Chapter 2: Introduction to Linear Polynomials

Step 1  Identify what one box of each colour contains
Red box → $$4$$ pens     Blue box → $$5$$ pencils

Step 2  Translate the numbers of boxes into algebra
Number of red boxes bought  =  $$x$$
Number of blue boxes bought =  $$y$$

Step 3  Write an expression for the pens
Each red box gives $$4$$ pens, so total pens from red boxes
$$\text{Pens}=4 \times x = 4x$$

Step 4  Write an expression for the pencils
Each blue box gives $$5$$ pencils, so total pencils from blue boxes
$$\text{Pencils}=5 \times y = 5y$$

Step 5  Total writing items without any offer
Adding the two quantities gives
\[4x + 5y\]

Step 6  Account for the special offer
Raju gets $$3$$ extra pens free, so new number of pens
$$4x + 3$$
Total pens + pencils now become
$$(4x + 3) + 5y = 4x + 5y + 3$$

Result
Pens  =  $$4x + 3$$
Pencils =  $$5y$$
Total items =  $$4x + 5y + 3$$

Step 1 : Understand the three different expenses

• Wire fence to be put up along each length of the rectangle.
  Cost = ₹100 per metre.
• Wooden fence to be built along each width of the rectangle.
  Cost = ₹80 per metre.
• Special seeds to be sown over the entire area of the garden.
  Cost = ₹50 per square metre.

The garden has length $$l$$ metres and width $$w$$ metres.

Step 2 : Cost of the wire fence (along the lengths)

There are two opposite sides each of length $$l$$ metres.
Total length to be fenced with wire = $$2l$$ metres.

Cost per metre = ₹100, hence

Step 3 : Cost of the wooden fence (along the widths)

There are two opposite sides each of width $$w$$ metres.
Total length to be fenced with wood = $$2w$$ metres.

Cost per metre = ₹80, hence

Step 4 : Cost of sowing special seeds (over the area)

Area of the rectangular garden = $$l\times w = lw\,\text{square metres}$$.

Cost per square metre = ₹50, hence

Step 5 : Total cost

Add all three costs:

This algebraic expression is a polynomial in the two variables $$l$$ and $$w$$.

First understand what “bending a wire into a rectangle” means.

A rectangle has two equal long sides (length) and two equal short sides (width). The perimeter of a rectangle is therefore

$$\text{Perimeter}=2(\text{length}+\text{width}).$$

The wire is 20 cm long, so after bending we must have

$$2(\text{length}+\text{width}) = 20.$$

Let us denote the length by $$x$$ cm.
Then

$$2(x+\text{width}) = 20 \;\;\Longrightarrow\;\; x+\text{width}=10.$$

Solve this for the width:

$$\text{width}=10-x.$$

Because the wire really does form a rectangle, both $$x$$ and $$10-x$$ must be positive, so $$0<x<10.$$

The area of a rectangle is

$$\text{Area}=\text{length}\times\text{width}=x(10-x).$$

Multiplying out gives a polynomial in one variable:

$$x(10-x)=10x-x^{2}.$$

Thus the area, expressed only in terms of the chosen length $$x,$$ is the quadratic polynomial

\[A(x)=10x-x^{2}.\]

Check with the examples supplied in the question:

• For a 7 cm × 3 cm rectangle: $$A(7)=10(7)-7^{2}=70-49=21 \text{ cm}^{2}.$$
• For a 5.5 cm × 4.5 cm rectangle: $$A(5.5)=10(5.5)-(5.5)^{2}=55-30.25=24.75 \text{ cm}^{2}.$$

Other possibilities (all adding to 10):

Every rectangle that can be made from this 20 cm wire has its area given by the polynomial $$10x-x^{2}$$, where $$x$$ is any number between 0 and 10 (exclusive).

Step 1 : Write the expression for perimeter
For a square, the perimeter $$P$$ is the sum of the lengths of all four equal sides:
$$P = \text{side} + \text{side} + \text{side} + \text{side}$$

Step 2 : Substitute the given side length
If each side measures $$x$$ units, then
$$P = x + x + x + x$$

Step 3 : Add the like terms
Because all four terms are like terms (each is simply $$x$$), add their coefficients:
$$P = (1 + 1 + 1 + 1)x = 4x$$

Step 4 : Identify the type of polynomial
A polynomial is classified by the highest (greatest) power of its variable.
• In $$4x$$ the exponent of $$x$$ is $$1$$.
• A polynomial whose highest power of the variable is $$1$$ is called a linear polynomial.
Therefore, $$4x$$ is a linear polynomial in the variable $$x$$.

Conclusion
The perimeter of the square is $$4x$$, and since its highest power of $$x$$ is $$1$$, it is indeed a linear polynomial.

Let $$m$$ denote the number of chess matches a member plays.

1. Fixed charge
   Joining fee = ₹200  =  $$200$$.
2. Variable charge
   Each match costs ₹50.
   For $$m$$ matches this is ₹50 × $$m$$ = $$50m$$.
3. Total amount
   Total amount to be paid
   \[A(m)=200+50m\]

The expression $$A(m)=200+50m$$ is a polynomial in one variable $$m$$. Its highest power of $$m$$ is $$1$$, so it is a linear polynomial.

Verification with the table

Thus the polynomial $$A(m)=200+50m$$ correctly gives the amount a player has to pay for any whole number of matches $$m$$.

Step 1 — Introduce the unknown.
Let the smaller number be $$x$$. Then the larger number is $$x+10$$ (because it is 10 more).

Step 2 — Form the equation from the given sum.
The sum of the two numbers is $$64$$:
$$x + (x+10) = 64$$

Step 3 — Solve for $$x$$.
Combine like terms: $$2x + 10 = 64$$
Subtract $$10$$ from both sides: $$2x = 64 - 10 = 54$$
Divide by $$2$$: $$x = \frac{54}{2} = 27$$

Step 4 — Determine the second number.
Second number $$= x + 10 = 27 + 10 = 37$$.

Step 5 — Verification.
Sum: $$27 + 37 = 64$$ (✓)
Difference: $$37 - 27 = 10$$ (✓)

Thus, the required numbers are $$27$$ and $$37$$.

Step 1 – Set up the relation
Bela starts with $$₹100$$ and spends $$₹5$$ each day.

Let $$n$$ be the number of days after which she has $$₹40$$ left.

Pocket money left after $$n$$ days
$$\text{Money left} = 100 - 5n$$

We need this to equal $$₹40$$, so

$$100 - 5n = 40$$

Step 2 – Solve the equation

• Subtract $$100$$ from both sides: $$100 - 5n - 100 = 40 - 100$$
• Simplify: $$-5n = -60$$
• Divide by $$-5$$: $$n = \dfrac{-60}{-5}$$
• Compute: $$n = 12$$

Conclusion
Bela will be left with $$₹40$$ after 12 days.

Step 1 : Understand the tariff

• The first 2 km cost a fixed amount of ₹25.
• For every kilometre beyond those 2 km, the fare increases by ₹15 per km.

Step 2 : Set up an algebraic expression

Let the total distance travelled be $$x$$ km  (with $$x \ge 2$$, because the rule changes only after 2 km).

Additional kilometres beyond the first 2 km are therefore $$x-2$$ km.

Additional cost  = ₹15 per km × number of additional km
$$15(x-2)$$

Total fare $$F(x)$$ is then

$$F(x)=25+15(x-2)$$

Step 3 : Simplify the expression

$$\begin{aligned}F(x)&=25+15(x-2)\\&=25+15x-30\\&=15x-5\end{aligned}$$

So, for any trip longer than 2 km, the polynomial giving the fare is $$F(x)=15x-5$$ (with $$x\ge2$$).

Step 4 : Substitute $$x=10$$ km

For a 10 km journey:

$$\begin{aligned}F(10)&=15(10)-5\\&=150-5\\&=145\end{aligned}$$

Step 5 : State the answer

The auto-rickshaw fare for 10 km is ₹145.

Given linear function: $$C(d)=100+60d$$, where

• $$C(d)$$ is the total cost (in rupees).
• $$d$$ is the distance travelled (in kilometres).

To find how the cost changes every kilometre from $$d=0$$ km to $$d=10$$ km, substitute each integer value of $$d$$ into the formula.

1. For $$d = 0$$: $$C(0)=100+60\times 0=100$$
2. For $$d = 1$$: $$C(1)=100+60\times 1=160$$
3. For $$d = 2$$: $$C(2)=100+60\times 2=220$$
4. For $$d = 3$$: $$C(3)=100+60\times 3=280$$
5. For $$d = 4$$: $$C(4)=100+60\times 4=340$$
6. For $$d = 5$$: $$C(5)=100+60\times 5=400$$
7. For $$d = 6$$: $$C(6)=100+60\times 6=460$$
8. For $$d = 7$$: $$C(7)=100+60\times 7=520$$
9. For $$d = 8$$: $$C(8)=100+60\times 8=580$$
10. For $$d = 9$$: $$C(9)=100+60\times 9=640$$
11. For $$d = 10$$: $$C(10)=100+60\times 10=700$$

Tabulating the results:

The table clearly shows that the cost increases by a constant amount of ₹60 for every additional kilometre, which is consistent with the coefficient of $$d$$ in the linear function $$C(d)=100+60d$$.

Step 1 : Identify the type of algebraic expression
The rule for the height is
$$h(t)=3-0.5t$$
This expression is a linear polynomial in the variable $$t$$ because the highest power of $$t$$ is 1.

Step 2 : Compare with the standard linear form
A linear polynomial in one variable is generally written as
$$y=mt+c$$,
where $$m$$ is the slope (rate of change) and $$c$$ is the y-intercept (initial value).
On comparing,
$$m=-0.5$$ and $$c=3$$.

Interpretation

• The initial height of water at the start of summer ($$t=0$$) is
  $$h(0)=3-0.5\times0=3\;\text{m}$$.
• The coefficient $$m=-0.5$$ tells us the level falls by $$0.5\,\text{m}$$ every month.

Step 3 : Find when the tank becomes empty
The tank is empty when $$h(t)=0$$.
$$3-0.5t=0$$
Add $$0.5t$$ to both sides:
$$3=0.5t$$
Divide by $$0.5$$:
$$t=\dfrac{3}{0.5}=6$$

Hence the tank will be empty 6 months after the start of summer.

Summary
\[\begin{aligned} \text{Initial height}&=3\,\text{m}\\[2pt] \text{Rate of fall}&=0.5\,\text{m per month}\\[2pt] \text{Tank empty after}&=6\,\text{months} \end{aligned}\]

The tariff is stated to follow a linear relation of the form $$y = ax + b$$, where

• $$x$$ = data used (in GB)
• $$y$$ = monthly bill (in rupees)

Two observations are supplied:

• When $$x = 10$$, $$y = 350$$.
• When $$x = 20$$, $$y = 550$$.

Substituting these ordered pairs into $$y = ax + b$$ gives two equations.

So we have the system

$$10a + b = 350 \qquad\text{(1)}$$
$$20a + b = 550 \qquad\text{(2)}$$

Step 1: Eliminate $$b$$ to find $$a$$.

Subtract (1) from (2):

$$\bigl(20a + b\bigr) - \bigl(10a + b\bigr) = 550 - 350$$

$$20a - 10a = 200$$

$$10a = 200$$

$$a = \frac{200}{10} = 20$$

Step 2: Substitute $$a = 20$$ into either equation to find $$b$$.

Using (1):

$$10(20) + b = 350$$

$$200 + b = 350$$

$$b = 350 - 200 = 150$$

Hence,

The required linear relation becomes

Thus the company charges a fixed fee of ₹150 per month plus ₹20 per GB of data.

We are given the five points

• $$A(-1,-3)$$
• $$B(0,0)$$
• $$C(1,3)$$
• $$D(3,9)$$
• $$E(4,12)$$

After plotting them on a graph sheet and joining the extreme points $$A$$ and $$E$$ with a ruler, we notice that all five points lie on the same straight line. A straight-line graph is always described by a linear equation of the form $$y = mx + c$$, where

• $$m$$ is the slope (gradient) of the line, and
• $$c$$ is the $$y$$-intercept.

1. Looking directly at the coordinates

Compare the $$x$$ and $$y$$ values of every point:

Except for the origin (where the ratio is not defined), every point gives the same constant value $$3$$ for $$\dfrac{y}{x}$$. This constant ratio tells us that $$y$$ is always three times $$x$$, i.e.

\[y = 3x\]

2. Confirming by slope–intercept form

If we pick any two of the points, say $$A(-1,-3)$$ and $$E(4,12)$$, the slope is

$$m = \dfrac{12 - (-3)}{4 - (-1)} = \dfrac{15}{5} = 3$$.

Using the point–slope form through $$A$$:

$$y - (-3) = 3\,(x - (-1)) \;\Longrightarrow\; y + 3 = 3(x + 1)$$

Simplifying gives $$y = 3x$$, which is identical to the result obtained above.

Conclusion

The single linear equation that passes through all five plotted points is therefore

\[y = 3x\]

Given points : $$(-3,6),\;(-2,4),\;(0,0),\;(1,-2),\;(2,-4),\;(3,-6)$$

Step 1 : Plot and join
Plot the six ordered pairs on graph paper. Join the extreme points $$(-3,6)$$ and $$(3,-6)$$ with a ruler; every other point falls exactly on this same straight line, confirming that all six points are collinear.

Step 2 : Detect the numerical pattern

For every point, $$y=-2x$$ holds true, suggesting that this is the required linear relation.

Step 3 : Verification by slope–intercept form

Take the two extreme points to compute the slope:

Using point $$(0,0)$$, the line in slope–intercept form is

Thus, the equation satisfied by every plotted point—and hence by the whole line—is

For any square the perimeter (P) is four times the length of one side (s): $$P = 4s$$.

Now increase every side length by $$0.5\,\text{cm}$$. The new side length is $$s+0.5$$ and its perimeter becomes

$$P' = 4(s+0.5) = 4s + 2.$$

Thus each perimeter rises by exactly $$2\,\text{cm}$$. Calculated one–by–one:

Observation: Adding $$0.5\,\text{cm}$$ to every side always adds $$2\,\text{cm}$$ to the perimeter of the square.

Let $$x$$ be the number of matches the player played.

From the earlier part of the question, the polynomial giving the total amount (in rupees) that has to be paid after playing $$x$$ matches is

\[A(x) = 500 + 50x\]

where
• ₹500 is the fixed (one-time) membership fee, and
• ₹50 is the charge for each match played.

The player actually paid ₹750. Hence

$$A(x) = 750$$

Substituting this in the above expression:

$$500 + 50x = 750$$

Subtract 500 from both sides:

$$50x = 750 - 500$$
$$50x = 250$$

Divide both sides by 50 to isolate $$x$$:

$$x = \frac{250}{50}$$
$$x = 5$$

Therefore, the player played 5 matches.

Step 1 : Understanding the “input–output” idea

Think of the algebraic expression (here, the area of the rectangle) $$10x - x^2$$ as a machine:

• We input a number for the variable $$x$$ (for example a possible breadth of the rectangle).
• The “machine” performs the operations “multiply by 10” and “subtract the square” in the exact order shown.
• The number that finally comes out is the output, which in this context represents the area (in $$ ext{cm}^2$$).

This is exactly what mathematicians call a function: every admissible input $$x$$ produces a unique output $$10x - x^2$$.

Step 2 : Substituting $$x = 6$$ cm

Insert the given value into the expression, showing each operation separately:

First, write the expression with brackets around the substituted value so there is no confusion about order:

$$10x - x^2 \;\longrightarrow\; 10(6) - (6)^2$$

Step 3 : Perform the multiplication

$$10(6) = 60$$, so the expression becomes

$$60 - (6)^2$$

Step 4 : Find the square of the input

$$6^2 = 6 \times 6 = 36$$. Thus

$$60 - 36$$

Step 5 : Complete the subtraction

$$60 - 36 = 24$$

Step 6 : Attach the correct unit

The original expression represents an area, so the unit is square centimetres (cm²).

\[ \boxed{24\;\text{cm}^2} \]

Thus, when the “function machine” receives the input $$x = 6$$ cm, the output (area of the rectangle) is $$24\;\text{cm}^2$$.

Step 1 : Observe the first three stages shown in Fig. 2.4
From the textbook figure we can read the number of unit (small) squares in each of the first three stages:

• Stage 1 → 1 square
• Stage 2 → 4 squares
• Stage 3 → 9 squares

The jump from 1 → 4 → 9 suggests that the figure in Stage n is an n × n big square made up of unit squares. Hence the count in Stage n should be $$n^{2}$$.

Step 2 : Use the rule $$n^{2}$$ to predict the next three stages

Therefore the next three stages contain 16, 25 and 36 squares respectively.

Step 3 : Write the full sequence up to Stage 7
Listing the values $$n^{2}$$ for $$n = 1,2,3,4,5,6,7$$ gives:

Stage 1 : 1     Stage 2 : 4     Stage 3 : 9     Stage 4 : 16     Stage 5 : 25     Stage 6 : 36     Stage 7 : 49

Hence the required sequence is
$$1,\;4,\;9,\;16,\;25,\;36,\;49.$$

The pattern described in the textbook has its number of tiles at the n-th stage given by the polynomial (or algebraic expression)

$$2n - 1.$$

This is the general term of the sequence of consecutive odd numbers: 1, 3, 5, 7, … . We shall use it in two different ways:

1. Substitution  (to find how many tiles are in a given stage).
2. Equation solving  (to find which stage has a given number of tiles).

1. Tiles in the 15^th and 26^th stages

(i) For the 15^th stage, put $$n = 15$$ in $$2n - 1$$:

$$2(15) - 1 = 30 - 1 = 29.$$

So, the 15^th stage has 29 tiles.

(ii) For the 26^th stage, put $$n = 26$$:

$$2(26) - 1 = 52 - 1 = 51.$$

Thus, the 26^th stage has 51 tiles.

2. Stage that contains 21 tiles

Let the required stage number be $$n$$. Equate the formula to 21:

$$2n - 1 = 21.$$

Add 1 to both sides:

$$2n = 22.$$

Divide by 2:

$$n = 11.$$

Therefore, the 11^th stage contains 21 tiles.

3. Stage that contains 47 tiles

Again, let the stage number be $$n$$ and equate:

$$2n - 1 = 47.$$

Add 1:

$$2n = 48.$$

Divide by 2:

$$n = 24.$$

Hence, the 24^th stage contains 47 tiles.

All results collected:

Let the amount  (in rupees)  left with the person after n days be represented by the polynomial

$$A(n)=400-20n$$

(That is, the person started with Rs 400 and spends Rs 20 every day.)

1. Amount left on the 15th day

Substitute n = 15 in $$A(n)$$ :

$$A(15)=400-20\times15$$

$$=400-300$$

$$=100$$

Hence, Rs 100 will be left on the 15^th day.

2. Number of days required to spend the entire amount

The entire amount is spent when the amount left becomes zero, i.e. when $$A(n)=0$$.

Set the polynomial equal to 0 and solve for n:

$$400-20n=0$$

$$20n=400$$

$$n=\frac{400}{20}=20$$

Therefore, it will take 20 days for the person to spend the whole amount.

Let $$x$$ denote the total distance (in km) covered by the taxi.

The fare pattern given in the question is :

• Fare for the first kilometre  =  ₹10
• Fare for every additional kilometre  =  ₹8

For a ride of $$x\;(x\ge 1)$$ kilometres, the number of additional kilometres is $$x-1$$. Hence the total fare, in rupees, is

$$F(x)=10+8(x-1).$$

Simplifying,

\[F(x)=10+8x-8=8x+2.\]

We are told that the fare is ₹130, so

$$8x+2 = 130.$$

Subtract 2 from both sides :

$$8x = 128.$$

Divide by 8 :

$$x = 16.$$

Therefore, the fare becomes ₹130 when the taxi is hired for 16 kilometres.

Let the total fare (in rupees) for travelling $$x$$ km be denoted by the polynomial $$C(x)$$ that was obtained in the earlier part of the question:

• a fixed charge (flag-down fare) of ₹50 is paid the moment one hires the taxi, and
• thereafter the charge is ₹10 per kilometre actually travelled.

Therefore

$$C(x)=50+10x\;.$$

(i) Cost of travelling 15 km

Substitute $$x=15$$ in $$C(x)$$:

$$C(15)=50+10\times15$$

$$=50+150$$

$$=200\;.$$

Hence the cost of a 15 km journey is ₹200.

(ii) Distance covered when the fare is ₹700

Set $$C(x)=700$$ and solve for $$x$$:

$$50+10x=700$$

$$10x=700-50$$

$$10x=650$$

$$x=\dfrac{650}{10}$$

$$x=65\;.$$

Therefore one can travel 65 km for a fare of ₹700.

The height of the water (in metres) after t months is represented by the polynomial

$$h(t) = 4t^{2} - 5t + 19$$

We have to find the height at the end of 5 months, i.e. for $$t = 5$$.

Step 1 – Substitute $$t = 5$$ in the polynomial.

$$h(5) = 4(5)^{2} - 5(5) + 19$$

Step 2 – Calculate the individual terms.

• $$(5)^{2} = 25$$
• So, $$4(5)^{2} = 4 \times 25 = 100$$
• and $$-5(5) = -5 \times 5 = -25$$

Step 3 – Add the three terms carefully.

$$h(5) = 100 - 25 + 19$$

First add the positive numbers: $$100 + 19 = 119$$

Now subtract 25 from 119:

$$119 - 25 = 94$$

Hence, the height of the water after 5 months is

We compare the given equation with the standard form of a linear equation in one variable:
\[ y = mx + c \]

Here

• $$m$$ is called the slope (or gradient) of the line. It tells us how much $$y$$ changes when $$x$$ increases by one unit.
• $$c$$ is the y‑intercept. It gives the value of $$y$$ when $$x = 0$$; in other words, the point where the line cuts the $$y$$-axis.

In the given equation

$$y = 20x + 150$$

we can immediately read

• $$m = 20$$, so for every increase of 1 in $$x$$, the value of $$y$$ rises by $$20\,(=m\times1)$$ units. Thus 20 represents the rate of change of $$y$$ with respect to $$x$$.
• $$c = 150$$, so when $$x = 0$$ we have
  \[ y = 20\times0 + 150 = 150. \]
  Hence 150 is the y-intercept; it is the value of $$y$$ when $$x$$ is zero.

Therefore:

1. 20 is the coefficient of $$x$$ (the slope), indicating the unit change in $$y$$ per unit change in $$x$$.
2. 150 is the constant term (the y-intercept), giving the initial value of $$y$$ when $$x = 0$$.

We are given the straight-line equation

$$y = 2x + 1$$

To find the missing entries in the table we substitute each given $$x$$–value into the formula and simplify.

1. For $$x = 1$$ (already filled):

   $$y = 2(1) + 1 = 2 + 1 = 3$$

2. For $$x = 2$$:

   $$y = 2(2) + 1 = 4 + 1 = 5$$

3. For $$x = 5$$:

   $$y = 2(5) + 1 = 10 + 1 = 11$$

4. For $$x = 7$$:

   $$y = 2(7) + 1 = 14 + 1 = 15$$

5. For $$x = 9$$:

   $$y = 2(9) + 1 = 18 + 1 = 19$$

6. For $$x = 12$$:

   $$y = 2(12) + 1 = 24 + 1 = 25$$

7. For $$x = 20$$:

   $$y = 2(20) + 1 = 40 + 1 = 41$$

Thus the completed table is:

Step 1 – Identify the form of each equation

Both equations are in the slope–intercept form $$y = mx + c$$ where $$m$$ is the slope and $$c$$ is the y-intercept.

Step 2 – Plot a few points for each line

Step 3 – Draw the graphs

• On graph paper mark the common point $$(0,1)$$ on the y-axis.
• Through the points $$(1,4)$$ and $$( -1,-2)$$ draw a straight line; this is the graph of $$y = 3x + 1$$. It rises to the right (upward slant) making an acute angle with the positive x-axis.
• Through the points $$(1,-2)$$ and $$( -1,4)$$ draw a straight line; this is the graph of $$y = -3x + 1$$. It falls to the right (downward slant) making an obtuse angle with the positive x-axis.

Diagram description: Draw a Cartesian plane. Mark the common intercept point (0, 1). Draw one line rising rightwards through (1, 4) and falling leftwards through (-1, -2). Draw the other line falling rightwards through (1, -2) and rising leftwards through (-1, 4). The two lines cross at (0, 1) and form a symmetrical ‘X’ shape.

Step 4 – Compare the two graphs

• Slope: $$y = 3x + 1$$ has positive slope $$3$$; $$y = -3x + 1$$ has negative slope $$-3$$.
• Direction: The first line slants upward from left to right; the second slants downward from left to right.
• Common y-intercept: Both lines cut the y-axis at the same point $$(0,1)$$.
• Symmetry: Because their slopes are equal in magnitude but opposite in sign and their intercepts are equal, the two graphs are mirror images of each other in the y-axis.

Conclusion

The graph of $$y = 3x + 1$$ is an upward-sloping straight line through $$(0,1)$$; the graph of $$y = -3x + 1$$ is a downward-sloping straight line through the same point. Thus, they differ only in the sign of their slope, giving opposite inclinations while sharing the same y-intercept.

Step 1 : Choose convenient points for each equation

For any straight-line equation of the form $$y=ax+b$$ it is enough to locate two points.

Step 2 : Plot and observe

• All three graphs rise by the same amount when $$x$$ increases by 1: the change in $$y$$ is always $$a=2$$. This identical “steepness” (slope) means the lines never meet; they are parallel.
• The only difference is their intercept on the $$y$$-axis: $$b=-1,\;1,\;5$$. Larger $$b$$ moves the whole line upward without altering its tilt.

Step 3 : General conclusion

If in $$y=ax+b$$ the coefficient $$a$$ (slope) is kept fixed while $$b$$ varies, every new value of $$b$$ simply shifts the graph up or down. Because the slopes are identical, the resulting lines are always parallel to one another.

Thus the effect of changing the constant term $$b$$ is a vertical translation; it does not change the direction of the line.

Step 1 – Recall the meaning of “degree of a polynomial”
The degree of a polynomial in one variable (say, $$x$$) is the highest power (exponent) of $$x$$ that actually occurs in the expression with a non-zero coefficient.

For example, in the polynomial $$5x^{4} - 2x^{2} + 7$$ the highest power of $$x$$ present is $$4$$, so its degree is $$4$$.

Step 2 – Construct a polynomial of degree 1
The general form of a degree 1 (linear) polynomial in one variable is
\[a_{1}x + a_{0}\]
where $$a_{1} \neq 0$$.
Choose convenient non-zero coefficients, e.g. $$a_{1}=2$$ and $$a_{0}=1$$. Thus one suitable example is
\[2x + 1\]

Step 3 – Construct a polynomial of degree 2
The general form of a degree 2 (quadratic) polynomial is
\[a_{2}x^{2} + a_{1}x + a_{0}\]
with $$a_{2} \neq 0$$.
Select, say, $$a_{2}=3,\; a_{1}=-5,\; a_{0}=4$$. Hence an example is
\[3x^{2} - 5x + 4\]

Step 4 – Construct a polynomial of degree 3
The general form of a degree 3 (cubic) polynomial is
\[a_{3}x^{3} + a_{2}x^{2} + a_{1}x + a_{0}\]
with $$a_{3} \neq 0$$.
Choose $$a_{3}=1,\; a_{2}=-2,\; a_{1}=1,\; a_{0}=-7$$. Therefore an example is
\[x^{3} - 2x^{2} + x - 7\]

Conclusion
Three polynomials that meet the required degrees are:

• Degree 1: $$2x + 1$$
• Degree 2: $$3x^{2} - 5x + 4$$
• Degree 3: $$x^{3} - 2x^{2} + x - 7$$

Step 1 – Write the polynomial.

We are given the polynomial

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

Step 2 – Locate the x² term and read its coefficient.

The term that contains $$x^2$$ is $$6x^2$$. The numerical part in front of $$x^2$$ is 6.

Therefore, the coefficient of $$x^2$$ is 6.

Step 3 – Locate the x³ term and read its coefficient.

The term that contains $$x^3$$ is $$-3x^3$$. The numerical part in front of $$x^3$$ is $$-3$$.

Therefore, the coefficient of $$x^3$$ is $$-3$$.

Step 4 – State the final answer.

\[ \text{Coefficient of }x^2 = 6, \qquad \text{Coefficient of }x^3 = -3 \]

Objective : Find the coefficient of $$z$$ (that is, the term with exponent 1) in the polynomial $$4z^3 + 5z^2 - 11$$.

Step 1 – List the terms and their exponents

• $$4z^3$$ has exponent $$3$$.
• $$5z^2$$ has exponent $$2$$.
• $$-11$$ can be written as $$-11z^0$$, so its exponent is $$0$$.

Step 2 – Look for the $$z^1$$ term

No term with exponent $$1$$ (i.e. $$z^1$$) appears in the expansion.

Step 3 – State the coefficient

When a power of the variable is missing, its coefficient is taken to be $$0$$. Hence the coefficient of $$z$$ is $$0$$.

The given polynomial is $$9x^3 + 5x^2 - 8x - 10$$.

A constant term is the term in a polynomial that is independent of the variable; in other words, it is the term for which the power of $$x$$ is zero, i.e. $$x^0 = 1$$.

Let us list every term showing its power of $$x$$ explicitly:

• $$9x^3$$ has power 3
• $$5x^2$$ has power 2
• $$-8x$$ has power 1
• $$-10$$ can be written as $$-10x^0$$, so its power is 0

Because the power is zero in the last term, that last term is the required constant term.

Hence, the constant term of the polynomial $$9x^3 + 5x^2 - 8x - 10$$ is $$-10$$.

Given information

• Salil's mother is three times as old as Salil now.
• Five years later, the sum of their ages will be 70 years.

Step 1 : Choose the variable

Let Salil's present age be $$x$$ years.

Step 2 : Express the mother's present age

Salil's mother's present age = three times $$x$$ = $$3x$$ years.

Step 3 : Write their ages after 5 years

• Salil's age after 5 years: $$x + 5$$ years.
• Mother's age after 5 years: $$3x + 5$$ years.

Step 4 : Translate the condition into an equation

Five years later, the sum of their ages is 70 years:

$$ (x + 5) + (3x + 5) = 70 $$

Step 5 : Solve the equation

Combine like terms:

$$ 4x + 10 = 70 $$

Subtract 10 from both sides:

$$ 4x = 60 $$

Divide by 4:

So, Salil's present age is 15 years.

Step 6 : Find the mother's present age

Mother's age = $$3x = 3\times15 = 45$$ years.

Step 7 : Check (optional)

After 5 years:

• Salil: $$15 + 5 = 20$$ years
• Mother: $$45 + 5 = 50$$ years

Sum: $$20 + 50 = 70$$ years ✓

Therefore, the solution is consistent.

Conclusion

Salil is 15 years old now, and his mother is 45 years old.

Let the required positive integers be $$a$$ and $$b$$ with $$a<b$$.

The ratio is given to be $$a:b = 2:5$$.
Therefore we can write

• $$a = 2k$$
• $$b = 5k$$

for some positive number $$k$$ (the common multiplier).

Their difference is given to be $$63$$, so

$$b - a = 63$$

Substituting $$a = 2k$$ and $$b = 5k$$, we get

$$5k - 2k = 63$$

$$3k = 63$$

Substitute $$k=21$$ back to find the integers:

• Smaller integer $$= 2k = 2\times21 = 42$$
• Larger integer $$= 5k = 5\times21 = 105$$

Verification

• Difference: $$105 - 42 = 63$$ ✅
• Ratio: $$\dfrac{42}{105} = \dfrac{2}{5}$$ ✅

Both conditions of the problem are satisfied, hence the two integers are $$42$$ and $$105$$.

Step 1 — Introduce a variable
Let the number of five-rupee coins be $$x$$.
According to the question, the number of two-rupee coins is 3 times this amount, so it is $$3x$$.

Step 2 — Translate the money statement into an equation
Money coming from the five-rupee coins = number of coins $$\times$$ value per coin
$$5 \text{ rupees} \times x = 5x \text{ rupees}$$
Money coming from the two-rupee coins = $$2 \text{ rupees} \times 3x = 6x \text{ rupees}$$
The total money Ruby has is given to be ₹88, therefore
$$5x + 6x = 88.$$

Step 3 — Solve the linear equation
Combine like terms:
$$11x = 88$$
Divide both sides by 11:
$$x = 8.$$

Step 4 — Interpret the result
Number of five-rupee coins = $$x = 8$$.
Number of two-rupee coins = $$3x = 3 \times 8 = 24$$.

Step 5 — Verification
Total money = money from five-rupee coins + money from two-rupee coins
$$5 \times 8 + 2 \times 24 = 40 + 48 = 88 \text{ rupees},$$ which matches the given total. Hence our solution is correct.

Step 1 · Introduce a variable
Let the length of the shorter piece be $$x\text{ feet}$$.

Step 2 · Translate the condition into an algebraic expression
The longer piece is four times the shorter piece, so its length is $$4x\text{ feet}$$.

Step 3 · Form the equation
The two pieces together form the whole fence of $$300\text{ feet}$$: $$x + 4x = 300$$

Step 4 · Simplify the left-hand side
$$x + 4x = 5x$$, so the equation becomes $$5x = 300$$.

Step 5 · Solve for $$x$$
Divide both sides by $$5$$: $$x = \dfrac{300}{5} = 60$$.

Step 6 · Find the length of the longer piece
Longer piece = $$4x = 4(60) = 240\text{ feet}$$.

Step 7 · Verification
Add the two lengths: $$60 + 240 = 300$$, which matches the original fence length. Hence the solution is correct.

Let the width of the rectangle be denoted by the variable $$x$$ centimetres (cm).

According to the statement of the problem, the length is three centimetres more than twice the width. Hence

$$\text{length}=2x+3\;(\text{cm}).$$

The perimeter $$P$$ of a rectangle having length $$l$$ and width $$w$$ is given by the formula

$$P = 2(l + w).$$

Here the perimeter is given as 24 cm, so we substitute $$l = 2x+3$$ and $$w = x$$ into the formula:

$$2\bigl((2x+3)+x\bigr)=24.$$

Now simplify step by step:

• Combine the like terms inside the parentheses:
  $$2\bigl(3x+3\bigr)=24.$$
• Distribute the 2:
  $$6x+6=24.$$
• Isolate the variable term by subtracting 6 from both sides:
  $$6x=18.$$
• Divide both sides by 6 to solve for $$x$$:
  $$x=3.$$

Thus the width of the rectangle is $$3\text{ cm}.$$

Substitute $$x=3$$ back into the expression for the length:

$$\text{length}=2(3)+3=6+3=9\text{ cm}.$$

Hence the rectangle has

• width = $$3\text{ cm}$$,
• length = $$9\text{ cm}$$.

Verification

Compute the perimeter using these dimensions:

$$2(9+3)=2(12)=24\text{ cm},$$

which matches the given perimeter, confirming that the solution is correct.

Step 1 : Understand the situation
Initially the student has ₹500 in her savings account.
Every month she receives an additional pocket money of ₹150, which is deposited in the same account.

Step 2 : Work out the first few months

From the second month onwards the amounts are:

• 2^nd month → ₹800
• 3^rd month → ₹950
• 4^th month → ₹1100
• and so on.

Step 3 : Derive the general (linear) expression

Let $$n$$ denote the number of months completed.
Each month contributes an extra ₹150, so after $$n$$ months the total addition is $$150n$$.

Therefore, the amount in the account at the end of the $$n^{\text{th}}$$ month is

\[ 500 + 150n \]

This is a linear expression in $$n$$ because it can be written as $$150n + 500$$, whose highest power of $$n$$ is 1.

Answer to the questions

1. Amounts from the second month onwards: ₹800, ₹950, ₹1100, … (increasing each month by ₹150).
2. Linear expression for the amount after the $$n^{\text{th}}$$ month: $$150n + 500$$.

Step 1 • Understand the situation
At the beginning (time 0 h) the rally has 120 members. Every hour, exactly 9 members leave.

Step 2 • Work out the first few cases

• After 1 hour: there are $$120-9$$ members → $$111$$ members.
• After 2 hours: another 9 leave, so $$120-2\times9$$ members → $$102$$ members.
• After 3 hours: $$120-3\times9$$ members → $$93$$ members.
• The pattern continues in the same way.

Step 3 • Look for the pattern
After each hour we subtract one more group of 9. Hence, after $$n$$ hours we must subtract $$n$$ groups of 9, i.e. $$9n$$ members in total.

Step 4 • Write the general (linear) expression

Number of rally members after $$n$$ hours \[120-9n\] This is a linear expression in $$n$$ (degree 1 polynomial).

Step 5 • State the final answer

The numbers of members remaining are 111, 102, 93, … and in general $$120-9n$$ members remain after the $$n^{\text{th}}$$ hour (as long as $$n\le13$$ so the number does not become negative).

Given information

• Length of the rectangle (fixed) : $$L = 13\,\text{cm}$$.
• Breadth (variable) : call it $$b\,\text{cm}$$.

The area $$A$$ of a rectangle equals the product of its length and breadth:

$$A = L \times b$$.

Substituting the given length $$L = 13\,\text{cm}$$, we obtain the general (linear) expression for the area:

$$A = 13 b$$.

This single formula lets us calculate the area for any specified breadth. We now evaluate it for the three required breadths.

(i) Breadth = 12 cm

$$A = 13 \times 12 = 156$$

Therefore, $$A = 156\,\text{cm}^2$$.

(ii) Breadth = 10 cm

$$A = 13 \times 10 = 130$$

Therefore, $$A = 130\,\text{cm}^2$$.

(iii) Breadth = 8 cm

$$A = 13 \times 8 = 104$$

Therefore, $$A = 104\,\text{cm}^2$$.

Linear pattern for the area

The area as a function of breadth is the linear expression

$$A(b) = 13 b \;\text{cm}^2$$.

Because the coefficient of $$b$$ is $$13$$, every $$1\,\text{cm}$$ increase in breadth increases the area by $$13\,\text{cm}^2$$.

For the three breadths given in this question, $$b = 12,\,10,\,8\,\text{cm}$$, the breadth decreases by $$2\,\text{cm}$$ at each step, so the corresponding areas decrease by $$13 \times 2 = 26\,\text{cm}^2$$ each time, giving the pattern

$$156,\;130,\;104,\ldots\;\text{cm}^2$$ (consecutive terms differ by $$26\,\text{cm}^2$$).

Equivalently, if the breadth itself stepped down by $$1\,\text{cm}$$ each time (i.e. $$b = 12, 11, 10, 9, 8,\ldots$$), the areas would be $$156, 143, 130, 117, 104,\ldots\,\text{cm}^2$$ (consecutive terms differing by $$13\,\text{cm}^2$$), in line with the linear rule $$A(b)=13b$$.

Given data

• Length of the rectangular box, $$l = 7\text{ cm}$$
• Breadth of the rectangular box, $$b = 11\text{ cm}$$

Step 1 · Write the general formula for volume

The volume $$V$$ of a rectangular (cuboid-shaped) box is obtained by multiplying its length, breadth and height:

$$V = l \times b \times h$$

Step 2 · Find the constant product $$l\,b$$

Because the length and breadth are fixed, first calculate their product:

$$l\,b = 7 \times 11 = 77$$

Hence

$$V = 77 \times h$$

This already shows that the volume varies linearly with the height.

Step 3 · Compute the volume for each required height

Step 4 · Describe the linear pattern

• The first term (when $$h = 5$$) is $$385\text{ cm}^3$$.
• The common difference between consecutive volumes is $$77\text{ cm}^3$$, because each time the height increases by 1 cm, the volume increases by $$77\text{ cm}^3$$.

Hence the linear pattern (rule) connecting height $$h$$ (in cm) and volume $$V$$ (in cm³) is

\[ V = 77h \]

or stated verbally: “Multiply the height by 77 to get the volume.”

Let $$n$$ be the number of days that have passed since Sarita started reading.

Step 1 – Pages read in $$n$$ days
She reads 20 pages every day, so in $$n$$ days she will read

$$\text{pages read}=20 n$$

Step 2 – Pages left after $$n$$ days
Total pages in the book = 500, therefore

$$\text{pages left}=500-20 n$$

This expression, $$500-20 n$$, is a linear polynomial in $$n$$ because its highest power of $$n$$ is 1. Hence it represents the required linear pattern:

Step 3 – Pages left after 15 days

Substitute $$n = 15$$ into the linear expression:

$$P(15)=500-20\times 15$$
$$\phantom{P(15)}=500-300$$
$$\phantom{P(15)}=200$$

Conclusion
After 15 days, Sarita will still have 200 pages left to read. The linear pattern for the pages remaining each day is $$P(n)=500-20 n$$.

The relationship between the monthly bill $$y$$ (in rupees) and the number of modules accessed $$x$$ is given to be linear:

$$y = ax + b$$     (1)

The two observations supplied are:

• When $$x = 10$$, the bill is $$y = 400$$.
• When $$x = 14$$, the bill is $$y = 500$$.

We substitute each pair $$(x, y)$$ into equation (1) to obtain two simultaneous linear equations in the unknowns $$a$$ and $$b$$.

First substitution (10 modules):

$$400 = a(10) + b$$

Re-write neatly:

$$10a + b = 400$$     (2)

Second substitution (14 modules):

$$500 = a(14) + b$$

Re-write:

$$14a + b = 500$$     (3)

Now solve equations (2) and (3) for $$a$$ and $$b$$.

Step 1 – Eliminate $$b$$: Subtract (2) from (3):

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

The $$b$$ terms cancel automatically:

$$14a - 10a = 100$$

$$4a = 100$$

Divide both sides by 4:

$$a = 25$$

Step 2 – Find $$b$$ using either equation (2) or (3). Substitute $$a = 25$$ in equation (2):

$$10(25) + b = 400$$

$$250 + b = 400$$

Subtract 250 from both sides:

$$b = 400 - 250 = 150$$

Hence, the numerical values are

\[a = 25,\qquad b = 150\]

The required linear relation therefore becomes

$$y = 25x + 150$$

Let x be the number of hours the badminton court is used in a month and y be the total monthly bill (in rupees).
According to the statement, the bill varies linearly with hours and is given by the polynomial
$$y = ax + b$$
where

• $$a$$ = charge per hour for the badminton court,
• $$b$$ = fixed monthly gym fee (irrespective of court usage).

Two observations are supplied:

1. For $$x = 10$$ hours, $$y = 800$$ rupees.
2. For $$x = 15$$ hours, $$y = 1100$$ rupees.

These give two simultaneous linear equations in $$a$$ and $$b$$.

Step 1: Eliminate $$b$$

Subtract the first equation from the second:

Simplify:

Hence

Step 2: Find $$b$$

Substitute $$a = 60$$ in, say, the first equation:

Verification (optional)

Insert $$a = 60$$, $$b = 200$$ into the second observation:

First recall what the question is asking:

• We must write a polynomial in one variable, namely $$x$$.
• The degree of the polynomial has to be 3. That means the highest power of $$x$$ that appears must be $$x^3$$ and its coefficient must be non-zero.
• The coefficient of the $$x^2$$ term must be $$-7$$.

Start with the most general form of a cubic (degree 3) polynomial in $$x$$:

$$ax^3 + bx^2 + cx + d$$, where $$a \neq 0$$ to make sure the degree really is 3.

We are free to choose any values for $$a,\,b,\,c,\,d$$ except that two conditions must hold:

1. $$a \neq 0$$ (to keep degree 3)
2. The coefficient of $$x^2$$, i.e. $$b$$, must equal $$-7$$.

Pick simple values that satisfy these:

• Let $$a = 1$$ (the simplest non-zero choice).
• Set $$b = -7$$, exactly as required.
• For ease, let $$c = 0$$ and $$d = 5$$ (you may choose any real numbers here).

Substituting these into the general form gives

$$1\,x^3 + (-7)\,x^2 + 0\,x + 5$$.

Simplifying the obvious 1’s and 0’s, the polynomial becomes

This cubic (degree 3) polynomial clearly has the $$x^2$$ coefficient $$-7$$, so it satisfies all the given conditions.

Let the required number be $$x$$.

Multiplying the number by $$\frac{5}{2}$$ gives the product $$\frac{5}{2}x$$.

Adding $$\frac{2}{3}$$ to this product, the result becomes $$\frac{5}{2}x + \frac{2}{3}$$, and the question states that this equals $$\frac{-7}{12}$$. Hence

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

Step 1: Isolate the term containing $$x$$

Subtract $$\frac{2}{3}$$ from both sides:

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

Step 2: Simplify the right-hand side

Convert $$\frac{2}{3}$$ to twelfths so that the denominators match:

$$\frac{2}{3} = \frac{2}{3}\times\frac{4}{4} = \frac{8}{12}$$

Therefore

$$\frac{5}{2}x = \frac{-7}{12} - \frac{8}{12} = \frac{-15}{12}$$

Simplify $$\frac{-15}{12}$$ by dividing numerator and denominator by 3:

$$\frac{-15}{12} = \frac{-5}{4}$$

So we have

$$\frac{5}{2}x = \frac{-5}{4}$$

Step 3: Solve for $$x$$

Multiply both sides by the reciprocal $$\frac{2}{5}$$ of $$\frac{5}{2}$$:

$$x = \frac{-5}{4}\times\frac{2}{5}$$

Simplify: the 5s cancel, giving

$$x = \frac{-1}{4}\times 2 = \frac{-2}{4}$$

Reduce $$\frac{-2}{4}$$ by dividing numerator and denominator by 2:

$$x = \frac{-1}{2}$$

\[\boxed{x = -\dfrac{1}{2}}\]

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

Let the smaller positive number be $$x$$. Then the larger number, being five times the smaller, is $$5x$$.

Add $$21$$ to each number:

• first new number: $$x+21$$;
• second new number: $$5x+21$$.

One of these new numbers is twice the other, so one of the following must hold:

1. $$5x+21 = 2(x+21)$$
2. $$x+21 = 2(5x+21)$$

Case 1: $$5x+21 = 2(x+21)$$

$$5x+21 = 2x+42$$
$$3x = 21$$
$$x = 7$$ (positive, admissible)

The corresponding larger number is $$5x = 35$$.

Case 2: $$x+21 = 2(5x+21)$$

$$x+21 = 10x+42$$
$$-9x = 21$$
$$x = -\dfrac{7}{3}$$ (negative, not allowed)

Thus only Case 1 is valid.

Therefore the required numbers are $$7$$ and $$35$$. (Check: $$7+21=28$$, $$35+21=56$$ and $$56 = 2\times 28$$).

Let us denote the number of whole months you continue to save by the variable $$n$$.

Step 1 · Form the algebraic (linear) pattern

You already have a fixed amount of ₹800 at the start. Every month you add ₹250 to this amount. Therefore the total amount of money you will have after $$n$$ months is

\[ A(n) = 800 + 250n \]

This rule is a linear polynomial in $$n$$ because the highest power of $$n$$ is 1.

Step 2 · Amount after 6 months

Put $$n = 6$$ in the formula:

$$A(6) = 800 + 250\times6$$

$$A(6) = 800 + 1500$$

\[ A(6) = 2300 \]

So, after 6 months you will have ₹2300.

Step 3 · Amount after 2 years

There are 12 months in a year, so 2 years contain $$12\times2 = 24$$ months. Put $$n = 24$$:

$$A(24) = 800 + 250\times24$$

$$A(24) = 800 + 6000$$

\[ A(24) = 6800 \]

So, after 2 years (24 months) you will have ₹6800.

Conclusion (linear pattern)

The money you possess after $$n$$ months follows the linear pattern

\[ \boxed{A(n) = 800 + 250n}\]

Substituting $$n = 6$$ gives ₹2300 and substituting $$n = 24$$ gives ₹6800, exactly as calculated.

Step 1 : Represent the two–digit number algebraically

Let the tens digit be $$x$$ and the ones (units) digit be $$y$$.
The value of the original two–digit number is therefore $$10x + y$$.

Step 2 : Translate each given condition into an equation

1. The digits differ by 3.
   This means their absolute difference is 3, so we can write either $$x - y = 3$$ or $$y - x = 3$$   …(i)
2. After interchanging the digits, the new number is $$10y + x$$.
   Adding this to the original number gives $$143$$, so
   
   
   $$(10x + y) + (10y + x) = 143$$
   
   Simplify: $$11x + 11y = 143$$
   Divide by 11: $$x + y = 13$$   …(ii)

Step 3 : Solve the simultaneous equations

We have the system
$$x + y = 13$$   …(ii)
and either $$x - y = 3$$ or $$x - y = -3$$ (from (i)).

Case A : $$x - y = 3$$
Add to (ii):
$$\bigl(x + y\bigr) + \bigl(x - y\bigr) = 13 + 3 \implies 2x = 16 \implies x = 8.$$
Substitute in (ii): $$8 + y = 13 \implies y = 5.$$
So the original number is $$10x + y = 10\times8 + 5 = 85$$ and the number with digits interchanged is $$58$$.

Case B : $$x - y = -3$$ (equivalently $$y - x = 3$$)
Add to (ii):
$$\bigl(x + y\bigr) + \bigl(x - y\bigr) = 13 - 3 \implies 2x = 10 \implies x = 5.$$
Then $$5 + y = 13 \implies y = 8.$$
Thus the original number is $$10x + y = 10\times5 + 8 = 58$$ and the interchanged number is $$85$$.

Step 4 : State the two numbers

Hence the two required numbers are 85 and 58. Depending on which one we start with as the “original”, its partner is obtained by interchanging the digits, and the pair always adds to 143.

Step 1 – Understand the physical relation
When a constant force $$F$$ acts on a body and the body moves a distance $$d$$ (in the direction of the force), the work done $$w$$ is given by

$$w = F d$$

Step 2 – Write the linear equation in two variables
The question fixes the force at 3 units, so put $$F = 3$$:

$$w = 3 d$$

This already shows a straight-line relation between the two variables $$w$$ and $$d$$. In the form linear equation in two variables we usually write all terms on one side:

$$w - 3 d = 0$$

Either form  $$w = 3d$$  or  $$w - 3d = 0$$  is acceptable as the required equation.

Step 3 – Prepare an ordered-pair table

The three ordered pairs are (0,0), (1,3) and (2,6).

Step 4 – Draw the graph

• Take a Cartesian plane with the horizontal axis labelled “Distance $$d$$ (units)” and the vertical axis labelled “Work $$w$$ (units)”.
• Mark the points obtained above: (0,0), (1,3) and (2,6).
• Join them with a ruler; all points lie on one straight line because the relation is linear.

What the student should see: a straight line through the origin that rises 3 units in $$w$$ for every 1 unit increase in $$d$$.

Step 5 – Find the work when the distance is 2 units

Substitute $$d = 2$$ in the equation:

$$w = 3 \times 2 = 6$$

Step 6 – Verify on the graph

• Locate $$d = 2$$ on the horizontal axis, draw a vertical line up to the straight line.
• The intersection point’s $$w$$-coordinate is read as 6, matching the algebraic result.

Hence the work done for a 2-unit displacement under a constant force of 3 units is 6 units, confirmed both algebraically and graphically.

Let the required linear polynomials be $$p(x)=ax+b$$ and $$q(x)=cx+d$$, where $$a,b,c,d\in\mathbb{R}$$.

(i) Using $$p(0)=5$$

$$p(0)=a\cdot0+b=b \implies b=5.$$

(ii) Using “$$p(x)-q(x)$$ cuts the x-axis at $$(3,0)$$”

The value of $$p(x)-q(x)$$ at $$x=3$$ is zero:

$$p(3)-q(3)=0.$$

Compute each value:

$$p(3)=3a+5,\quad q(3)=3c+d.$$

\[3a+5-(3c+d)=0\]

Simplifying,

$$3a-3c+5-d=0 \;\;\;\;\; (A).$$

(iii) Using “$$p(x)+q(x)=6x+4$$ for all real $$x$$”

First write the left-hand side:

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

Since two polynomials are equal for every $$x$$, their corresponding coefficients must be equal:

Coefficient of $$x$$:  $$a+c=6 \;\;\;\;\; (1)$$
Constant term:    $$b+d=4 \;\;\;\;\; (2)$$

But $$b=5$$, so from (2)

$$5+d=4 \implies d=-1.$$

Substituting $$d=-1$$ in equation (A)

$$3a-3c+5-(-1)=0\;\Rightarrow\;3a-3c+6=0$$

Divide by $$3$$:

$$a-c+2=0 \implies a-c=-2 \;\;\;\;\; (3)$$

Solve the simultaneous equations (1) and (3):

$$\begin{aligned} a+c &= 6\\ a-c &= -2 \end{aligned}$$

Add the two equations:

$$2a=4 \implies a=2.$$

Substitute $$a=2$$ into $$a+c=6$$:

$$2+c=6 \implies c=4.$$

Hence

Verification (optional):

• $$p(0)=5$$ ✔
• $$p(3)=2\times3+5=11,\; q(3)=4\times3-1=11 \Rightarrow p(3)-q(3)=0$$ ✔
• $$p(x)+q(x)=(2x+5)+(4x-1)=6x+4$$ ✔

All conditions are satisfied, so the obtained polynomials are correct.

Let

$$p(x)=ax+b\quad\text{and}\quad q(x)=cx+d$$

where $$a,b,c,d$$ are real numbers still to be determined.

1. Use the two given points on the graph of $$p(x)$$

The point $$(2,3)$$ lies on the line $$y=p(x)$$, so

$$p(2)=a\,(2)+b=3\;\;\;\Longrightarrow\;\;2a+b=3.\quad(1)$$

The point $$(6,11)$$ also lies on the same line, hence

$$p(6)=a\,(6)+b=11\;\;\Longrightarrow\;\;6a+b=11.\quad(2)$$

Subtract equation (1) from equation (2):

$$(6a+b)-(2a+b)=11-3\quad\Longrightarrow\quad4a=8\quad\Longrightarrow\quad a=2.$$

Substitute $$a=2$$ back into (1):

$$2(2)+b=3\;\;\Longrightarrow\;\;4+b=3\;\;\Longrightarrow\;\;b=-1.$$

Thus

\[p(x)=2x-1.\]

2. Determine the coefficients of $$q(x)$$

(i) Parallelism to $$p(x)$$ means both lines have the same slope, so

$$c=a=2.$$

(ii) The point $$(4,-1)$$ lies on $$y=q(x)$$, giving

$$q(4)=2\,(4)+d=-1\quad\Longrightarrow\quad8+d=-1\quad\Longrightarrow\quad d=-9.$$

Hence

\[q(x)=2x-9.\]

3. x–intercepts of the two lines

• For $$p(x)$$: set $$p(x)=0\;\Rightarrow\;2x-1=0\;\Rightarrow\;x=\dfrac12$$. Thus the graph of $$p(x)$$ meets the x-axis at $$\bigl(\tfrac12,\,0\bigr)$$.
• For $$q(x)$$: set $$q(x)=0\;\Rightarrow\;2x-9=0\;\Rightarrow\;x=\dfrac92$$. Thus the graph of $$q(x)$$ meets the x-axis at $$\bigl(\tfrac92,\,0\bigr)$$.

Final results

\[p(x)=2x-1,\qquad q(x)=2x-9.\]

Their x-intercepts are $$\bigl(\tfrac12,0\bigr)$$ for $$p(x)$$ and $$\bigl(\tfrac92,0\bigr)$$ for $$q(x)$$.

Given: Every linear function in the family is
$$f(x)=ax+a,$$
where $$a>0$$ (so the graph is an upward–sloping straight line).

To discover what they all have in common, let us find the point(s) that every such line must pass through.

1. Find the x–intercept (where the graph cuts the x-axis)

The x-intercept is obtained by setting the y-value equal to $$0$$:

$$f(x)=0 \Longrightarrow ax+a=0.$$

Factor the left side:

$$a(x+1)=0.$$

The constant $$a$$ is positive and therefore cannot be zero. Hence the only way the product can be $$0$$ is

$$x+1=0 \Longrightarrow x=-1.$$

So every member of this family meets the x-axis at the point $$(-1,0).$$

2. Check that no other common point exists

The y-intercept of each line is obtained by putting $$x=0$$:

$$f(0)=a(0)+a=a.$$

This value depends on $$a$$, so the y-intercept varies from one line to another; the only shared point is therefore the one already found.

Conclusion

All lines of the form $$f(x)=ax+a$$ with $$a>0$$ pass through the single, common point

\[(-1,\,0).\]

No other point is common to the whole family.