern of square tiles where each stage is obtained by adding two more tiles to the previous stage. Stage 1: 1 tile Stage 2: 3 tiles Stage 3: 5 tiles Stage 4: 7 tiles

To find: The sequence of numbers up to Stage 7 and a linear expression for the number of squares at Stage $n$.

Solution:

• We observe that the difference between consecutive terms is $3-1=2$, $5-3=2$, $7-5=2$. This constant difference is 2.
• To find the next terms:
• Stage 5: $7+2=9$ tiles
• Stage 6: $9+2=11$ tiles
• Stage 7: $11+2=13$ tiles
• The sequence of the number of squares for the first 7 stages is: 1, 3, 5, 7, 9, 11, 13.
• To find the general expression, we note that the number of squares at each stage is one less than twice the stage number. For Stage $n$, this can be written as $2n-1$.
• For Stage 2: $2\times 2-1=3$
• For Stage 5: $2\times 5-1=9$
• The polynomial $2n-1$ has a degree of 1, indicating it is a linear polynomial.

Answer: The sequence up to Stage 7 is 1, 3, 5, 7, 9, 11, 13. The linear expression for the number of squares at Stage $n$ is $2n-1$.

Think and Reflect

Using the expression $2n-1$, find the number of tiles in the 15th stage and the 26th stage. Also, determine which stage will contain 21 tiles and 47 tiles.

• 15th stage: $2\times 15-1=30-1=29$ tiles.
• 26th stage: $2\times 26-1=52-1=51$ tiles.
• Stage with 21 tiles: $2n-1=21⟹2n=22⟹n=11$. So, the 11th stage.
• Stage with 47 tiles: $2n-1=47⟹2n=48⟹n=24$. So, the 24th stage.

Example 2: Pocket Money Spending

Given: Bela has 100 for pocket money and spends 5 every day.

To find: After how many days will she be left with `40? Also, find a linear expression for the amount left on the $n$th day.

Solution:

• The initial amount is `100.
• Amount spent per day is `5.
• The amount of money left after $n$ days can be represented by a linear expression.
• After 1 day: 100 - 1 × 5 = 95
• After 2 days: 100 - 2 × 5 = 90
• After $n$ days: 100 - 5n
• The linear expression for the amount left on the $n$th day is $100-5n$.
• To find when she will be left with `40:

$n=12$ days.

Answer: Bela will be left with `40 after 12 days. The linear expression for the amount left on the $n$th day is $100-5n$.

Think and Reflect

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 15th day: $100-5\times 15=100-75=25$. So, `25 will be left.
• Days to spend entire amount: $100-5n=0⟹5n=100⟹n=20$ days.

Example 3: Auto-Rikshaw Fare

Given: An auto-rikshaw fare starts at 25 for the initial 2 km. After that, it increases by 15 per km.

To find: The fare for a travel of 10 km.

Solution:

• The fare for the first 2 km is `25.
• The distance travelled beyond the initial 2 km is $10-2=8$ km.
• The additional fare for these 8 km is $8\times 15=120$.
• The total fare for 10 km is the initial fare plus the additional fare: 25 + 120 = 145.
• Let $n$ be the distance in km. For $n\ge 2$, the fare can be expressed as: Fare = (Fare for first 2 km) + (Fare for remaining $n-2$ km) Fare = $25+15\times (n-2)$ Fare = $25+15n-30$ Fare = $15n-5$

Answer: The fare for a travel of 10 km will be `145.

Think and Reflect

For how many kilometers will the fare be `130?

• Using the expression $15n-5$:

$n=9$ km. So, the fare will be `130 for 9 km.

Relationship between Linear Polynomials and Linear Patterns

A linear pattern is a sequence of numbers where the difference between consecutive terms is constant. The general term of such a sequence can be represented by a linear polynomial of the form $an+b$, where $a$ is the constant difference and $b$ is a constant. The variable $n$ usually represents the position of the term in the sequence (e.g., stage number, day number, term number).

• The constant difference ($a$) in the linear pattern corresponds to the coefficient of the variable in the linear polynomial.
• The linear polynomial describes the rule that generates the terms of the linear pattern.

Remember

The defining characteristic of a linear pattern is a constant rate of change. This means for every unit increase in the independent variable (like time or stage number), the dependent variable (like quantity or amount) changes by the same fixed amount.

Exam-smart

You might be given a table of values or a description of a situation. Your task will be to identify if it represents a linear pattern (by checking for a constant difference) and then to derive the linear expression that models it. For example, you might be given a scenario about savings or a growing object, and asked to find how much will be saved after a certain number of months or how large it will be after a certain number of years. You'll need to find the starting value (like an initial amount or the value at stage 0/1) and the rate of change.

Linear growth and linear decay

Linear growth describes a situation where a quantity increases by a constant amount over equal intervals. This constant increase is also known as the rate of growth.

Linear decay describes a situation where a quantity decreases by a constant amount over equal intervals. This constant decrease is known as the rate of decay.

These patterns can be represented by linear functions, which are typically in the form of $y=mx+c$, where:

• $y$ is the dependent variable (the quantity that changes).
• $x$ is the independent variable (the quantity that changes in equal intervals).
• $m$ is the slope or the constant rate of change (growth or decay).
• $c$ is the y-intercept or the initial value of $y$ when $x$ is 0.

Example 9: Linear Growth

The cost of a journey is given by the linear function $C(d)=100+60d$, where $C$ is the total cost in rupees and $d$ is the distance travelled in km.

Given: $C(d)=100+60d$ To find: A table showing how the cost increases for $d$ from 0 to 10 km, and the cost for 15 km.

Solution: The table shows the cost for increasing distances:

In this example, for every 1 km increase in distance ($d$), the cost ($C$) increases by a fixed amount of `60. This is linear growth.

To find the cost for 15 km:

Answer: The cost for travelling 15 km will be `1000.

Example 10: Linear Decay

The height of water in a cylindrical tank is 3 m at the start of summer. The height $h$ metres at the end of $t$ months is given by the linear function $h(t)=3-0.5t$.

Given: $h(t)=3-0.5t$ To find: A table showing how the height decreases for $t$ from 0 to 4 months, and the height at the end of 5 months.

Solution: The table shows the height for increasing months:

In this example, for every 1 month increase in time ($t$), the height ($h$) decreases by a fixed amount of 0.5 m. This is linear decay.

To find the height at the end of 5 months:

Answer: The height of the water at the end of 5 months will be 0.5 m.

To find how many kilometres the cost will be `700:

Answer: The cost of the journey will be `700 for 10 kilometres.

Remember

The slope ($m$) in a linear function $y=mx+c$ determines whether there is growth or decay. If $m>0$, it represents linear growth. If $m<0$, it represents linear decay.

Exam-smart

Questions will often describe a scenario (like savings, depreciation, or distance travelled) and ask you to identify whether it shows linear growth or decay, or to write the linear expression that models it. Look for a constant rate of increase or decrease over time or some other unit.

Exercise 2.4, Question 1

Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month.

(i) Find the height after 7 months. Given: Initial height = 1.75 feet, Growth rate = 0.5 feet/month To find: Height after 7 months. Solution: The linear expression for height $h$ after $t$ months is $h(t)=1.75+0.5t$.

Answer: The height after 7 months will be 5.25 feet.

(ii) Make a table of values for $t$ varying from 0 to 10 months and show how the height, $h$, increases every month. Given: $h(t)=1.75+0.5t$ To find: Table of values for $h(t)$ for $t=0,1,\dots ,10$. Solution:

(iii) Find an expression that relates $h$ and $t$, and explain why it represents linear growth. Given: Initial height = 1.75 feet, growth rate = 0.5 feet/month. To find: Expression relating $h$ and $t$, and explanation for linear growth. Solution: The expression is $h(t)=1.75+0.5t$. This represents linear growth because the height ($h$) increases by a constant amount (0.5 feet) for every unit increase in the number of months ($t$). The rate of growth (0.5) is positive.

Try this: A mobile phone's value decreases by 500 each year. If its initial value was 12,000, write an expression for its value $v(t)$ after $t$ years and state why it is linear decay. Answer: $v(t)=12000-500t$. This represents linear decay because the value ($v$) decreases by a constant amount (`500) for each year ($t$).

Linear Relationships

A linear relationship describes how two variables, typically denoted as $x$ and $y$, are connected. This connection can always be represented by an equation of the form $y=ax+b$, where $a$ and $b$ are constants. This equation signifies that for every unit increase in $x$, $y$ changes by a fixed amount, $a$.

The term $ax$ represents a variable part of the relationship, which changes proportionally with $x$. The term $b$ is the constant part, also known as the y-intercept (as discussed in Visualising Linear Relationships), which is the value of $y$ when $x$ is zero.

This form is particularly useful for modelling situations where there's a starting value (or a value at zero) and a constant rate of change.

Example 1: Telecom Bill

A telecom company charges a fixed monthly fee plus an additional cost per GB of internet data used. A student observed:

• When she used 10 GB, her bill was ₹350.
• When she used 20 GB, her bill was ₹550.

The monthly bill $y$ (in ₹) depends on the amount of data used $x$ (in GB) according to the linear relationship $y=ax+b$. Find the values of $a$ and $b$.

Given:

• When $x=10$ GB, $y=350$ ₹.
• When $x=20$ GB, $y=550$ ₹.
• The relationship is of the form $y=ax+b$.

To find: The values of $a$ and $b$.

Solution: Substitute the given values into the equation $y=ax+b$:

1. For $x=10,y=350$:

2. For $x=20,y=550$:

Now, we solve these two linear equations for $a$ and $b$. Subtract equation (1) from equation (2):

Divide by 10 to find $a$:

Substitute the value of $a$ into equation (1):

Subtract 200 from both sides to find $b$:

The linear relationship is $y=20x+150$.

Answer: $a=20$ and $b=150$.

In this context, $a=20$ represents the cost per GB of data (₹20/GB), and $b=150$ represents the fixed monthly fee (₹150).

Example 2: Learning Platform Charges

A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed.

• When a student 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 $b$.

Given:

• When $x=10$ modules, $y=400$ ₹.
• When $x=14$ modules, $y=500$ ₹.
• The relationship is $y=ax+b$.

To find: The values of $a$ and $b$.

Solution: Substitute the given values into the equation $y=ax+b$:

1. For $x=10,y=400$:

2. For $x=14,y=500$:

Subtract equation (1) from equation (2):

Divide by 4 to find $a$:

Substitute the value of $a$ into equation (1):

Subtract 250 from both sides to find $b$:

The linear relationship is $y=25x+150$.

Answer: $a=25$ and $b=150$.

Here, $a=25$ is the cost per module (₹25/module), and $b=150$ is the fixed monthly fee (₹150).

Example 3: Temperature Conversion

Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F). This relationship is linear and can be expressed as °C = $a$ °F + $b$. Find $a$ and $b$, given that:

• Ice melts at 0°C and 32°F.
• Water boils at 100°C and 212°F.

Given:

• When °C = 0, °F = 32.
• When °C = 100, °F = 212.
• The relationship is °C = $a$ °F + $b$.

To find: The values of $a$ and $b$.

Solution: Substitute the given temperature points into the equation °C = $a$ °F + $b$:

1. For °C = 0, °F = 32:

2. For °C = 100, °F = 212:

From equation (1), we can express $b$ in terms of $a$:

Substitute this expression for $b$ into equation (2):

Divide by 180 to find $a$:

Simplify the fraction:

Now substitute the value of $a$ back into the expression for $b$:

The linear relationship is °C = $\frac{5}{9}$ °F - $\frac{160}{9}$.

Answer: $a=\frac{5}{9}$ and $b=-\frac{160}{9}$.

Remember

When finding a linear relationship $y=ax+b$ from two given points $(x_1,y_1)$ and $(x_2,y_2)$, the constant $a$ represents the slope or rate of change, calculated as $a=\frac{y_2-y_1}{x_2-x_1}$. The constant $b$ is the y-intercept, the value of $y$ when $x=0$.

Exam-smart

This topic often appears in problems where you're given two data points (like costs at different usage levels, or temperatures) and asked to find the specific linear equation governing the relationship. You might also be asked to interpret what the constants $a$ and $b$ represent in the given real-world context.

Visualising Linear Relationships

A linear relationship between two variables, typically denoted as $x$ and $y$, can be represented graphically as a straight line on a coordinate plane. The general form of a linear equation is $y=ax+b$, where $a$ and $b$ are constants. Understanding how to plot these equations and interpret their graphical features is crucial for visualising linear relationships.

To plot a linear equation $y=ax+b$, we need to identify at least two distinct points that satisfy the equation. Once these points are plotted on the coordinate plane, a straight line can be drawn by connecting them and extending it indefinitely in both directions.

Slope and y-intercept

In the linear equation $y=ax+b$:

• The constant 'a' is known as the slope of the line.
• The constant 'b' is known as the y-intercept of the line.

Slope

The slope of a line indicates its steepness and direction. It quantifies how much the $y$-value changes for a unit change in the $x$-value.

• A positive slope ($a>0$) means that as $x$ increases, $y$ also increases (linear growth). The line rises from left to right.
• A negative slope ($a<0$) means that as $x$ increases, $y$ decreases (linear decay). The line falls from left to right.
• A slope of zero ($a=0$) results in a horizontal line ($y=b$).
• The magnitude of the slope determines how steep the line is. A larger absolute value of $a$ means a steeper line.

The slope can be calculated if two points $(x_1,y_1)$ and $(x_2,y_2)$ on the line are known, using the formula:

y-intercept

The y-intercept is the y-coordinate of the point where the line crosses the y-axis. This occurs when $x=0$. Substituting $x=0$ into the equation $y=ax+b$, we get $y=a(0)+b$, which simplifies to $y=b$. Therefore, the line always intersects the y-axis at the point (0, b).

• If $b>0$, the line intersects the y-axis above the origin.
• If $b<0$, the line intersects the y-axis below the origin.
• If $b=0$, the line passes through the origin (0, 0).

Relationship between Slope, y-intercept, and Graph

• Fixed $b$, varying $a$: When the y-intercept ($b$) is fixed, changing the slope ($a$) results in lines with different steepness but all intersecting the y-axis at the same point. Lines with the same slope are parallel.
• Fixed $a$, varying $b$: When the slope ($a$) is fixed, changing the y-intercept ($b$) results in a family of parallel lines, each shifted vertically from the others.

Example 1: Finding Slope and y-intercept

Given: The equation of a line is $y=-3x+4$.

To find: The slope and y-intercept of the line.

Solution:

1. Compare the given equation $y=-3x+4$ with the standard form $y=ax+b$.
2. Identify the coefficient of $x$ as the slope.
3. Identify the constant term as the y-intercept.

Answer: The slope ($a$) is $-3$, and the y-intercept ($b$) is $4$.

Example 2: Plotting a Linear Equation

Given: The linear equation $y=2x+1$.

To find: The graph of the line.

Solution:

1. Find two points:

• When $x=0$, $y=2(0)+1=1$. So, point A is (0, 1).
• When $x=3$, $y=2(3)+1=6+1=7$. So, point B is (3, 7).

2. Plot the points: Mark A (0, 1) and B (3, 7) on a coordinate plane.
3. Draw the line: Connect points A and B with a straight line and extend it in both directions.

Remember

To plot any linear equation, finding just two points is sufficient. The origin (0,0) is a convenient point if it lies on the line (i.e., when $b=0$).

Exam-smart

You might be given two points and asked to find the equation of the line, or given an equation and asked to identify its slope and y-intercept from a graph. You may also be asked to determine if a given point lies on a line by substituting its coordinates into the equation.

Try this: Plot the graph of the equation $y=-x+2$. Answer: The line passes through (0, 2) and (2, 0).

Example 3: Relating Graphs to Equations

Given: A graph showing points (–1, –3), (0, 0), (1, 3), (3, 9), (4, 12).

To find: The equation of the line passing through these points.

Solution:

1. Observe the relationship: For each point $(x,y)$, notice how $y$ relates to $x$.

• For (0, 0): $y=0$, $x=0$.
• For (1, 3): $y=3$, $x=1$. $y$ is 3 times $x$.
• For (3, 9): $y=9$, $x=3$. $y$ is 3 times $x$.
• For (4, 12): $y=12$, $x=4$. $y$ is 3 times $x$.
• For (–1, –3): $y=-3$, $x=-1$. $y$ is 3 times $x$.

2. Formulate the equation: The relationship appears to be $y=3x$.
3. Verify: Check if all points satisfy $y=3x$. Yes, they do.

Answer: The equation of the line is $y=3x$.

Example 4: Comparing Linear Equations

Given: The graphs of $y=x$ and $y=2x$ on the same axes.

To find: The conclusions about these graphs.

Solution:

1. Identify common features: Both lines pass through the origin (0, 0) because their y-intercept ($b$) is 0.
2. Compare slopes: The slope of $y=x$ is $a=1$, and the slope of $y=2x$ is $a=2$.
3. Interpret steepness: Since $2>1$, the line $y=2x$ is steeper than the line $y=x$.
4. Generalize: For equations of the form $y=ax$ where $a>0$:

• If $a>1$, the line is steeper than $y=x$.
• If $0<a<1$, the line is less steep than $y=x$.

Remember

Lines of the form $y=ax$ always pass through the origin. The value of $a$ dictates how sharply the line rises.

Exam-smart

You may be shown graphs of lines like $y=x$, $y=2x$, $y=\frac{1}{2}x$ and asked to match them to their equations, or asked to describe the difference in steepness between $y=ax$ and $y=cx$ given $a>c>0$.

Try this: Compare the graphs of $y=-x$ and $y=-3x$. Answer: Both lines pass through the origin. $y=-3x$ is steeper than $y=-x$ and falls more rapidly from left to right.

Example 5: Parallel Lines

Given: The graphs of $y=2x-1$, $y=2x+1$, and $y=2x+5$ on the same axes.

To find: Conclusions about these graphs.

Solution:

1. Identify common features: All three equations have the same slope, $a=2$.
2. Identify differing features: The y-intercepts are different: $b=-1$, $b=1$, and $b=5$.
3. Interpret the graphs: The lines are parallel to each other because they have the same slope. They are vertically shifted from each other due to their different y-intercepts.

Remember

Lines with the same slope are parallel. The y-intercept determines their position relative to each other.

Exam-smart

You might be given several linear equations and asked to identify which lines are parallel, or given a graph with several parallel lines and asked to deduce the common slope.

Try this: Identify the parallel lines among the following: $y=5x+2$, $y=-5x+1$, $y=5x-3$, $y=x+4$. Answer: $y=5x+2$ and $y=5x-3$ are parallel.

Example 6: Celsius to Fahrenheit Conversion

Given: The relationship between Celsius (°C) and Fahrenheit (°F) is a linear equation: °C = a °F + b. Ice melts at 0°C (32°F) and water boils at 100°C (212°F).

To find: The values of $a$ and $b$, and the linear relationship.

Solution:

1. Set up equations using the given points:

• When °C = 0, °F = 32: $0=a(32)+b$ (Equation 1)
• When °C = 100, °F = 212: $100=a(212)+b$ (Equation 2)

2. Solve the system of equations:

• From Equation 1, $b=-32a$.
• Substitute this into Equation 2: $100=212a-32a$
• $100=180a$
• $a=\frac{100}{180}=\frac{10}{18}=\frac{5}{9}$

3. Find b: $b=-32a=-32\times \frac{5}{9}=-\frac{160}{9}$
4. Write the equation: °C = $\frac{5}{9}$ °F - $\frac{160}{9}$

This equation can also be rearranged to express Fahrenheit in terms of Celsius: °F = $\frac{9}{5}$ °C + 32.

Answer: $a=\frac{5}{9}$ and $b=-\frac{160}{9}$. The linear relationship is °C = $\frac{5}{9}$ °F - $\frac{160}{9}$.

Remember

The formula °F = $\frac{9}{5}$ °C + 32 is the standard conversion. The example uses °C = a°F + b, so $a$ and $b$ are derived accordingly.

Exam-smart

You could be given a real-world scenario that can be modeled by a linear equation and asked to find the equation or predict values. For instance, a question might involve cost based on a fixed charge and a per-item rate.

Try this: If a linear relationship is given by $y=3x+5$, what is the value of $y$ when $x=-2$? Answer: $y=-1$.