Chapter 2 Introduction to Linear Polynomials (Class 9 - Latest Maths NCERT (Ganita Manjari I) Solutions)

To Find: The degree of the given polynomials.

The degree of a polynomial is the highest power of the variable present in the expression.

---

Solution:

(i) In $2x^2 - 5x + 3$, the variable is $x$. The powers of $x$ are $2$ and $1$. The highest power is $2$.

---

(ii) In $y^3 + 2y - 1$, the variable is $y$. The powers of $y$ are $3$ and $1$. The highest power is $3$.

---

(iii) The given expression is $-9$. This is a constant. It can be written as $-9x^0$.

---

(iv) In $4z - 3$, the variable is $z$. The highest power of $z$ is $1$.

Solution:

The following are examples of polynomials with the requested degrees:

---

(i) Polynomial of Degree 1 (Linear Polynomial):

$P(x) = 3x + 5$

---

(ii) Polynomial of Degree 2 (Quadratic Polynomial):

$P(y) = y^2 - 4y + 7$

---

(iii) Polynomial of Degree 3 (Cubic Polynomial):

$P(z) = 2z^3 + 5z^2 - z + 10$

Given: Polynomial $P(x) = x^4 – 3x^3 + 6x^2 – 2x + 7$.

---

Solution:

A coefficient is the numerical factor of a term.

1. In the term $-3x^3$, the numerical factor multiplied by $x^3$ is $-3$.

2. In the term $6x^2$, the numerical factor multiplied by $x^2$ is $6$.

Given: Polynomial $P(z) = 4z^3 + 5z^2 – 11$.

---

Solution:

We need to find the coefficient of the term containing $z$. Observing the polynomial, we see there is no term with $z$.

We can rewrite the polynomial including the missing term as:

$P(z) = 4z^3 + 5z^2 + 0 \cdot z - 11$

Therefore, the coefficient of $z$ is $0$.

Solution:

A constant term in a polynomial is the term that does not contain any variable (or contains the variable with power $0$).

In the given polynomial $9x^3 + 5x^2 – 8x – 10$:

The terms are $9x^3$, $5x^2$, $-8x$, and $-10$.

The term without the variable $x$ is $-10$.

Given:

Linear polynomial $P(x) = 5x - 3$.

---

Solution:

(i) For $x = 0$:

$P(0) = 5(0) - 3$

$P(0) = 0 - 3$

$P(0) = -3$

---

(ii) For $x = -1$:

$P(-1) = 5(-1) - 3$

$P(-1) = -5 - 3$

$P(-1) = -8$

---

(iii) For $x = 2$:

$P(2) = 5(2) - 3$

$P(2) = 10 - 3$

$P(2) = 7$

Given:

Quadratic polynomial $P(s) = 7s^2 - 4s + 6$.

---

Solution:

(i) For $s = 0$:

$P(0) = 7(0)^2 - 4(0) + 6$

$P(0) = 0 - 0 + 6$

$P(0) = 6$

---

(ii) For $s = -3$:

$P(-3) = 7(-3)^2 - 4(-3) + 6$

$P(-3) = 7(9) + 12 + 6$

$P(-3) = 63 + 18$

$P(-3) = 81$

---

(iii) For $s = 4$:

$P(4) = 7(4)^2 - 4(4) + 6$

$P(4) = 7(16) - 16 + 6$

$P(4) = 112 - 16 + 6$

$P(4) = 102$

Given:

1. Salil's mother's age is 3 times Salil's age.

2. Sum of their ages after 5 years = 70.

---

To Find:

Their present ages.

---

Solution:

Let the present age of Salil be $x$ years.

After 5 years:

Salil's age = $(x + 5)$ years

Mother's age = $(3x + 5)$ years

According to the problem:

$4x + 10 = 70$

$4x = 70 - 10$

$4x = 60$

Present Ages:

Salil's present age = $x = \mathbf{15 \text{ years}}$

Mother's present age = $3x = 3 \times 15 = \mathbf{45 \text{ years}}$

Given:

Ratio of two integers = $2:5$

Difference between them = 63

---

Solution:

Let the two positive integers be $2x$ and $5x$.

According to the condition:

$3x = 63$

The Integers are:

First integer = $2x = 2 \times 21 = \mathbf{42}$

Second integer = $5x = 5 \times 21 = \mathbf{105}$

---

Verification:

$105 - 42 = 63$

To Find:

Number of $\textsf{₹}2$ and $\textsf{₹}5$ coins.

---

Solution:

Let the number of five-rupee coins be $x$.

Total amount Ruby has = $\textsf{₹}88$

$\text{Value of } \textsf{₹}5 \text{ coins} = 5 \times x = 5x$

$\text{Value of } \textsf{₹}2 \text{ coins} = 2 \times 3x = 6x$

Total value equation:

$11x = 88$

Number of Coins:

Number of five-rupee coins = $x = \mathbf{8}$

Number of two-rupee coins = $3x = 3 \times 8 = \mathbf{24}$

Given:

Total length of fence = 300 ft

Longer piece = $4 \times$ Shorter piece

---

Solution:

Let the length of the shorter piece be $x$ feet.

Then, the length of the longer piece = $4x$ feet.

Since the total length is 300 feet:

$5x = 300$

$x = \frac{300}{5} = 60$

Lengths:

Shorter piece = $\mathbf{60 \text{ feet}}$

Longer piece = $4 \times 60 = \mathbf{240 \text{ feet}}$

Given:

1. Perimeter of the rectangle = $24\text{ cm}$

2. The length of the rectangle is $3$ more than twice its width.

---

To Find:

The dimensions (Length and Width) of the rectangle.

---

Solution:

Let the width of the rectangle be $x$ cm.

According to the problem, the length ($L$) is $3$ more than twice the width ($x$).

We know that the formula for the perimeter ($P$) of a rectangle is:

Substituting the given perimeter and the expressions for length and width in equation (ii):

$24 = 2((2x + 3) + x)$

Dividing both sides by 2:

$12 = 3x + 3$

Now, we solve for $x$ by subtracting 3 from both sides:

$3x = 12 - 3$

$x = \frac{\cancel{9}^{3}}{\cancel{3}_{1}}$

$x = 3 \text{ cm}$

Now, we find the length by substituting the value of $x$ back into equation (i):

$L = 2(3) + 3$

$L = 6 + 3$

$L = 9 \text{ cm}$

---

Final Dimensions:

The width of the rectangle is $3\text{ cm}$ and the length is $9\text{ cm}$.

Given:

Initial savings = $\textsf{₹} 500$

Monthly addition (pocket money) = $\textsf{₹} 150$

---

To Find:

1. Money at the end of every month from the second month onwards.

2. Linear expression for the amount in the $n^{th}$ month.

---

Solution:

Let $n$ represent the number of months passed.

Amount at the end of $1^{\text{st}}$ month = $500 + 150 = \textsf{₹} 650$

Amount at the end of $2^{\text{nd}}$ month = $650 + 150 = \textsf{₹} 800$

Amount at the end of $3^{\text{rd}}$ month = $800 + 150 = \textsf{₹} 950$

Following this pattern, the money at the end of every month from the second month onwards is $\textsf{₹} 800$, $\textsf{₹} 950$, $\textsf{₹} 1100, \dots$

---

Linear Expression:

The total amount $A$ after $n$ months is the initial balance plus the product of the number of months and the monthly addition.

Where $A$ is the amount in $\textsf{₹}$ and $n$ is the number of months.

Given:

Initial members = $120$

Drop out rate per hour = $9$

---

To Find:

1. Remaining members after $1, 2, 3, \dots$ hours.

2. Linear expression for the $n^{th}$ hour.

---

Solution:

Let $n$ be the number of hours.

Members after 1 hour = $120 - 9 = 111$

Members after 2 hours = $111 - 9 = 102$

Members after 3 hours = $102 - 9 = 93$

The number of members remaining after $1, 2, 3 \dots$ hours will be $111, 102, 93, \dots$

---

Linear Expression:

Let $M$ be the number of remaining members.

Here, $120$ is the starting value and $9n$ represents the total members dropped after $n$ hours.

Given:

Length ($l$) = $13 \text{ cm}$

---

Solution:

The formula for the area of a rectangle is $\text{Area} = \text{Length} \times \text{Breadth}$.

(i) If breadth ($b$) = $12 \text{ cm}$:

$\text{Area} = 13 \times 12 = \mathbf{156 \text{ cm}^2}$

---

(ii) If breadth ($b$) = $10 \text{ cm}$:

$\text{Area} = 13 \times 10 = \mathbf{130 \text{ cm}^2}$

---

(iii) If breadth ($b$) = $8 \text{ cm}$:

$\text{Area} = 13 \times 8 = \mathbf{104 \text{ cm}^2}$

---

Linear Pattern:

Since the length is fixed at $13 \text{ cm}$, the area ($A$) varies linearly with the breadth ($b$).

This linear pattern shows that for every $1 \text{ cm}$ increase in breadth, the area increases by $13 \text{ cm}^2$.

Given:

Length ($l$) = $7 \text{ cm}$

Breadth ($b$) = $11 \text{ cm}$

---

Solution:

The formula for the volume of a rectangular box (cuboid) is $\text{Volume} = \text{Length} \times \text{Breadth} \times \text{Height}$.

The area of the base is $7 \times 11 = 77 \text{ cm}^2$.

(i) If height ($h$) = $5 \text{ cm}$:

$\text{Volume} = 77 \times 5 = \mathbf{385 \text{ cm}^3}$

---

(ii) If height ($h$) = $9 \text{ cm}$:

$\text{Volume} = 77 \times 9 = \mathbf{693 \text{ cm}^3}$

---

(iii) If height ($h$) = $13 \text{ cm}$:

$\text{Volume} = 77 \times 13 = \mathbf{1001 \text{ cm}^3}$

---

Linear Pattern:

Let $V$ represent the volume and $h$ represent the height.

This linear expression represents the volume as a function of height.

Given:

Total pages in book = $500$

Reading rate = $20$ pages/day

Time passed = $15$ days

---

Solution:

Pages read in 15 days = $20 \times 15 = 300$ pages

Pages left = Total pages $-$ Pages read

$\text{Pages left} = 500 - 300 = 200$

Therefore, 200 pages will be left after 15 days.

---

Linear Pattern:

Let $P$ be the number of pages left and $d$ be the number of days elapsed.

Where $500$ is the initial count and $20d$ is the total pages read after $d$ days.

Given:

Initial height of the plant ($h_0$) = $1.75\text{ feet}$

Growth rate per month ($r$) = $0.5\text{ feet/month}$

---

To Find:

(i) Height ($h$) after $t = 7\text{ months}$.

(ii) Table of values for $t = 0$ to $10$.

(iii) Linear expression relating $h$ and $t$.

---

Solution:

(i) To find the height after $7$ months, we add the total growth to the initial height.

$\text{Total growth} = \text{Growth rate} \times \text{Time}$

$\text{Total growth} = 0.5 \times 7 = 3.5\text{ feet}$

---

(ii) Table of values:

Time $t$ (months)	Height $h$ (feet) ($h = 1.75 + 0.5t$)
0	1.75
1	2.25
2	2.75
3	3.25
4	3.75
5	4.25
6	4.75
7	5.25
8	5.75
9	6.25
10	6.75

---

(iii) Expression and Explanation:

The relationship between height $h$ and time $t$ can be expressed as:

This represents linear growth because the height increases by a constant amount ($0.5\text{ feet}$) for every unit increase in time (each month). In a graph, this would form a straight line with a constant slope.

Given:

Initial value of the phone ($v_0$) = $\textsf{₹}10,000$

Depreciation rate per year ($d$) = $\textsf{₹}800$

---

To Find:

(i) Value ($v$) after $t = 3\text{ years}$.

(ii) Table of values for $t = 0$ to $8$.

(iii) Expression relating $v$ and $t$.

---

Solution:

(i) To find the value after $3$ years, we subtract the total depreciation from the initial price.

$\text{Total depreciation} = 800 \times 3 = \textsf{₹}2,400$

---

(ii) Table of values:

Time $t$ (years)	Value $v$ ($\textsf{₹}$) ($v = 10000 - 800t$)
0	10,000
1	9,200
2	8,400
3	7,600
4	6,800
5	6,000
6	5,200
7	4,400
8	3,600

---

(iii) Expression and Explanation:

The relationship between value $v$ and time $t$ is:

This represents linear decay (or depreciation) because the value of the phone decreases by a fixed constant amount ($\textsf{₹}800$) every year. The rate of change is constant and negative.

Given:

Initial population ($P_0$) = $750$

Increase per year ($i$) = $50$ people

---

To Find:

(i) Population ($P$) after $t = 6\text{ years}$.

(ii) Table of values for $t = 0$ to $10$.

(iii) Expression relating $P$ and $t$.

---

Solution:

(i) To find the population after $6$ years:

$\text{Total increase} = 50 \times 6 = 300$

---

(ii) Table of values:

Time $t$ (years)	Population $P$ ($P = 750 + 50t$)
0	750
1	800
2	850
3	900
4	950
5	1000
6	1050
7	1100
8	1150
9	1200
10	1250

---

(iii) Expression and Explanation:

The relationship between population $P$ and time $t$ is:

This represents linear growth because the population increases by a constant rate of $50$ individuals per year. The addition is independent of the current population size, leading to a constant slope.

Given:

Initial balance = $\textsf{₹}600$

Daily reduction = $\textsf{₹}15$

---

To Find:

(i) Equation for $b(x)$ and reason for linear decay.

(ii) Number of days until balance becomes zero.

(iii) Table for $x = 1$ to $10$.

---

Solution:

(i) Equation:

The remaining balance $b(x)$ after $x$ days is calculated by subtracting the total reduction from the initial recharge amount.

This represents linear decay because the balance reduces by a fixed amount ($\textsf{₹}15$) every day. This constant rate of reduction results in a linear relationship with a negative slope.

---

(ii) Balance running out:

The balance runs out when $b(x) = 0$.

$15x = 600$

$x = \frac{600}{15}$

Therefore, the balance will run out after $40$ days.

---

(iii) Table of values:

Days $x$	Balance $b(x)$ ($\textsf{₹}$)
1	585
2	570
3	555
4	540
5	525
6	510
7	495
8	480
9	465
10	450

Given:

Monthly bill relation: $y = ax + b$

Case 1: When $x = 10$, $y = \textsf{₹}400$

Case 2: When $x = 14$, $y = \textsf{₹}500$

---

To Find:

The values of $a$ (cost per module) and $b$ (fixed monthly fee).

---

Solution:

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

Subtracting equation (i) from equation (ii):

$(14a + b) - (10a + b) = 500 - 400$

$4a = 100$

Now, substitute the value of $a = 25$ in equation (i):

$400 = 10(25) + b$

$400 = 250 + b$

$b = 400 - 250$

Therefore, the values are $a = 25$ and $b = 150$.

Given:

Linear relation: $y = ax + b$

For $x = 10$ hours, $y = \textsf{₹}800$

For $x = 15$ hours, $y = \textsf{₹}1100$

---

To Find:

The values of $a$ (hourly rate) and $b$ (fixed fee).

---

Solution:

Forming equations based on the given data:

Subtracting equation (i) from equation (ii):

$(15a + b) - (10a + b) = 1100 - 800$

$5a = 300$

Substituting $a = 60$ into equation (i):

$800 = 10(60) + b$

$800 = 600 + b$

$b = 800 - 600$

Thus, the constants for the gym bill are $a = 60$ and $b = 200$.

Given:

Linear equation: $C = aF + b$

Freezing point: $C = 0$ when $F = 32$

Boiling point: $C = 100$ when $F = 212$

---

To Find:

The values of coefficients $a$ and $b$.

---

Solution:

Using the melting point values in $C = aF + b$:

Using the boiling point values in $C = aF + b$:

Subtracting equation (i) from equation (ii):

$100 - 0 = (212a + b) - (32a + b)$

$100 = 180a$

Now, substitute $a = \frac{5}{9}$ into equation (i):

$0 = 32 \left( \frac{5}{9} \right) + b$

$0 = \frac{160}{9} + b$

Therefore, the values are $a = \frac{5}{9}$ and $b = -\frac{160}{9}$.

---

Note: The final linear relationship can be written as $C = \frac{5}{9}F - \frac{160}{9}$ or $C = \frac{5}{9}(F - 32)$.

To Find:

The solutions (coordinates) for each set of equations and their respective graphical representations to understand the role of $a$ and $b$ in $y = ax + b$.

---

(i) Equations: $y = 4x, y = 2x, y = x$

For these equations, the $y$-intercept $b$ is $0$, meaning they all pass through the origin $(0, 0)$.

$x$	$y = 4x$	$y = 2x$	$y = x$
-1	-4	-2	-1
0	0	0	0
1	4	2	1

Reflection: As '$a$' (the coefficient of $x$) increases, the line becomes steeper and rotates towards the $y$-axis.

---

(ii) Equations: $y = -6x, y = -3x, y = -x$

In this set, the slope '$a$' is negative. These lines pass through the second and fourth quadrants.

$x$	$y = -6x$	$y = -3x$	$y = -x$
-1	6	3	1
0	0	0	0
1	-6	-3	-1

Reflection: A negative '$a$' results in a downward slope. As the absolute value of '$a$' increases, the line becomes steeper.

---

(iii) Equations: $y = 5x, y = -5x$

These lines have slopes of equal magnitude but opposite signs.

$x$	$y = 5x$	$y = -5x$
-1	-5	5
0	0	0
1	5	-5

Reflection: The line $y = -5x$ is a mirror image (reflection) of $y = 5x$ across the $y$-axis.

---

(iv) Equations: $y = 3x - 1, y = 3x, y = 3x + 1$

Here, the slope '$a$' is constant ($a = 3$), making the lines parallel. The intercept '$b$' varies.

$x$	$y = 3x - 1$	$y = 3x$	$y = 3x + 1$
-1	-4	-3	-2
0	-1	0	1
1	2	3	4

Reflection: When '$a$' is constant, changing '$b$' shifts the line vertically. These are called parallel lines.

---

(v) Equations: $y = -2x - 3, y = -2x, y = 2x + 3$

This set compares parallel lines with negative slopes and a line with a positive slope.

$x$	$y = -2x - 3$	$y = -2x$	$y = 2x + 3$
-1	-1	2	1
0	-3	0	3
1	-5	-2	5

Reflection: The first two lines are parallel. The third line intersects them. The point $(0, -3)$ is the $y$-intercept for the first line, while $(0, 3)$ is the intercept for the third.

---

Final Conclusion:

1. The coefficient '$a$': Represents the Slope. It controls the steepness and direction (increase/decrease) of the line.

2. The constant '$b$': Represents the $y$-intercept. It determines the point where the line crosses the vertical axis and indicates a vertical shift.

To Find:

A polynomial of degree $3$ in $x$ with the coefficient of $x^2$ as $-7$.

---

Solution:

A general cubic polynomial (degree 3) in variable $x$ is given by:

$P(x) = ax^3 + bx^2 + cx + d$

According to the given conditions:

1. The degree must be $3$, so the highest power of $x$ is $3$. We can choose any non-zero value for $a$. Let $a = 1$.

2. The coefficient of $x^2$ must be $-7$. Thus, $b = -7$.

3. Other coefficients $c$ and $d$ can be any real numbers. Let $c = 5$ and $d = 10$.

One such polynomial is:

$P(x) = x^3 - 7x^2 + 5x + 10$

Given:

(i) $P(x) = 5x^2 – 3x + 7$

(ii) $Q(t) = 4t^3 – t^2 + 6$

---

Solution:

(i) For $x = 1$:

Substituting $x = 1$ in the polynomial:

$P(1) = 5(1)^2 - 3(1) + 7$

$P(1) = 5(1) - 3 + 7$

$P(1) = 5 - 3 + 7$

$P(1) = 9$

---

(ii) For $t = a$:

Substituting the variable $t$ with $a$:

$Q(a) = 4a^3 - a^2 + 6$

To Find:

A number that satisfies the given algebraic condition.

---

Solution:

Let the required number be $x$.

According to the problem:

Transposing $\frac{2}{3}$ to the right side:

$\frac{5}{2}x = -\frac{7}{12} - \frac{2}{3}$

Taking the L.C.M. of 12 and 3, which is 12:

$\frac{5}{2}x = \frac{-7 - 8}{12}$

$\frac{5}{2}x = -\frac{15}{12}$

Solving for $x$:

$x = -\frac{15}{12} \times \frac{2}{5}$

$x = \frac{-\cancel{15}^{3} \times \cancel{2}^{1}}{\cancel{12}_{6} \times \cancel{5}_{1}}$

$x = -\frac{3}{6}$

Therefore, the required number is $-\frac{1}{2}$.

To Find:

The two initial positive numbers.

---

Solution:

Let the smaller number be $x$.

Then the other number is $5x$.

If $21$ is added to both:

New smaller number = $x + 21$

New larger number = $5x + 21$

According to the condition, the larger new number is twice the smaller new number:

$5x - 2x = 42 - 21$

$3x = 21$

$x = \frac{21}{3}$

$x = 7$

Finding the numbers:

Smaller number = $x = \mathbf{7}$

Larger number = $5x = 5 \times 7 = \mathbf{35}$

The numbers are $7$ and $35$.

Given:

Initial balance = $\textsf{₹}800$

Monthly savings = $\textsf{₹}250$

---

Solution:

Let $n$ be the number of months.

The total amount ($A$) can be calculated using the pattern:

---

(i) After 6 months ($n = 6$):

$A = 800 + 250(6)$

$A = 800 + 1500$

$A = \textsf{₹} 2300$

---

(ii) After 2 years:

Since 1 year = 12 months, 2 years = $2 \times 12 = 24$ months.

$A = 800 + 250(24)$

$A = 800 + 6000$

$A = \textsf{₹} 6800$

---

Summary Table:

Duration (n)	Calculation ($800 + 250n$)	Total Amount ($\textsf{₹}$)
6 months	$800 + 1500$	2300
24 months	$800 + 6000$	6800

To Find: The two-digit numbers satisfying the given conditions.

---

Solution:

Let the tens digit be $x$ and the units digit be $y$.

The original number can be expressed as $10x + y$.

According to the first condition:

When digits are interchanged, the new number is $10y + x$.

According to the second condition:

$(10x + y) + (10y + x) = 143$

$11x + 11y = 143$

Dividing by $11$ on both sides:

Adding equations (i) and (ii):

$(x - y) + (x + y) = 3 + 13$

$2x = 16$

$x = 8$

Substituting $x = 8$ in (ii):

$8 + y = 13 \implies y = 5$

The original number is $10(8) + 5 = 85$.

---

Case 2: If the difference is $y - x = 3$:

Adding $y - x = 3$ and $x + y = 13$:

$2y = 16 \implies y = 8$

Then $x = 5$.

The original number is $10(5) + 8 = 58$.

Therefore, the two possible numbers are $85$ and $58$.

Given:

Four linear equations are provided:

(i) $y = –3x + 4$

(ii) $2y = 4x + 7$

(iii) $5y = 6x – 10$

(iv) $3y = 6x – 11$

---

To Find:

1. Slope ($m$) and $y$-intercept ($c$) for each equation.

2. Coordinates of the points where the lines cut the $y$-axis.

3. Identify if any of the lines are parallel.

4. Draw the graph of the equations.

---

Solution:

We use the Slope-Intercept form of a linear equation, which is given by:

Where $m$ is the slope and $c$ is the $y$-intercept. The point where the line cuts the $y$-axis is always $(0, c)$.

---

Analysis of Equation (i):

Comparing with $y = mx + c$, we get:

Slope ($m_1$) = $-3$

$y$-intercept ($c_1$) = $4$

Point on $y$-axis = $(0, 4)$

---

Analysis of Equation (ii):

$2y = 4x + 7$

Comparing with $y = mx + c$, we get:

Slope ($m_2$) = $2$

$y$-intercept ($c_2$) = $3.5$

Point on $y$-axis = $(0, 3.5)$

---

Analysis of Equation (iii):

$5y = 6x - 10$

Comparing with $y = mx + c$, we get:

Slope ($m_3$) = $1.2$

$y$-intercept ($c_3$) = $-2$

Point on $y$-axis = $(0, -2)$

---

Analysis of Equation (iv):

$3y = 6x - 11$

Comparing with $y = mx + c$, we get:

Slope ($m_4$) = $2$

$y$-intercept ($c_4$) = $-3.67$ (approx)

Point on $y$-axis = $(0, -3.67)$

---

Table of Values for Graphing:

Equation	Slope ($m$)	$y$-intercept ($c$)	Point on $y$-axis
$y = -3x + 4$	$-3$	$4$	$(0, 4)$
$2y = 4x + 7$	$2$	$3.5$	$(0, 3.5)$
$5y = 6x - 10$	$1.2$	$-2$	$(0, -2)$
$3y = 6x - 11$	$2$	$-3.67$	$(0, -3.67)$

---

Conclusion on Parallel Lines:

Two lines are parallel if and only if their slopes are equal ($m_a = m_b$).

From our calculations:

Slope of line (ii) = $2$

Slope of line (iv) = $2$

Since $m_2 = m_4$, the lines (ii) and (iv) are parallel to each other.

---

Given Equation:

---

Solution:

(i) When $x = 313\text{ K}$:

$y = \frac{9}{5}(313 - 273) + 32$

$y = \frac{9}{5}(40) + 32$

$y = 9 \times 8 + 32$

$y = 72 + 32$

$y = 104\ ^\circ\text{F}$

---

(ii) When $y = 158\ ^\circ\text{F}$:

$158 = \frac{9}{5}(x - 273) + 32$

$158 - 32 = \frac{9}{5}(x - 273)$

$126 = \frac{9}{5}(x - 273)$

$x - 273 = 126 \times \frac{5}{9}$

$x - 273 = 14 \times 5$

$x - 273 = 70$

$x = 70 + 273$

$x = 343\text{ K}$

Given:

Constant Force ($F$) = $3$ units.

Formula: $\text{Work } (w) = \text{Force } (F) \times \text{Distance } (d)$.

---

Solution:

The linear equation representing the situation is:

Let's find some points for the graph:

Distance (d)	Work (w)	Point (d, w)
0	0	(0, 0)
1	3	(1, 3)
2	6	(2, 6)

Work done when $d = 2$ units:

Substituting $d = 2$ in equation (i):

$w = 3 \times 2 = 6 \text{ units}$.

This corresponds to the point $(2, 6)$ on the graph.

Given:

Points on the graph: $(1, 5)$ and $(3, 11)$.

---

Solution:

(i) A linear polynomial is of the form $p(x) = ax + b$.

Substituting $(1, 5)$: $5 = a(1) + b \implies a + b = 5$     ... (i)

Substituting $(3, 11)$: $11 = a(3) + b \implies 3a + b = 11$     ... (ii)

Subtracting (i) from (ii):

$(3a + b) - (a + b) = 11 - 5$

$2a = 6 \implies a = 3$

Substituting $a = 3$ in (i): $3 + b = 5 \implies b = 2$.

The polynomial is $p(x) = 3x + 2$.

---

(ii) Coordinates where it cuts the axes:

For the $y$-axis, set $x = 0$:

$y = 3(0) + 2 = 2$. Point is $(0, 2)$.

For the $x$-axis, set $y = 0$:

$0 = 3x + 2 \implies 3x = -2 \implies x = -\frac{2}{3}$. Point is $(-\frac{2}{3}, 0)$.

---

(iii) Verification:

Plotting the line through $(1, 5)$ and $(3, 11)$ shows that it crosses the vertical axis at exactly $2$ and the horizontal axis at $-\frac{2}{3}$ (approx $-0.67$).

Given:

1. Two linear polynomials $p(x) = ax + b$ and $q(x) = cx + d$.

2. $p(0) = 5$.

3. $p(x) + q(x) = 6x + 4$ for all real $x$.

4. $p(x) - q(x)$ has an $x$-intercept at $(3, 0)$, which means at $x = 3$, $p(3) - q(3) = 0$.

---

To Find:

The explicit expressions for polynomials $p(x)$ and $q(x)$.

---

Solution:

Step 1: Finding $b$ and $d$ using constant terms.

From the definition of $p(x) = ax + b$:

$p(0) = a(0) + b = b$

Now, we use the sum of the polynomials:

$p(x) + q(x) = (ax + b) + (cx + d) = (a + c)x + (b + d)$

Comparing this with the given $6x + 4$, we equate the constant terms:

Substituting $b = 5$ in equation (i):

$5 + d = 4$

$d = 4 - 5$

Therefore, $d = -1$.

---

Step 2: Relating $a$ and $c$ using the $x$-coefficient.

Equating the coefficients of $x$ from the sum $p(x) + q(x) = 6x + 4$:

---

Step 3: Finding another relation for $a$ and $c$ using the $x$-intercept.

The polynomial $p(x) - q(x)$ passes through $(3, 0)$, so $p(3) - q(3) = 0$.

$(a(3) + b) - (c(3) + d) = 0$

$(3a + 5) - (3c - 1) = 0$

$3a - 3c + 5 + 1 = 0$

$3a - 3c + 6 = 0$

Dividing the whole equation by 3:

---

Step 4: Solving for $a$ and $c$.

To find $a$, let's add equation (ii) and equation (iii):

$\begin{array}{cc} & a & + & c & = & 6 \\ + & a & - & c & = & -2 \\ \hline & 2a & & & = & 4 \\ \hline \end{array}$

$2a = 4 \implies a = 2$

Now, substitute $a = 2$ into equation (ii):

$2 + c = 6$

$c = 6 - 2 = 4$

---

Final Polynomials:

By substituting the values of $a, b, c,$ and $d$, we get:

$p(x) = 2x + 5$

$q(x) = 4x - 1$

To Find:

1. The number of matchsticks required for the $4^{th}$ and $5^{th}$ stages.

2. A linear rule (expression) for the $n^{th}$ stage.

3. Matchsticks for the $15^{th}$ stage.

4. Check if $200$ matchsticks can form a complete stage.

---

Solution:

(i) Next Stages and Matchstick Count:

In the given pattern, Stage 1 has $1$ hexagon ($6$ sticks). Every subsequent stage adds $1$ hexagon that shares $1$ side with the previous one, thus adding only $5$ new sticks.

For Stage 4: $16 + 5 = \mathbf{21 \text{ matchsticks}}$.

For Stage 5: $21 + 5 = \mathbf{26 \text{ matchsticks}}$.

---

(ii) Table of Values:

Stage Number ($n$)	1	2	3	4	5	$\dots$	$n$
Number of matchsticks	6	11	16	21	26	$\dots$	$5n + 1$

---

(iii) Rule for the $n^{th}$ stage:

The first stage requires $6$ sticks. Each additional stage adds $5$ sticks. For $n$ stages, the number of additional hexagons is $(n-1)$.

$\text{Total sticks } (S) = 6 + (n - 1)5$

$S = 6 + 5n - 5$

---

(iv) Matchsticks for the $15^{th}$ stage:

Substitute $n = 15$ in the rule derived in (i):

$S = 5(15) + 1$

$S = 75 + 1$

$S = 76 \text{ matchsticks}$

---

(v) Can $200$ matchsticks form a stage?

Let $S = 200$. We need to check if $n$ is a whole number (integer).

$5n + 1 = 200$

$5n = 200 - 1$

$n = \frac{199}{5}$

$n = 39.8$

Justification: Since the stage number $n$ must be a natural number ($1, 2, 3, \dots$), and $39.8$ is not an integer, $200$ matchsticks cannot form a complete stage in this pattern. A stage would require either $196$ sticks ($n=39$) or $201$ sticks ($n=40$).

Given:

1. Polynomial $p(x) = ax + b$ passes through $(2, 3)$ and $(6, 11)$.

2. Polynomial $q(x) = cx + d$ passes through $(4, -1)$.

3. The graph of $q(x)$ is parallel to the graph of $p(x)$.

---

To Find:

Polynomials $p(x)$ and $q(x)$, and their $x$-intercepts.

---

Solution:

Step 1: Find the slope and equation of $p(x)$.

The slope $a$ is given by the formula $a = \frac{y_2 - y_1}{x_2 - x_1}$:

$a = \frac{11 - 3}{6 - 2} = \frac{8}{4}$

Substituting $a = 2$ and the point $(2, 3)$ into $p(x) = ax + b$:

$3 = 2(2) + b$

$3 = 4 + b$

Thus, the polynomial $p(x)$ is $p(x) = 2x - 1$.

---

Step 2: Find the equation of $q(x)$.

Since the graph of $q(x)$ is parallel to $p(x)$, their slopes must be equal.

Substituting $c = 2$ and the point $(4, -1)$ into $q(x) = cx + d$:

$-1 = 2(4) + d$

$-1 = 8 + d$

Thus, the polynomial $q(x)$ is $q(x) = 2x - 9$.

---

Step 3: Find the $x$-intercepts.

To find where the lines meet the $x$-axis, we set $p(x) = 0$ and $q(x) = 0$.

For $p(x)$:

$2x - 1 = 0 \Rightarrow 2x = 1 \Rightarrow x = 0.5$

Point: $(0.5, 0)$.

For $q(x)$:

$2x - 9 = 0 \Rightarrow 2x = 9 \Rightarrow x = 4.5$

Point: $(4.5, 0)$.

To Find:

Identify common characteristics of the family of functions $f(x) = ax + a$ where $a > 0$.

---

Solution:

Let's analyze the given function $f(x) = ax + a$. We can factor out '$a$':

$f(x) = a(x + 1)$

We can check for key points on the graph:

1. The $x$-intercept:

Set $f(x) = 0$:

$a(x + 1) = 0$

Since it is given that $a > 0$, we can divide by $a$:

$x + 1 = 0$

---

Conclusion:

All linear functions of the form $f(x) = ax + a$ share the following features:

1. They all pass through the fixed point $(-1, 0)$ on the $x$-axis.

2. Since $a > 0$, they all have a positive slope, meaning they are all increasing functions.

3. The $y$-intercept of each line is $(0, a)$, which varies depending on the value of $a$, but it will always be on the positive $y$-axis.