Grade 9 Mathematics – Part I
Chapter 2: Introduction to Linear Polynomials
Colorful Concepts • Complete Book Questions • Easy & Kids-Friendly Solutions
Chapter 2 – Introduction to Linear Polynomials
Textbook pp. 16–40
Algebraic Expression
A combination of numbers, variables and operation symbols. Example: 4x + 5y + 3.
A combination of numbers, variables and operation symbols. Example: 4x + 5y + 3.
Variable
A letter-number such as x, y or z whose value can vary.
A letter-number such as x, y or z whose value can vary.
Coefficient
The number multiplying a variable. In 5y, the coefficient of y is 5.
The number multiplying a variable. In 5y, the coefficient of y is 5.
Constant
A number without a variable, such as 3 or −7.
A number without a variable, such as 3 or −7.
Polynomial
An algebraic expression in one variable with non-negative integer powers of the variable.
An algebraic expression in one variable with non-negative integer powers of the variable.
Degree
The highest power of the variable in a polynomial.
The highest power of the variable in a polynomial.
Linear Polynomial
A polynomial of degree 1, such as 2x + 3.
A polynomial of degree 1, such as 2x + 3.
Linear Pattern
A sequence in which the difference between consecutive terms is constant.
A sequence in which the difference between consecutive terms is constant.
Linear Growth
A quantity increases by a fixed amount over equal intervals.
A quantity increases by a fixed amount over equal intervals.
Linear Decay
A quantity decreases by a fixed amount over equal intervals.
A quantity decreases by a fixed amount over equal intervals.
Linear Relationship
Usually written as y = ax + b.
Usually written as y = ax + b.
Slope & y-intercept
In y = ax + b, a is the slope and b is the y-intercept.
In y = ax + b, a is the slope and b is the y-intercept.
Remember:
Lines with equal slopes and different y-intercepts are parallel.
A line y = ax passes through the origin.
Think and Reflect
Page 17
1. Can you identify the terms, variables and coefficients of this algebraic expression?
200l + 160w + 50lw
Easy Solution
Terms: 200l, 160w and 50lw.Variables: l and w.
Coefficients: 200 for l, 160 for w and 50 for lw.
2. How is it different from the algebraic expression in Example 1?
Easy Solution
Example 1 has 4x + 5y + 3, where each variable occurs only to the first power and there is no product of variables.The expression 200l + 160w + 50lw contains the product lw, so it is not a linear polynomial in one variable.Think and Reflect
Page 17
1. Can you identify the terms, variables and coefficients of this algebraic expression?
10x − x²
Easy Solution
Terms: 10x and −x².Variable: x.
Coefficient of x is 10.
Coefficient of x² is −1.
2. Can you point out any similarity or difference between the algebraic expressions obtained in Examples 1 and 3?
Easy Solution
Both are algebraic expressions involving variables.Example 1 has two variables, x and y, while Example 3 has only one variable, x.Also, Example 3 contains x², so it is quadratic rather than linear.Exercise Set 2.1
Page 18
1. Find the degrees of the following polynomials:
(i) 2x² − 5x + 3
(ii) y³ + 2y − 1
(iii) − 9
(iv) 4z − 3
(i) 2x² − 5x + 3
(ii) y³ + 2y − 1
(iii) − 9
(iv) 4z − 3
Solution
(i) Highest power of x is 2 → degree 2.(ii) Highest power of y is 3 → degree 3.
(iii) −9 is a constant polynomial → degree 0.
(iv) Highest power of z is 1 → degree 1.
2. Write polynomials of degrees 1, 2 and 3.
Easy Solution
Degree 1:
2x + 3Degree 2: x² + 2x + 1
Degree 3: x³ + x + 1
3. What are the coefficients of x² and x³ in the polynomial
x⁴ − 3x³ + 6x² − 2x + 7?
Answer
Coefficient of x² = 6.Coefficient of x³ = −3.
4. What is the coefficient of z in the polynomial 4z³ + 5z² − 11?
Easy Solution
There is no z-term.We can write it as 0z.Therefore, the coefficient of z is 0.
5. What is the constant term of the polynomial 9x³ + 5x² − 8x −10?
Answer
The term without x is −10.Therefore, the constant term is −10.Think and Reflect
Page 19
Find the perimeter of squares with sides 1 cm, 1.5 cm, 2 cm, 2.5 cm and 3 cm. What will happen to the perimeters if the sides increase by 0.5 cm?
Easy Solution
Perimeter = 4 × side.For sides 1, 1.5, 2, 2.5 and 3 cm, the perimeters are:4 cm, 6 cm, 8 cm, 10 cm and 12 cm.Each time the side increases by 0.5 cm:4 × 0.5 = 2 cmSo the perimeter increases by 2 cm each time.
If a player paid `750, how many matches did he play?
Easy Solution
Amount = 200 + 50m.750 = 200 + 50m550 = 50mm = 11So, the player played 11 matches.Exercise Set 2.2
Page 21
1. Find the value of the linear polynomial 5x − 3 if:
(i) x = 0 (ii) x = −1 (iii) x = 2
(i) x = 0 (ii) x = −1 (iii) x = 2
Solution
(i) 5(0) − 3 = −3.(ii) 5(−1) − 3 = −5 − 3 = −8.
(iii) 5(2) − 3 = 10 − 3 = 7.
2. Find the value of the quadratic polynomial 7s² − 4s + 6 if:
(i) s = 0 (ii) s = −3 (iii) s = 4
(i) s = 0 (ii) s = −3 (iii) s = 4
Solution
(i) 7(0)² − 4(0) + 6 = 6.(ii) 7(−3)² − 4(−3) + 6 = 63 + 12 + 6 = 81.
(iii) 7(4)² − 4(4) + 6 = 112 − 16 + 6 = 102.
3. The present age of Salil’s mother is three times Salil’s present age. After 5 years, their ages will add up to 70 years. Find their present ages.
Easy Solution
Let Salil’s present age = x years.Mother’s age = 3x years.After 5 years:(x + 5) + (3x + 5) = 704x + 10 = 704x = 60x = 15So Salil is 15 years old.Mother’s age = 3 × 15 = 45 years.
4. The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.
Easy Solution
Let the integers be 2x and 5x.5x − 2x = 633x = 63x = 21Numbers:2 × 21 = 425 × 21 = 105
5. Ruby has 3 times as many two-rupee coins as she has five rupee-coins. If she has a total `88, how many coins does she have of each type?
Easy Solution
Let the number of five-rupee coins be x.Two-rupee coins = 3x.Total value:5x + 2(3x) = 8811x = 88x = 8Five-rupee coins = 8.Two-rupee coins = 3 × 8 = 24.
6. A farmer cuts a 300 feet fence into two pieces of different sizes. The longer piece is four times as long as the shorter piece. How long are the two pieces?
Easy Solution
Let shorter piece = x feet.Longer piece = 4x feet.x + 4x = 3005x = 300x = 60Longer piece = 4 × 60 = 240 feet.Therefore, the pieces are 60 feet and 240 feet.
7. If the length of a rectangle is three more than twice its width and its perimeter is 24 cm, what are the dimensions of the rectangle?
Easy Solution
Let width = x cm.Length = 2x + 3 cm.Perimeter:2(length + width) = 242(2x + 3 + x) = 246x + 6 = 246x = 18x = 3Width = 3 cm.Length = 2(3) + 3 = 9 cm.Think and Reflect
Page 20
We have learnt that to evaluate the value of an algebraic expression, we substitute a value of the variable in the given expression. Consider Example 3, where the wire is bent to form a rectangle. Here, the area of the rectangle, 10x − x², is a function of x. Can you interpret this as an input-output process? What value does the expression take when x = 6 cm?
Easy Solution
Input is x.Output is 10x − x².When x = 6:10(6) − 6²= 60 − 36= 24 cm².So the input 6 cm gives output 24 cm².Exploring Linear Patterns – Think and Reflect
Page 22
Predict the number of squares in the next three stages of the pattern and write the sequence of numbers up to Stage 7 of the pattern.
Solution
The pattern increases by 2 each time:1, 3, 5, 7, …Next three stages:9, 11, 13Sequence up to Stage 7:1, 3, 5, 7, 9, 11, 13
Using the expression 2n − 1, can you find out how many tiles will be there in the 15th stage and the 26th stage of the pattern? Also, which stage will contain 21 tiles and 47 tiles?
Easy Solution
15th stage:2(15) − 1 = 29 tiles.26th stage:2(26) − 1 = 51 tiles.For 21 tiles:2n − 1 = 212n = 22n = 11.For 47 tiles:2n − 1 = 472n = 48n = 24.Think and Reflect
Pages 22–23
What amount will be left on the 15th day? How many days will it take for the entire amount to be spent?
Easy Solution
Amount after n days:100 − 5nAfter 15 days:100 − 5(15)= 100 − 75= `25.For the entire amount to be spent:100 − 5n = 05n = 100n = 20 days.
For how many km will the fare be `130?
Easy Solution
Fare = 15n − 5.130 = 15n − 515n = 135n = 9 km.Exercise Set 2.3
Pages 23–24
1. A student has `500 in her savings bank account. She gets `150 every month as pocket money. How much money will she have at the end of every month from the second month onwards? Find a linear expression to represent the amount she will have in the nthmonth.
Easy Solution
Starting amount = `500.Each month she gets `150.End of 2nd month = `800.3rd month = `950.
4th month = `1100.
5th month = `1250.After n months:500 + 150n
2. A rally starts with 120 members. Each hour, 9 members drop out of the group. How many members will remain after 1, 2, 3, … hours? Find a linear expression to represent the number of members at the end of the nth hour.
Easy Solution
After 1 hour = 111 members.After 2 hours = 102 members.
After 3 hours = 93 members.
After 4 hours = 84 members.Each hour 9 members leave.After n hours:120 − 9n
3. Suppose the length of a rectangle is 13 cm. Find the area if the breadth is (i) 12 cm, (ii) 10 cm, (iii) 8 cm. Find the linear pattern representing the area of the rectangle.
Solution
Area = length × breadth.(i) 13 × 12 = 156 cm².(ii) 13 × 10 = 130 cm².(iii) 13 × 8 = 104 cm².If breadth is b:A = 13bThis is a linear relationship.
4. Suppose the length of a rectangular box is 7 cm and breadth is 11 cm. Find the volume if the height is (i) 5 cm, (ii) 9 cm, (iii) 13 cm. Find the linear pattern representing the volume of the rectangular box.
Solution
Volume = length × breadth × height= 7 × 11 × h= 77h(i) h = 5:77 × 5 = 385 cm³.(ii) h = 9:77 × 9 = 693 cm³.(iii) h = 13:77 × 13 = 1001 cm³.Linear pattern:V = 77h
5. Sarita is reading a book of 500 pages. She reads 20 pages every day. How many pages will be left after 15 days? Express this as a linear pattern.
Solution
Pages read in 15 days:20 × 15 = 300Pages left:500 − 300 = 200 pages.After n days:500 − 20nExercise Set 2.4
Pages 25–26
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.
(ii) Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month.
(iii) Find an expression that relates h and t, and explain why it represents linear growth.
(i) Find the height after 7 months.
(ii) Make a table of values for t varying from 0 to 10 months and show how the height, h, increases every month.
(iii) Find an expression that relates h and t, and explain why it represents linear growth.
Solution
(i)h = 1.75 + 0.5(7)= 5.25 feet.(ii)| t | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| h | 1.75 | 2.25 | 2.75 | 3.25 | 3.75 | 4.25 | 4.75 | 5.25 | 5.75 | 6.25 | 6.75 |
2. A mobile phone is bought for `10,000. Its value decreases by `800 every year.
(i) Find the value of the phone after 3 years.
(ii) Make a table of values for t varying from 0 to 8 years and show how the value of the phone, v, depreciates with time.
(iii) Find an expression that relates v and t, and explain why it represents linear decay.
(i) Find the value of the phone after 3 years.
(ii) Make a table of values for t varying from 0 to 8 years and show how the value of the phone, v, depreciates with time.
(iii) Find an expression that relates v and t, and explain why it represents linear decay.
Solution
(i)Value after 3 years:10000 − 800(3)= `7600.(ii)| t | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| v (`) | 10000 | 9200 | 8400 | 7600 | 6800 | 6000 | 5200 | 4400 | 3600 |
3. The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.
(i) Find the population of the village after 6 years.
(ii) Make a table of values for t varying from 0 to 10 years and show how the population, P, increases every year.
(iii) Find an expression that relates P and t, and explain why it represents linear growth.
(i) Find the population of the village after 6 years.
(ii) Make a table of values for t varying from 0 to 10 years and show how the population, P, increases every year.
(iii) Find an expression that relates P and t, and explain why it represents linear growth.
Solution
(i)P = 750 + 50(6)= 1050.(ii)| t | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| P | 750 | 800 | 850 | 900 | 950 | 1000 | 1050 | 1100 | 1150 | 1200 | 1250 |
4. A telecom company charges `600 for a certain recharge scheme. This prepaid balance is reduced by `15 each day after the recharge.
(i) Write an equation that models the remaining balance b(x) after using the scheme for x days. Explain why it represents linear decay.
(ii) After how many days will the balance run out?
(iii) Make a table of values for x varying from 1 to 10 days and show how the balance b(x), reduces with time.
(i) Write an equation that models the remaining balance b(x) after using the scheme for x days. Explain why it represents linear decay.
(ii) After how many days will the balance run out?
(iii) Make a table of values for x varying from 1 to 10 days and show how the balance b(x), reduces with time.
Solution
(i)b(x) = 600 − 15xIt is linear decay because the balance decreases by a fixed `15 each day.(ii)600 − 15x = 015x = 600x = 40 days.(iii)| x | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| b(x) (`) | 585 | 570 | 555 | 540 | 525 | 510 | 495 | 480 | 465 | 450 |
Think and Reflect
Page 24
What is the cost for travelling 15 km? For how many kilometres will the cost of the journey be `700?
Easy Solution
C(d) = 100 + 60d.For 15 km:C = 100 + 60(15)= `1000.For cost `700:700 = 100 + 60d600 = 60dd = 10 km.Think and Reflect
Page 25
What will be the height of the water at the end of 5 months?
Answer
h(t) = 3 − 0.5th(5) = 3 − 0.5(5)= 3 − 2.5= 0.5 m.Exercise Set 2.5
Page 27
1. A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed. A student observes that when she accessed 10 modules, her bill was `400. When she accessed 14 modules, her bill was `500. If the monthly bill y depends on the number of modules accessed, x, according to the relation y = ax + b, find the values of a and b.
Solution
400 = 10a + b500 = 14a + bSubtract:100 = 4aa = 25Now:400 = 10(25) + b400 = 250 + bb = 150Therefore:y = 25x + 150.
2. A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hours, her bill was `800. When she used it for 15 hours, her bill was `1100. If the monthly bill y depends on the hours of the use of the badminton court, x, according to the relation y = ax + b, find the values of a and b.
Solution
800 = 10a + b1100 = 15a + bSubtract:300 = 5aa = 60Now:800 = 60(10) + b800 = 600 + bb = 200Therefore:y = 60x + 200.
3. Consider the relationship between temperature measured in degrees Celsius (°C) and degrees Fahrenheit (°F), which is given by °C = a °F + b. Find a and b, given that ice melts at 0 degrees Celsius and 32 degrees Fahrenheit, and water boils at 100 degrees Celsius and 212 degrees Fahrenheit.
(Hint: When °C = 0, °F = 32 and when °C = 100, °F = 212. Use this information to find a and b, and thus, the linear relationship between °C and °F.)
(Hint: When °C = 0, °F = 32 and when °C = 100, °F = 212. Use this information to find a and b, and thus, the linear relationship between °C and °F.)
Solution
Using:°C = a°F + bAt 32°F:0 = 32a + bAt 212°F:100 = 212a + bSubtract:100 = 180aa = 5/9Now:0 = 32(5/9) + bb = −160/9Therefore:°C = (5/9)°F − 160/9.Visualising Linear Relationships – Think and Reflect
Page 28
Identify other points on the line by completing the following table.
x 1 2 5 7 9 12 20
y 3 15
x 1 2 5 7 9 12 20
y 3 15
Easy Solution
The line is:y = 2x + 1For x = 2:y = 2(2) + 1 = 5For x = 5:y = 2(5) + 1 = 11For x = 9:y = 2(9) + 1 = 19For x = 12:y = 2(12) + 1 = 25For x = 20:y = 2(20) + 1 = 41| x | 1 | 2 | 5 | 7 | 9 | 12 | 20 |
|---|---|---|---|---|---|---|---|
| y | 3 | 5 | 11 | 15 | 19 | 25 | 41 |
Exercise Set 2.6
Page 36
1. Draw the graphs of the following sets of lines. In each case, reflect on the role of ‘a’ and ‘b’.
(i) y = 4x, y = 2x, y = x
(ii) y = − 6x, y = − 3x, y = − x
(iii) y = 5x, y = −5x
(iv) y = 3x − 1, y = 3x, y = 3x + 1
(v) y = −2x − 3, y = −2x, y = 2x + 3
(i) y = 4x, y = 2x, y = x
(ii) y = − 6x, y = − 3x, y = − x
(iii) y = 5x, y = −5x
(iv) y = 3x − 1, y = 3x, y = 3x + 1
(v) y = −2x − 3, y = −2x, y = 2x + 3
Easy Solution / Graph Guide
(i)All pass through (0,0).Slopes are 4, 2 and 1.Greater positive slope → steeper line.
(ii)
All pass through (0,0).Slopes are −6, −3 and −1.More negative slope → steeper downward line.
(iii)
Slopes are 5 and −5.Both pass through the origin and have equal steepness in opposite directions.
(iv)
All slopes are 3, so the lines are parallel.Their y-intercepts are −1, 0 and 1.
(v)
Slopes are −2, −2 and 2.y-intercepts are −3, 0 and 3.The first two lines are parallel; the third has the opposite slope.
Graph Practice:
Plot at least two points for each equation, join the points with a ruler, and extend the line.
Plot at least two points for each equation, join the points with a ruler, and extend the line.
End-of-Chapter Exercises
Pages 36–40
1. Write a polynomial of degree 3 in the variable x, in which the coefficient of the x² term is −7.
Answer
One possible polynomial is:x³ − 7x² + 2x + 1Its degree is 3 and the coefficient of x² is −7.
2. Find the values of the following polynomials at the indicated values of the variables.
(i) 5x² − 3x + 7 if x = 1
(ii) 4t³ − t² + 6 if t = a
(i) 5x² − 3x + 7 if x = 1
(ii) 4t³ − t² + 6 if t = a
Solution
(i)5(1)² − 3(1) + 7= 5 − 3 + 7= 9(ii)Substitute t = a:4a³ − a² + 6
3. If we multiply a number by 5/2 and add 2/3 to the product, we get −7/12. Find the number.
Solution
Let the number be x.(5/2)x + 2/3 = −7/12(5/2)x = −7/12 − 8/12(5/2)x = −15/12(5/2)x = −5/4x = (−5/4) × (2/5)x = −1/2
4. A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
Solution
Let the smaller number be x.Larger number = 5x.After adding 21:Smaller = x + 21Larger = 5x + 21The larger is twice the smaller:5x + 21 = 2(x + 21)5x + 21 = 2x + 423x = 21x = 7Therefore, the numbers are:7 and 35.
5. If you have `800 and you save `250 every month, find the amount you have after (i) 6 months (ii) 2 years. Express this as a linear pattern.
Solution
After n months:800 + 250n(i) After 6 months:800 + 250(6)= 800 + 1500= `2300(ii) 2 years = 24 months.800 + 250(24)= 800 + 6000= `6800
*6. The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. Find both the numbers.
Solution
Let the number be 10a + b.The reversed number is 10b + a.Their sum:10a + b + 10b + a = 14311a + 11b = 143a + b = 13The digits differ by 3.The digits are therefore 5 and 8.Hence the two possible numbers are:58 and 85Both give 58 + 85 = 143.
*7. Draw the graph of the following equations, and identify their slopes and y-intercepts. Also, find the coordinates of the points where these lines cut the y-axis.
(i) y = –3x + 4
(ii) 2y = 4x + 7
(iii) 5y = 6x – 10
(iv) 3y = 6x – 11
Are any of the lines parallel?
(i) y = –3x + 4
(ii) 2y = 4x + 7
(iii) 5y = 6x – 10
(iv) 3y = 6x – 11
Are any of the lines parallel?
Solution
(i)y = −3x + 4Slope = −3y-intercept = 4Point on y-axis = (0,4)(ii)2y = 4x + 7y = 2x + 7/2Slope = 2y-intercept = 7/2Point on y-axis = (0,7/2)
(iii)5y = 6x − 10y = 6/5x − 2Slope = 6/5y-intercept = −2Point on y-axis = (0,−2)
(iv)3y = 6x − 11y = 2x − 11/3Slope = 2y-intercept = −11/3Point on y-axis = (0,−11/3)
Lines (ii) and (iv) have equal slope 2.Therefore, they are parallel.
*8. If the temperature of a liquid can be measured in Kelvin units as x K and in Fahrenheit units as y °F, the relation between the two systems of measurement of temperature is given by the linear equation y = 9/5 (x – 273) + 32.
(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 K.
(ii) If the temperature is 158 °F, then find the temperature in Kelvin.
(i) Find the temperature of the liquid in Fahrenheit if the temperature of the liquid is 313 K.
(ii) If the temperature is 158 °F, then find the temperature in Kelvin.
Solution
(i)y = 9/5(313 − 273) + 32= 9/5 × 40 + 32= 72 + 32= 104°F(ii)158 = 9/5(x − 273) + 32126 = 9/5(x − 273)126 × 5/9 = x − 27370 = x − 273x = 343 K
*9. The work done by a body on the application of a constant force is the product of the constant force and the distance travelled by the body in the direction of the force. Express this in the form of a linear equation in two variables (work w and distance d), and draw its graph by taking the constant force as 3 units. What is the work done when the distance travelled is 2 units? Verify it by plotting it on the graph.
Solution
Work = Force × Distance.Force = 3 units.Therefore:w = 3dFor d = 2:w = 3 × 2= 6 unitsGraph points can be:(0,0), (1,3), (2,6), (3,9).The point (2,6) verifies the answer.
*10. The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11).
(i) Find the polynomial p(x).
(ii) Find the coordinates where the graph of p(x) cuts the axes.
(iii) Draw the graph of p(x) and verify your answers.
(i) Find the polynomial p(x).
(ii) Find the coordinates where the graph of p(x) cuts the axes.
(iii) Draw the graph of p(x) and verify your answers.
Solution
Slope:(11 − 5)/(3 − 1)= 6/2= 3So:p(x) = 3x + bUsing (1,5):5 = 3 + bb = 2Therefore:p(x) = 3x + 2(ii) y-axis:x = 0y = 2Point = (0,2)
x-axis:y = 03x + 2 = 0x = −2/3Point = (−2/3,0)
(iii)Plot points such as:(0,2), (1,5), (3,11)and join them with a straight line.
*11. Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:
(i) p(0) = 5.
(ii) The polynomial p(x) – q(x) cuts the x-axis at (3, 0).
(iii) The sum p(x) + q(x) is equal to 6x + 4 for all real x.
Find the polynomials p(x) and q(x).
(i) p(0) = 5.
(ii) The polynomial p(x) – q(x) cuts the x-axis at (3, 0).
(iii) The sum p(x) + q(x) is equal to 6x + 4 for all real x.
Find the polynomials p(x) and q(x).
Solution
From p(0) = 5:b = 5Also:p(x) + q(x) = 6x + 4Therefore:a + c = 6b + d = 4Since b = 5:d = −1Now:p(x) − q(x)= (a − c)x + (b − d)The graph cuts the x-axis at (3,0), so:3(a − c) + 5 − (−1) = 03(a − c) + 6 = 0a − c = −2Now solve:a + c = 6a − c = −2Adding:2a = 4a = 2Therefore:c = 4Hence:p(x) = 2x + 5q(x) = 4x − 1
*12. Look at the first three stages of a growing pattern of hexagons made using matchsticks. A new hexagon gets added at every stage which shares a side with the last hexagon of the previous stage.
(i) Draw the next two stages of the pattern. How many matchsticks will be required at these stages?
(ii) Complete the following table.
Stage
Number
1 2 3 4 5 … n
Number of matchsticks
(iii) Find a rule to determine the number of matchsticks required for the nth stage.
(iv) How many matchsticks will be required for the 15th stage of the pattern?
(v) Can 200 matchsticks form a stage in this pattern? Justify your answer.
(i) Draw the next two stages of the pattern. How many matchsticks will be required at these stages?
(ii) Complete the following table.
Stage
Number
1 2 3 4 5 … n
Number of matchsticks
(iii) Find a rule to determine the number of matchsticks required for the nth stage.
(iv) How many matchsticks will be required for the 15th stage of the pattern?
(v) Can 200 matchsticks form a stage in this pattern? Justify your answer.
Solution
Each hexagon needs 6 matchsticks.When a new hexagon is added, it shares one side with the previous hexagon.Therefore each new hexagon adds 5 matchsticks.(i)Stage 4:6 + 5 + 5 + 5 = 21 matchsticksStage 5:6 + 5 + 5 + 5 + 5 = 26 matchsticks
(ii)
| Stage | 1 | 2 | 3 | 4 | 5 | … | n |
|---|---|---|---|---|---|---|---|
| Number of matchsticks | 6 | 11 | 16 | 21 | 26 | … | ? |
(iv)For n = 15:5(15) + 1= 75 + 1= 76 matchsticks
(v)5n + 1 = 2005n = 199n = 199/5This is not a whole number.Therefore, 200 matchsticks cannot form a complete stage in this pattern.
*13. Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:
(i) The graph of p(x) passes through the points (2, 3) and (6, 11).
(ii) The graph of q(x) passes through the point (4, −1).
(iii) The graph of q(x) is parallel to the graph of p(x).
Find the polynomials p(x) and q(x). Also, find the coordinates of the point where these lines meet the x-axis.
(i) The graph of p(x) passes through the points (2, 3) and (6, 11).
(ii) The graph of q(x) passes through the point (4, −1).
(iii) The graph of q(x) is parallel to the graph of p(x).
Find the polynomials p(x) and q(x). Also, find the coordinates of the point where these lines meet the x-axis.
Solution
Slope of p:(11 − 3)/(6 − 2)= 8/4= 2So:p(x) = 2x + bUsing (2,3):3 = 2(2) + b3 = 4 + bb = −1Therefore:p(x) = 2x − 1Since q is parallel to p, its slope is also 2.q(x) = 2x + dUsing (4,−1):−1 = 2(4) + d−1 = 8 + dd = −9Therefore:q(x) = 2x − 9
For p(x), x-axis point:0 = 2x − 1x = 1/2Point = (1/2,0)
For q(x):0 = 2x − 9x = 9/2Point = (9/2,0)
*14. What do all linear functions of the form f(x) = ax + a, a > 0, have in common?
Easy Solution
For:f(x) = ax + aThe slope is a.The y-intercept is also a.Therefore every graph is a straight line with slope a and y-intercept a.It passes through:(0,a)Also:f(−1) = a(−1) + a = 0So every line passes through:(−1,0)Chapter Summary
Pages 39–40
An algebraic expression combines numbers, variables and operation symbols.
A univariate polynomial has one variable. Its highest power gives its degree.
A polynomial of degree one is a linear polynomial.
Linear growth increases by a fixed amount over equal intervals.
Linear decay decreases by a fixed amount over equal intervals.
A linear pattern has a constant difference between consecutive terms.
A linear relationship is represented by a straight line y = ax + b.
In y = ax + b, a is the slope and b is the y-intercept.
When b = 0, y = ax passes through the origin.
Positive slope represents linear growth; negative slope represents linear decay.
Parallel lines have the same slope but different y-intercepts.
Study Tip:First identify the variable, then substitute values carefully.For graphs, put the equation into y = ax + b form before identifying slope and y-intercept.