Class 9 Maths Ganita Manjari
Chapter 2: Introduction to Linear Polynomials
Hello students! I have shared all the solutions for Chapter 2: Introduction to Linear Polynomials below. Click on any section to jump directly to it:
Just as we used coordinates to find exact locations in the previous chapter, we can use linear polynomials to represent consistent patterns, constant growth, and straight lines on that very same coordinate plane.
Core Concepts
- Polynomial Degree: The highest power of the variable in a polynomial is its degree. Linear polynomials have a degree of 1.
- Linear Growth/Decay: A pattern where a quantity increases (growth, positive slope) or decreases (decay, negative slope) by a constant amount over equal intervals.
- Linear Relationship: Expressed as \(y = ax + b\), where \(a\) is the slope (constant difference) and \(b\) is the y-intercept.
Exercise Set 2.1
1.
Find the degrees of the following polynomials:
(i) \(2x^2 - 5x + 3\)
(ii) \(y^3 + 2y - 1\)
(iii) \(-9\)
(iv) \(4z - 3\)
Answer
The degree of a polynomial is the highest power of the variable in the expression.
-
(i) \( 2x^2 - 5x + 3\) , Degree = 2.
(ii) \( y^3 + 2y - 1\), Degree = 3.
(iii) \( -9\) , Degree = 0.
(iv) \( 4z - 3\), Degree = 1.
2. Write polynomials of degrees 1, 2 and 3.
Answer
Degree = 1 (Linear polynomial): \(3x + 7\)
Degree = 2 (Quadratic polynomial): \(4y^2 - 2y + 1\)
Degree = 3 (Cubic polynomial): \(z^3 + 7z^2 - 4\)
3. What are the coefficients of \(x^2\) and \(x^3\) in the polynomial \(x^4 - 3x^3 + 6x^2 - 2x + 7\)?
Answer
The coefficient of \(x^2\) is 6.
The coefficient of \(x^3\) is -3.
4. What is the coefficient of 𝓏 in the polynomial \(4z^3 + 5z^2 - 11\)?
Solution
The polynomial can be written as \(4z^3 + 5z^2 + 0z - 11\).
Coefficient of 𝓏 is 0.
5. What is the constant term of the polynomial \(9x^3 + 5x^2 - 8x - 10\)?
Answer
The constant term is the term without a variable.
Constant term = -10.
Exercise Set 2.2
1.
Find the value of the linear polynomial \(5x - 3\) if:
(i) 𝑥 = 0
Answer
\[ \begin{align*} &= 5x - 3 \\ &= 5(0) - 3 \\ &= 0 - 3 \\ &= \color{red} - 3 \end{align*} \]
(ii) 𝑥 = -1
Answer
\[ \begin{align*} &= 5x - 3 \\ &= 5(-1) - 3 \\ &= -5 - 3 \\ &= \color{red} -8 \end{align*} \]
(iii) 𝑥 = 2
Answer
\[ \begin{align*} &= 5x - 3 \\ &= 5(2) - 3 \\ &= 10 - 3 \\ &= \color{red} 7 \end{align*} \]
2.
Find the value of the quadratic polynomial \(7s^2 - 4s + 6\) if
(i) s = 0
Answer
\[ \begin{align*} &= 7s^2 - 4s + 6 \\ &= 7(0)^2 - 4(0) + 6 \\ &= 0-0 + 6 \\ &= \color{red} 6 \\ \end{align*} \]
(ii) s = -3
Answer
\[ \begin{align*} &= 7s^2 - 4s + 6 \\ &= 7(-3)^2 - 4(-3) + 6 \\ &= 7(9) +12 + 6 \\ &= 63 + 18 \\ &= \color{red} 81 \end{align*} \]
(iii) s = 4
Answer
\[ \begin{align*} &= 7s^2 - 4s + 6 \\ &= 7(4)^2 - 4(4) + 6 \\ &= 7(16) - 16 + 6 \\ &= 112 - 10 \\ &= \color{red} 102 \end{align*} \]
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.
Solution
\[ \begin{aligned} \text{Let Salil's present age} & \implies x \ years \\ \text{His mother's present age } & \implies 3x \ years \\ \\ \textbf{After 5 years} & \\ \text{Salil's age} & = x + 5 \\ \text{Salil's mother's age} & = 3x + 5 \\ \text{Sum of their ages} & = 70 \ years \\ \\ \textbf{ATQ} & \\ (x + 5) + (3x + 5) &= 70 \\ 4x + 10 &= 70 \\ 4x &= 70 - 10 \\ 4x &= 60 \\[4pt] x &= \frac{\cancel{60}^{15}}{\cancel4_1} \\[4pt] \color{green} x &= \color{green} 15 \\ \\ \color{magenta} \textbf{Salil's present age} & = 15 \ years\\ \color{magenta} \textbf{His mother's present age} & = 3x \\ &= 3(15)\\ &= 45 \ years \\ \\ \end{aligned} \]
Answer Salil is 15 years old; his mother is 45 years old.
4. The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.
Solution
\[ \begin{aligned} \text{Let the two positive integers be } & 2x \text{ and } 5x \\ \text{Difference} & = 63 \\ \\ \textbf{ATQ} & \\ 5x - 2x &= 63 \\ 3x &= 63 \\ x &= \frac{\cancel{63}^{21}}{\cancel3_1} \\[4pt] \color{green} x &= \color{green} 21 \\ \\ \color{magenta} \textbf{First integer} & = 2x \\ &= 2(21) \\ & = 42 \\[5pt] \color{magenta} \textbf{Second integer} & = 5x \\ & = 5(21) \\ & = 105\\ \end{aligned} \]
Answer The integers are 42 and 105.
5. Ruby has 3 times as many two-rupee coins as she has five-rupee coins. If she has a total of ₹88, how many coins does she have of each type?
Solution
\[ \begin{aligned} \text{Let no. of 5-rupee coins} & \implies x \\ \text{Let no. of 2-rupee coins} & \implies 3x \\ \\ \text{Value 5-rupee coins} & = 5(x) \implies 5x \\ \text{Value 2-rupee coins} & = 2(3x) \implies 6x \\ \text{Total value} & = 88 \\ \\ \textbf{ATQ} & \\ 5x + 6x &= 88 \\ 11x &= 88 \\ x &= \frac{\cancel{88}^{8}}{\cancel{11}_1} \\[4pt] \color{green} x &= \color{green} 8 \\ \\ \color{magenta} \textbf{No. of 5-rupee coins} & = 8 \ coins \\ \color{magenta} \textbf{No. of 2-rupee coins} & = 3x \\ &= 3(8)\\ &= 24 \ coins \\ \\ \end{aligned} \]
Answer 5-rupee coins = 8 and 2-rupee coins = 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?
Solution
\[ \begin{aligned} \text{Let shorter piece be} & \implies x \\ \text{Longer piece be} & \implies 4x \\ \text{Total length} & = 300 \ feet \\ \\ \textbf{ATQ} & \\ x + 4x &= 300 \\ 5x &= 300 \\ x &= \frac{\cancel{300}^{60}}{\cancel{5}_1} \\[4pt] \color{green} x &= \color{green} 60 \\ \\ \color{magenta} \textbf{Shorter piece} & = 60 \ feet \\ \color{magenta} \textbf{Longer piece} & = 4x \\ &= 4(60)\\ &= 240 \ feet \\ \\ \end{aligned} \]
Answer Shorter piece = 60 feet and Longer piece = 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?
Solution
\[ \begin{aligned} \text{Let width of the rectangle} & \implies x \\ \text{Length of the rectangle} & \implies 2x + 3 \\ \text{Perimeter of rectangle} & \implies 24 \ cm \\ \color{magenta} \text{Perimeter of rectangle} & \implies \color{magenta} 2(l+b) \\ \\ \textbf{ATQ} & \\ 2(l+b) &= 24 \\ 2(2x + 3 + x) &= 24 \\ 2(3x + 3) &= 24 \\ 6x + 6 &= 24 \\ 6x &= 24 - 6 \\ 6x &= 18 \\ x &= \frac{\cancel{18}^{3}}{\cancel{6}_1} \\[4pt] \color{green} x &= \color{green} 3 \\ \\ \color{magenta} \textbf{Width} & = 3 \ cm \\ \color{magenta} \textbf{Length} & = 2x + 3 \\ &= 2(3) + 3 \\ &= 6 + 3 \\ &= 9 \ cm \\ \\ \end{aligned} \]
Answer Width is 3 cm and Length is 9 cm.
Exercise Set 2.3
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 \(n^{th}\) month.
Solution
\[ \begin{aligned} \textbf{Initial amount} & = \text{₹}500 \\ \textbf{Monthly pocket money} & = \text{₹}150 \\ \\ \text{End of } 1^{st} \text{ month} & = 500 + 150(1) \\ & = 500 + 150 \\ & \implies \color{magenta} \text{₹} 650 \\[4pt] \text{End of } 2^{nd} \text{ month} & = 500 + 150(2) \\ & = 500 + 300 \\ & \implies \color{magenta} \text{₹} 800 \\[4pt] \text{End of } 3^{rd} \text{ month} & = 500 + 150(3) \\ & = 500 + 450 \\ & \implies \color{magenta} \text{₹} 950 \\[4pt] \text{End of } n^{th} \text{ month} & = 500 + 150(n) \\ & \implies \color{magenta} 500 + 150n \\ \end{aligned} \]
Answer At the end of the \(\color{red} n^{th}\) month, she will have 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 \(n^{th}\) hour.
Solution
\[ \begin{aligned} \textbf{Initial members} & = 120 \\ \textbf{Members dropping per hour} & = 9 \\ \\ \text{End of } 1^{st} \text{ hour} & = 120 - 9(1) \\ & = 120 - 9 \\ & \implies \color{magenta} 111 \\[4pt] \text{End of } 2^{nd} \text{ hour} & = 120 - 9(2) \\ & = 120 - 18 \\ & \implies \color{magenta} 102 \\[4pt] \text{End of } 3^{rd} \text{ hour} & = 120 - 9(3) \\ & = 120 - 27 \\ & \implies \color{magenta} 93 \\[4pt] \text{End of } n^{th} \text{ hour} & = 120 - 9(n) \\ & \implies \color{magenta} 120 - 9n \\ \end{aligned} \]
Answer At the end of the \(\color{red} n^{th}\) hour, 120 - 9n members remain.
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
\[ \begin{aligned} \textbf{Length} & = 13 \ cm \\ \color{magenta} Area & = \color{magenta} Length \times Breadth \\ \\ \textbf{(i) } b & = 12 \ cm \\ \text{Area} & = 13 \times 12 \\ & \implies \color{magenta} 156 \ cm^2 \\[4pt] \textbf{(ii) } b & = 10 \ cm \\ \text{Area} & = 13 \times 10 \\ & \implies \color{magenta} 130 \ cm^2 \\[4pt] \textbf{(iii) } b & = 8 \ cm \\ \text{Area} & = 13 \times 8 \\ & \implies \color{magenta} 104 \ cm^2 \\[4pt] \textbf{Linear } & \textbf{pattern}\\ Breadth & = b \\ \textbf{Area} & = 13 \times b \\ & \implies \color{magenta} 13b \end{aligned} \]
Answer The linear pattern representing the area: A = 13b.
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
\[ \begin{aligned} \textbf{Length} & = 7 \ cm \\ \textbf{Breadth} & = 11 \ cm \\ \textbf{Base area} & = 77 \ cm^2 \\ \color{magenta} Volume & = \color{magenta} Length \times Breadth \times Height \\ \color{magenta} Volume & = \color{magenta} 77 \times Height \\ \\ \textbf{(i) } h & = 5 \ cm \\ \text{Volume} & = 77 \times 5 \\ & \implies \color{magenta} 385 \ cm^3 \\[4pt] \textbf{(ii) } h & = 9 \ cm \\ \text{Area} & = 77 \times 9 \\ & \implies \color{magenta} 693 \ cm^3 \\[4pt] \textbf{(iii) } h & = 13 \ cm \\ \text{Area} & = 77 \times 13 \\ & \implies \color{magenta} 1001 \ cm^3 \\[4pt] \textbf{Linear } & \textbf{pattern}\\ Height & = h \\ \textbf{Volume} & = 77 \times h \\ & \implies \color{magenta} 77h \end{aligned} \]
Answer The linear pattern representing the volume of the rectangular box is 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
\[ \begin{aligned} \textbf{Total pages} & = 500 \ pages \\ \textbf{Pages read per day} & = 20 \ pages \\ \\ \textbf{Pages read in 15 days} & = 20 \times 15 \\ & = 300 \ pages \\[5pt] \textbf{Remaining pages} & = 500 - 300 \\ & \implies \color{magenta} 200 \ pages \\ \\ \textbf{Linear } & \textbf{pattern}\\ \textbf{Let no. of days} & = n \\ & = 500 - 20(n) \\ & \implies \color{magenta} 500 - 20n \end{aligned} \]
Answer 200 pages left after 15 days. The linear pattern = 500 - 20n
Exercise Set 2.4
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.
Solution
| \(t\) (months) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(h\) (feet) | 1.75 | 2.25 | 2.75 | 3.25 | 3.75 | 4.25 | 4.75 | 5.25 | 5.75 | 6.25 | 6.75 |
Expression \( \implies \color{magenta} h = 1.75 + 0.5t\)
It represents linear growth because the height increases by a constant value (0.5 feet) every month.
Answer Height of the plant after 7 days = 5.25 feet. Expression: h(t) = 1.75 + 0.5t
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.
Solution
| \(t\) (years) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| \(v\) (₹) | 10000 | 9200 | 8400 | 7600 | 6800 | 6000 | 5200 | 4400 | 3600 |
Expression \( \implies \color{magenta} v = 10000 - 800t\).
It represents linear decay because the value decreases by a constant amount (₹800) every year.
Answer Value of the phone after 3 years = ₹7600. Expression: v = 10000 - 800t
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.
Solution
| \(t\) (years) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(P\) (population) | 750 | 800 | 850 | 900 | 950 | 1000 | 1050 | 1100 | 1150 | 1200 | 1250 |
Expression: \( \implies \color{magenta} P= 750 + 50t\).
It represents linear growth because the population increases by a constant 50 people every year.
Answer Population of the village after 6 years = 1050 people. Expression: P= 750 + 50t
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.
Solution
It represents linear decay because the balance drops by a constant ₹15 every single day.
| \(x\) (days) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| \(b(x)\) (balance) | 585 | 570 | 555 | 540 | 525 | 510 | 495 | 480 | 465 | 450 |
Answer Expression: b(𝑥) = 600 - 15𝑥, Balance will run out in 40 days.
Exercise Set 2.5
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
\[ \begin{aligned} \textbf{Linear relationship: } y & = ax + b \\[5pt] x = 10 \ , \ y = 400 \implies 400 & = 10a + b \color{magenta} \longrightarrow \enclose{circle}{1} \\[5pt] x = 14 \ , \ y = 500 \implies 500 & = 14a + b \color{magenta} \longrightarrow \enclose{circle}{2} \\[5pt] \end{aligned} \] \[ \begin{aligned} \textbf{Subtract} \ & \ eq \ \enclose{circle}{2} - eq \ \enclose{circle}{1} \\[5pt] 500 & = + 14a + b \\ \overset{\color{red} \scriptsize (-)}+400 & = \overset{\color{red} \scriptsize (-)}+10a \overset{\color{red} \scriptsize (-)}+ b \\ \hline 100 & = 4a \\ \hline \\ 4a &= 100 \\ a &= \frac{\cancel{100}^{25}}{\cancel4_1} \\ \color{green} a &= \color{green} 25 \end{aligned} \] \[ \begin{aligned} \textbf{Substitute: } &{\color{magenta} a = 25} \textbf{ in } \enclose{circle}{1} \\ 400 &= 10a + b \\ 400 &= 10(25) + b \\ 400 &= 250 + b \\ b &= 400 - 250 \\ \color{green} b &= \color{green} 150 \end{aligned} \]
Answer a = 25, b = 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
\[ \begin{aligned} \textbf{Linear relationship: } y & = ax + b \\[5pt] x = 10 \ , \ y = 800 \implies 800 & = 10a + b \color{magenta} \longrightarrow \enclose{circle}{1} \\[5pt] x = 15 \ , \ y = 1100 \implies 1100 & = 15a + b \color{magenta} \longrightarrow \enclose{circle}{2} \\[5pt] \end{aligned} \] \[ \begin{aligned} \textbf{Subtract} \ & \ eq \ \enclose{circle}{2} - eq \ \enclose{circle}{1} \\[5pt] 1100 & = + 15a + b \\ \overset{\color{red} \scriptsize (-)}+800 & = \overset{\color{red} \scriptsize (-)}+10a \overset{\color{red} \scriptsize (-)}+ b \\ \hline 300 & = 5a \\ \hline \\ 5a &= 300 \\ a &= \frac{\cancel{300}^{60}}{\cancel5_1} \\ \color{green} a &= \color{green} 60 \end{aligned} \] \[ \begin{aligned} \textbf{Substitute: } &{\color{magenta} a = 60} \textbf{ in } \enclose{circle}{1} \\ 800 &= 10a + b \\ 800 &= 10(60) + b \\ 800 &= 600 + b \\ b &= 800 - 600 \\ \color{green} b &= \color{green} 200 \end{aligned} \]
Answer \(a = 60\), \(b = 200\)
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.
Solution
\[ \begin{aligned} \textbf{Linear relationship: } ^\circ C & = a ^\circ F + b \\[5pt] ^\circ F = 32 \ , \ ^\circ C = 0 \implies 0 & = 32a + b \color{magenta} \longrightarrow \enclose{circle}{1} \\[5pt] ^\circ F = 212 \ , \ ^\circ C = 100 \implies 100 & = 212a + b \color{magenta} \longrightarrow \enclose{circle}{2} \\ \end{aligned} \] \[ \begin{aligned} \textbf{Subtract} \ & \ eq \ \enclose{circle}{2} - eq \ \enclose{circle}{1} \\[5pt] 100 & = + 212a + b \\ \overset{\color{red} \scriptsize (-)}0 & = \overset{\color{red} \scriptsize (-)}+32a \overset{\color{red} \scriptsize (-)}+ b \\ \hline 100 & = 180a \\ \hline \\ 180a &= 100 \\[5pt] a &= \frac{\cancel{100}^{5}}{\cancel{180}_9} \\[5pt] \color{green} a &= \color{green} \frac{5}{9} \end{aligned} \] \[ \begin{aligned} \textbf{Substitute: } &{\color{magenta} a = \frac{5}{9}} \textbf{ in } \enclose{circle}{1} \\[5pt] 0 &= 32a + b \\[5pt] 0 &= 32\left(\frac{5}{9}\right) + b \\[5pt] 0 &= \frac{160}{9} + b \\[5pt] \color{green} b &= \color{green} -\frac{160}{9} \end{aligned} \]
Answer \(a = \dfrac{5}{9}\), \(b = -\dfrac{160}{9}\)
Exercise Set 2.6
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
Answer
\[
\begin{aligned}
\color{blue} y = 4x \ \ \ \\
\begin{array}{|c|c|c|}
\hline
x & 0 & 1 \\
\hline
y & 0 & 4 \\
\hline
\end{array}
\end{aligned}
\ \ \ \
\begin{aligned}
\color{red}y = 2x \ \ \ \\
\begin{array}{|c|c|c|}
\hline
x & 0 & 2 \\
\hline
y & 0 & 4 \\
\hline
\end{array}
\end{aligned}
\ \ \ \
\begin{aligned}
\color{magenta}y = x \ \ \ \ \ \\
\begin{array}{|c|c|c|}
\hline
x & 0 & 3 \\
\hline
y & 0 & 3 \\
\hline
\end{array}
\end{aligned}
\]
\[
\begin{aligned}
\color{green}{y = ax + b} & =
\begin{cases}
\color{green} a = \textbf{Slope of the line} \\
\color{green} b = \textbf{𝑦-intercept}\\
\end{cases}\\[6pt]
\color{blue}y = 4x & \implies \color{blue}a = 4 \color{black}, b = 0 \\
\color{red}y = 2x & \implies \color{red}a = 2 \color{black}, b = 0 \\
\color{magenta}y = x & \implies \color{magenta}a = 1 \color{black}, b = 0
\end{aligned}
\]
Role of a: All the values of 'a' are positive. As the value of 'a' increases, the line becomes steeper.
Role of b: b = 0, so all lines pass through the origin.
(ii) 𝑦 = -6𝑥 , 𝑦 = -3𝑥 , 𝑦 = -𝑥
Answer
\[
\begin{aligned}
\color{blue} y = -6x \ \ \ \\
\begin{array}{|c|c|c|}
\hline
x & 0 & 1 \\
\hline
y & 0 & -6 \\
\hline
\end{array}
\end{aligned}
\ \ \ \
\begin{aligned}
\color{red}y = -3x \ \ \ \\
\begin{array}{|c|c|c|}
\hline
x & 0 & 1 \\
\hline
y & 0 & -3 \\
\hline
\end{array}
\end{aligned}
\ \ \ \
\begin{aligned}
\color{magenta}y = -x \ \ \ \ \ \\
\begin{array}{|c|c|c|}
\hline
x & 0 & 2 \\
\hline
y & 0 & -2 \\
\hline
\end{array}
\end{aligned}
\]
\[
\begin{aligned}
\color{green}{y = ax + b} & =
\begin{cases}
\color{green} a = \textbf{Slope of the line} \\
\color{green} b = \textbf{y-intercept}\\
\end{cases}\\[6pt]
\color{blue}y = -6x & \implies \color{blue}a = -6 \color{black}, b = 0 \\
\color{red}y = -3x & \implies \color{red}a = -3 \color{black}, b = 0 \\
\color{magenta}y = -x & \implies \color{magenta}a = -1 \color{black}, b = 0
\end{aligned}
\]
Role of a: All the values of 'a' are negative. The line becomes steeper but slopes downwards.
Role of b: b = 0, so all lines pass through the origin.
(iii) y = 5𝑥, y = -5𝑥
Answer
\[
\begin{aligned}
\color{blue} y = 5x \ \ \ \\
\begin{array}{|c|c|c|}
\hline
x & 0 & 1 \\
\hline
y & 0 & 5 \\
\hline
\end{array}
\end{aligned}
\ \ \ \ \ \ \ \
\begin{aligned}
\color{red}y = -5x \ \ \ \\
\begin{array}{|c|c|c|}
\hline
x & 0 & 1 \\
\hline
y & 0 & -5 \\
\hline
\end{array}
\end{aligned}
\]
\[
\begin{aligned}
\color{green}{y = ax + b} & =
\begin{cases}
\color{green} a = \textbf{Slope of the line} \\
\color{green} b = \textbf{y-intercept}\\
\end{cases}\\[6pt]
\color{blue}y = 5x & \implies \color{blue}a = 5 \color{black}, b = 0 \\
\color{red}y = -5x & \implies \color{red}a = -5 \color{black}, b = 0
\end{aligned}
\]
Role of a: The values of 'a' have opposite signs. They have the same steepness in opposite direction, one slopes upwards and the other slopes downwards.
Role of b: b = 0, so both lines pass through the origin.
(iv) y = 3𝑥 - 1, y = 3𝑥, y = 3𝑥 + 1
Answer
\[
\begin{aligned}
\color{blue} y = 3x - 1 \ \\
\begin{array}{|c|c|c|}
\hline
x & 0 & 1 \\
\hline
y & -1 & 2 \\
\hline
\end{array}
\end{aligned}
\ \ \ \
\begin{aligned}
\color{red}y = 3x \ \ \ \\
\begin{array}{|c|c|c|}
\hline
x & 0 & 1 \\
\hline
y & 0 & 3 \\
\hline
\end{array}
\end{aligned}
\ \ \ \
\begin{aligned}
\color{magenta}y = 3x + 1 \\
\begin{array}{|c|c|c|}
\hline
x & 0 & 1 \\
\hline
y & 1 & 4 \\
\hline
\end{array}
\end{aligned}
\]
\[
\begin{aligned}
\color{green}{y = ax + b} & =
\begin{cases}
\color{green} a = \textbf{Slope of the line} \\
\color{green} b = \textbf{y-intercept}\\
\end{cases}\\[6pt]
\color{blue}y = 3x - 1 & \implies \color{blue}a = 3 \color{black}, b = -1 \\
\color{red}y = 3x & \implies \color{red}a = 3 \color{black}, b = 0 \\
\color{magenta}y = 3x + 1 & \implies \color{magenta}a = 3 \color{black}, b = 1
\end{aligned}
\]
Role of a: All the values of 'a' are equal (a = 3), all three lines are parallel to each other.
Role of b: The different values of 'b' (-1, 0, 1) change the y-intercept. The lines intercepts y-axis at (0,-1) , (0,0) and (0,1).
(v) y = -2𝑥 - 3, y = -2𝑥, y = 2𝑥 + 3
Answer
\[
\begin{aligned}
\color{blue} y = -2x - 3 \ \ \ \\
\begin{array}{|c|c|c|}
\hline
x & 0 & 1 \\
\hline
y & -3 & -5 \\
\hline
\end{array}
\end{aligned}
\ \ \ \
\begin{aligned}
\color{red}y = -2x \ \ \ \\
\begin{array}{|c|c|c|}
\hline
x & 0 & 2 \\
\hline
y & 0 & -4 \\
\hline
\end{array}
\end{aligned}
\ \ \ \
\begin{aligned}
\color{magenta}y = 2x + 3 \ \ \ \ \ \\
\begin{array}{|c|c|c|}
\hline
x & 0 & 1 \\
\hline
y & 3 & 5 \\
\hline
\end{array}
\end{aligned}
\]
\[
\begin{aligned}
\color{green}{y = ax + b} & =
\begin{cases}
\color{green} a = \textbf{Slope of the line} \\
\color{green} b = \textbf{y-intercept}\\
\end{cases}\\[6pt]
\color{blue}y = -2x - 3 & \implies \color{blue}a = -2 \color{black}, b = -3 \\
\color{red}y = -2x & \implies \color{red}a = -2 \color{black}, b = 0 \\
\color{magenta}y = 2x + 3 & \implies \color{magenta}a = 2 \color{black}, b = 3
\end{aligned}
\]
Role of a: The first two lines have the same 'a' (a = -2), so they are parallel and slope downwards. The third line (a = 2) has the opposite slope direction.
Role of b: The different values of 'b' (-3, 0, 3) change the y-intercept. The lines intercepts y-axis at (0,-3) , (0,0) and (0,3).
End-of-Chapter Exercises
1. Write a polynomial of degree 3 in the variable 𝑥, in which the coefficient of the \(x^2\) term is -7.
Solution \(x^3 - 7x^2 + 2x + 5\)
2. Find the values of the following polynomials at the indicated values of the variables.
(i) \(5x^2 - 3x + 7\) if 𝑥 = 1
Solution
\[ \begin{aligned} & = 5x^2 - 3x + 7 \\ & = 5(1)^2 - 3(1) + 7\\ & = 5 - 3 + 7 \\ & = 2+7 \\ & = 9 \end{aligned} \]
(ii) \(4t^3 - t^2 + 6\) if t = a
Solution \(4a^3 - a^2 + 6\)
3. If we multiply a number by \(\dfrac{5}{2}\) and add \(\dfrac{2}{3}\) to the product, we get \(-\dfrac{7}{12}\). Find the number.
Solution
\[ \begin{aligned} \text{Let the number be } & = x \\[5pt] \frac{5}{2}x + \frac{2}{3} &= -\frac{7}{12} \\[5pt] \frac{5}{2}x &= -\frac{7}{12} - \frac{2}{3} \\[5pt] \frac{5}{2}x &= \frac{-7-8}{12} \\[5pt] \frac{5}{2}x &= -\frac{\cancel{15}^{5}}{\cancel{12}_4} \\[5pt] x &= -\frac{\cancel5}{\cancel4_2} \times \frac{\cancel2^1}{\cancel5} \\[5pt] x &= -\frac{1}{2} \end{aligned} \]
Answer The number is \(-\dfrac{1}{2}\)
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
\[ \begin{aligned} \text{Let smaller number} & = x \\ \text{Bigger number} & = 5x \\[5pt] \text{New smaller number} & = x + 21 \\ \text{New bigger number} & = 5x + 21 \\[5pt] \textbf{ATQ} & \\ \color{magenta} \textbf{New bigger number} & = \color{magenta} 2 \times \textbf{New smaller number} \\[5pt] 5x + 21 &= 2(x + 21) \\ 5x + 21 &= 2x + 42 \\ 5x - 2x &= 42 - 21 \\ 3x &= 21 \\ x &= 7 \\ \text{Smaller number } &= \color{green} 7 \\ \text{Bigger number } &= 5(7) \implies \color{green} 35 \end{aligned} \]
Answer 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
\[ \begin{aligned} \textbf{Initial amount} & = \text{₹}800 \\ \textbf{Monthly savings} & = \text{₹}250 \\[5pt] \textbf{Let months be } t & \textbf{ Amount be } A \\ \textbf{Linear pattern: } \color{magenta} A & = \color{magenta} 800 + 250t \\[5pt] (i) \textbf{ Amount after 6 months} & = 800 + 250(6) \\ & = 800 + 1500 \\ & = \color{red} \text{₹}2300 \\[5pt] (ii) \textbf{ Amount after 2 years} & = 800 + 250(24) \\ & = 800 + 6000 \\ & = \color{red} \text{₹}6800 \\ \end{aligned} \]
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
\[ \begin{aligned} \text{Let the ones digit be } & = y \\ \text{Let the tens digit be } & = x \\ x - y & = 3 \color{magenta} \longrightarrow \enclose{circle}{1} \\[5pt] \text{Original number } &= 10x + y \\ \text{Interchanged number } &= 10y + x \\[5pt] \textbf{ATQ}\\ (10x + y) + (10y + x) &= 143 \\ 11x + 11y &= 143 \\ 11(x + y) &= 143 \\ x + y &= 13 \color{magenta} \longrightarrow \enclose{circle}{2} \\[5pt] \textbf{ADD: } eq \ \enclose{circle}{1} &+ eq \ \enclose{circle}{2} \\[5pt] \begin{aligned} x - y & = 3 \\ x + y & = 13 \\ \hline 2x & = 16 \\ \hline \color{green} x & = \color{green} 8 \\ \end{aligned} \\ \begin{aligned} \textbf{Substitute } x \text{ in } eq \ \enclose{circle}{2} \\ x + y &= 13 \\ 8 + y &= 13 \\ y &= 13 - 8\\ \color{green} y &= \color{green} 5 \\[5pt] \color{green} \text{Original number } &= \color{green} 85 \\ \color{green} \text{Interchanged number } &= \color{green} 58 \end{aligned} \end{aligned} \]
Answer The numbers are 85 and 58.
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?
Answer
To find the coordinates and draw the graphs , convert all equations into the standard format:
\[
\begin{aligned}
\color{green}{y = ax + b} & =
\begin{cases}
\color{green} a = \textbf{Slope of the line} \\
\color{green} b = \textbf{y-intercept}\\
\end{cases}\\[6pt]
\end{aligned}
\]
\[
\begin{aligned}
(i) \ \color{blue} y = -3x + 4 \\
\begin{array}{|c|c|c|}
\hline
x & 0 & 1 \\
\hline
y & 4 & 1 \\
\hline
\end{array}
\\[5pt]
\textbf{Slope } \color{blue} (a) &= \color{blue}-3 \\
y\textbf{-intercept } \color{blue} (b) &= \color{blue}4 \\
\textbf{Lines cut } \color{blue} y-axis &= \color{blue} (0,4) \\
\end{aligned}
\]
\[
\begin{aligned}
(ii) \ 2y = 4x + 7 \\
y = \frac{4x}{2} + \frac{7}{2} \\[5pt]
\color{red}y = 2x + 3.5 \\[5pt]
\begin{array}{|c|c|c|}
\hline
x & 0 & 1 \\
\hline
y & 3.5 & 5.5 \\
\hline
\end{array}
\\[5pt]
\textbf{Slope } \color{red}(a) &= \color{red}2 \\
y\textbf{-intercept } \color{red}(b) &= \color{red} 3.5 \\
\textbf{Lines cut } \color{red} y-axis &= \color{red} (0,3.5) \\
\end{aligned}
\]
\[
\begin{aligned}
(iii) \ 5y = 6x - 10 \\[5pt]
y = \frac{6x}{5} - \frac{10}{5} \\[5pt]
\color{magenta}y = 1.2x - 2 \\
\begin{array}{|c|c|c|}
\hline
x & 0 & 5 \\
\hline
y & -2 & 4 \\
\hline
\end{array}
\\[5pt]
\textbf{Slope } \color{magenta}(a) &= \color{magenta} 1.2 \\
y\textbf{-intercept } \color{magenta}(b) &= \color{magenta} -2 \\
\textbf{Lines cut } \color{magenta} y-axis &= \color{magenta} (0,-2) \\
\end{aligned}
\]
\[
\begin{aligned}
(iv) \ 3y = 6x - 11 \\[5pt]
y = \frac{6x}{3} - \frac{11}{3} \\[5pt]
\color{brown}y = 2x - \frac{11}{3} \ \ \ \\
\begin{array}{|c|c|c|}
\hline
x & 0 & 2 \\
\hline
y & -\frac{11}{3} & \frac{1}{3} \\
\hline
\end{array}
\\[5pt]
\textbf{Slope } \color{brown}(a) &= \color{brown} 2 \\[5pt]
y\textbf{-intercept } \color{brown}(b) &= \color{brown} - \frac{11}{3} \\[5pt]
\textbf{Lines cut } \color{brown} y-axis &= \color{brown} \left(0,- \frac{11}{3}\right)
\end{aligned}
\]
Are any of the lines parallel?
(2y = 4x + 7) and (3y = 6x - 11) are parallel to each other, as they have the same slope ('a') value of 2.
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 \(y = \dfrac{9}{5}(x - 273) + 32\).
(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.
Solution
\[ \begin{aligned} \textbf{Linear equation: } \color{magenta} y &= \color{magenta} \frac{9}{5}(x - 273) + 32 \\[5pt] (i) \textbf{Given: } x &= 313 \\ y & = \frac{9}{5}(313 - 273) + 32 \\[5pt] y & = \frac{9}{5}(40) + 32 \\[5pt] y & = 72 + 32 \\ y & = \color{red} 104 ^\circ F \\ \\ (ii) \textbf{Given: } y &= 158 \\ 158 & = \frac{9}{5}(x - 273) + 32 \\[5pt] 158 - 32 & = \frac{9}{5}(x - 273) \\[5pt] 126 & = \frac{9}{5}(x - 273) \\[5pt] \cancel{126}^{14} \times \frac{5}{\cancel9_1} & = x - 273 \\[5pt] 14 \times 5 & = x - 273 \\ 70 & = x - 273 \\ 70 + 273 & = x \\ 343 & = x \\ x & = \color{red} 343 \ k \end{aligned} \]
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
\[ \begin{aligned} \textbf{Given: } \ w &= fd \\ \textbf{Constant force: } \ f &= 3 \\ \textbf{Linear equation: } \ \color{magenta} w & = \color{magenta} 3d \\ \textbf{When: }d & = 2 \\ w & = 3 \times 2 \\ w & = 6 \ units \\ \begin{array}{|c|c|c|c|} \hline d \ (distance) & 0 & 1 & 2 \\ \hline w \ (work) & 0 & 3 & 6 \\ \hline \end{array} \end{aligned} \]
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.
Solution
\[ \begin{aligned} \textbf{Linear equation: } y & = ax + b \\[5pt] \textbf{Passing through} & = (1,5) \\[5pt] x = 1 \ , \ y = 5 \implies 5 & = a + b \color{magenta} \longrightarrow \enclose{circle}{1} \\[5pt] \textbf{Passing through} & = (3,11) \\[5pt] x = 3 \ , \ y = 11 \implies 11 & = 3a + b \color{magenta} \longrightarrow \enclose{circle}{2} \\[5pt] \end{aligned} \] \[ \begin{aligned} \textbf{Subtract} \ & \ eq \ \enclose{circle}{2} - eq \ \enclose{circle}{1} \\[5pt] 11 & = + 3a + b \\ \overset{\color{red} \scriptsize (-)}+5 & = \overset{\color{red} \scriptsize (-)}+1a \overset{\color{red} \scriptsize (-)}+ b \\ \hline 6 & = 2a \\ \hline \\ 2a &= 6 \\ \color{green} a &= \color{green} 3 \end{aligned} \] \[ \begin{aligned} \textbf{Substitute: } &{\color{magenta} a = 3} \textbf{ in } \enclose{circle}{2} \\ 11 &= 3a + b \\ 11 &= 3(3) + b \\ 11 &= 9 + b \\ 11 - 9 &= b \\ 2 &= b \\ \color{green} b &= \color{green} 2 \end{aligned} \] \[ \begin{aligned} (i) \ \textbf{Polynomial: } & p(x) \\ p(x) &= ax + b\\ p(x) &= (3)x + 2\\ \color{red} p(x) &= \color{red} 3x + 2 \end{aligned} \] (ii) Coordinates where the graph of p(𝑥) cuts the axes \[ \begin{aligned} p(x): y &= 3x + 2 \\ \textbf{When } x & = 0 \\ y &= 3(0) + 2 \\ y &= 2 \\ \textbf{Line cuts } (y-axis) & = (0,2) \\[5pt] \textbf{When } y & = 0 \\ 0 &= 3x + 2 \\ -3x &= 2 \\ x &= \frac{-2}{3} \\ \textbf{Line cuts } (x-axis) & = \left(\frac{-2}{3},0 \right) \\[5pt] \end{aligned} \] (iii) Draw the graph and verify: \[ \begin{aligned} \color{blue} y = 3x + 2 \ \ \ \\ \begin{array}{|c|c|c|c|} \hline x & 0 & 1 & 2 \\ \hline y & 2 & 5 & 8 \\ \hline \end{array} \end{aligned} \]
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).
Solution
\[ \begin{aligned} \textbf{Given: } p(x) &= ax + b \\ q(x) &= cx + d \\[10pt] \textbf{(i) } p(0) &= 5 \\ a(0) + b &= 5 \\ \color{green} b &= \color{green} 5 \color{magenta} \longrightarrow \enclose{circle}{1} \\[10pt] \textbf{(ii) } p(x) - q(x) \textbf{ cuts 𝑥-axis at } & (3, 0) \\ \therefore \textbf{When } x & = 3\\ p(3) - q(3) &= 0 \\ [a(3) + b] - [c(3) + d] &= 0 \\ 3a + b - 3c - d &= 0 \\ 3a + 5 - 3c - d &= 0 \\ 3a - 3c - d &= -5 \color{magenta} \longrightarrow \enclose{circle}{2} \\[10pt] \textbf{(iii)} \ p(x) + q(x) &= 6x + 4 \\ (ax + b) + (cx + d) &= 6x + 4 \\ ax + 5 + cx + d &= 6x + 4 \\ ax + cx + d &= 6x + 4 -5 \\ (a + c)x + d &= 6x - 1 \\[5pt] \textbf{Comparing} & \textbf{ coefficients}\\ \textbf{Constant} & \textbf{ terms} \\ \color{green} d &= \color{green} -1 \color{magenta} \longrightarrow \enclose{circle}{3} \\[10pt] x & \textbf{ terms} \\ (a + c)x &= 6x \\ a + c &= 6 \\ c &= 6 - a \color{magenta} \longrightarrow \enclose{circle}{4} \\[10pt] \textbf{Substitute } d = -1, \ c &= 6 - a \textbf{ in } \enclose{circle}{2} \\[5pt] 3a - 3c - d &= -5 \\ 3a - 3(6-a) - (-1) &= -5 \\ 3a - 18 + 3a +1 &= -5 \\ 6a - 17 &= -5 \\ 6a &= -5 + 17 \\ 6a &= 12 \\ a &= \frac{12}{6} \\ \color{green} a &= \color{green} 2 \\[10pt] \textbf{Substitute } & a = 2 \textbf{ in } \enclose{circle}{4} \\ c &= 6 - 2 \\ \color{green} c &= \color{green} 4 \\[10pt] \textbf{Polynomials} \\ p(x) = ax + b \implies \color{red} p(x) &= \color{red} 2x + 5 \\ q(x) = cx + d \implies \color{red} q(x) &= \color{red} 4x - 1 \end{aligned} \]
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.
\[
\begin{array}{|l|c|c|c|c|c|c|c|}
\hline
\textbf{Stage Number} & 1 & 2 & 3 & 4 & 5 & \dots & n \\
\hline
\textbf{Number of matchsticks} & \ \ \ \ \ & \ \ \ \ \ & \ \ \ \ \ & \ \ \ \ \ & \ \ \ \ \ & \ \ \ \ \ & \ \ \ \ \ \\
\hline
\end{array}
\]
(iii) Find a rule to determine the number of matchsticks required for the \(n^{th}\) 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
(i) Next two stages: \[ \begin{aligned} \textbf{Stage 1} &= 6 \text{ matchsticks} \\ \textbf{Stage 2} &= 11 \text{ matchsticks} \\ \textbf{Stage 3} &= 16 \text{ matchsticks} \\ \textbf{Stage 4} &= 16 + 5 = \color{red} 21 \text{ matchsticks} \\ \textbf{Stage 5} &= 21 + 5 = \color{red} 26 \text{ matchsticks} \end{aligned} \]
Stage 4:
Stage 5:
(ii) Completed Table:
\[ \begin{aligned} \begin{array}{|l|c|c|c|c|c|c|c|} \hline \textbf{Stage Number} & 1 & 2 & 3 & 4 & 5 & \dots & n \\ \hline \textbf{No. of matchsticks} & 6 & 11 & 16 & \color{blue} 21 & \color{blue} 26 & \dots & \color{magenta} 5n+1 \\ \hline \end{array} \end{aligned} \](iii) Rule for the \( n^{th} \) stage:
\[ \begin{aligned} \text{Let stage no. be } & = n \\ \text{Initial matchsticks} & = 6 \\ \text{Additional matchsticks} & = 5 \\[5pt] n^{th} \textbf{ stage} &= 6 + 5(n - 1) \\ &= 6 + 5n - 5 \\ \color{magenta} n^{th} \textbf{ stage} &\color{magenta}= 5n + 1 \end{aligned} \](iv) Matchsticks for the \( 15^{th} \) stage:
\[ \begin{aligned} 15^{th} \textbf{ stage} &= 5n + 1 \\ &= 5(15) + 1 \\ &= 75 + 1 \\ &= 76 \\ \color{red} 15^{th} \textbf{ stage} &= \color{red} 76 \text{ matchsticks} \end{aligned} \](v) Can 200 matchsticks form a stage?
\[ \begin{aligned} Matchsticks &= 200 \\ 5n + 1 &= 200 \\ 5n &= 200 - 1 \\ 5n &= 199 \\ n &= \frac{199}{5} \\ n &= 39.8 \end{aligned} \]Justification: No, 200 matchsticks cannot form a complete stage in this pattern because, 'n' is not a natural number.
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.
Solution
(i) p(𝑥) passes through the points (2, 3) and (6, 11):
\[ \begin{aligned} p(x) & = ax + b \\[5pt] \textbf{Passes through} & = (2,3) \\[5pt] x = 2 \ , \ y = 3 \implies & {\color{green} 2a + b = 3} \color{magenta} \longrightarrow \enclose{circle}{1} \\[5pt] \textbf{Passes through} & = (6,11) \\[5pt] x = 6 \ , \ y = 11 \implies & {\color{green} 6a + b = 11} \color{magenta} \longrightarrow \enclose{circle}{2} \\[5pt] \textbf{From } eq \ \enclose{circle}{1} \implies b & = 3 - 2a \\[5pt] \textbf{Substitute } & b \textbf{ in } eq \ \enclose{circle}{2} \\ 6a + b & = 11 \\ 6a + 3 - 2a & = 11 \\ 4a & = 11 - 3 \\ 4a & = 8 \\ \color{green} a & = \color{green} 2 \\ \textbf{Substitute } & a \textbf{ in } eq \ \enclose{circle}{1} \\ 2a + b &= 3 \\ 2(2) + b &= 3 \\ 4 + b &= 3 \\ b &= 3 - 4 \\ \color{green} b & = \color{green} -1 \\ \implies \color{red} p(x) & = \color{red} 2x - 1 \\[5pt] \end{aligned} \](ii) q(𝑥) passes through the point (4, -1):
\[ \begin{aligned} q(x) & = cx + d \\[5pt] \textbf{Passes through} & = (4,-1) \\[5pt] x = 4 \ , \ y = -1 \implies & {\color{green} 4c + d = -1} \color{magenta} \longrightarrow \enclose{circle}{3} \\[5pt] \end{aligned} \](iii) q(𝑥) is parallel to the graph of p(𝑥):
\[ \begin{aligned} c = a & \implies 2 \\ \textbf{Substitute } & c \textbf{ in } eq \ \enclose{circle}{3} \\ 4c + d &= -1 \\ 4(2) + d &= -1 \\ 8 + d &= -1 \\ d &= -1 -8 \\ \color{green} d &= \color{green} -9 \\ \implies \color{red} q(x) & = \color{red} 2x - 9 \\[5pt] \end{aligned} \]Points meeting at 𝑥-axis (y = 0):
\[ \begin{aligned} p(𝑥) & = 0 \\ 2x - 1 & = 0 \\ 2x & = 1 \\ x & = \frac{1}{2} \\[5pt] \color{red} p(x) \textbf{ meets X-axis at } & = \color{red} \left(\frac{1}{2},0 \right)\\[5pt] q(𝑥) & = 0 \\ 2x - 9 & = 0 \\ 2x & = 9 \\ x & = \frac{9}{2} \\[5pt] \color{red} q(x) \textbf{ meets X-axis at } & = \color{red} \left(\frac{9}{2},0 \right) \\[5pt] \end{aligned} \]14. What do all linear functions of the form f(𝑥) = a𝑥 + a, a > 0, have in common?
Solution
\[ \begin{aligned} f(𝑥) & = ax + a \\ x-\textbf{intercept, } f(𝑥) & = 0 \\ ax + a & = 0 \\ ax & = -a \\ x & = \frac{-a}{a} \\ x & = -1 \\[5pt] \end{aligned} \]
All linear functions pass through the point (-1, 0) on X-axis.
Since a > 0, all have positive slope.
Also the y-intercept will be above the origin.