NCERT Solutions · Class 9 Mathematics Introduction to Linear Polynomials

Name the unknown with one letter, write everything else in terms of it.Let Salil's present age be \(\displaystyle x\) years. His mother's present age is three times this, so it is \(\displaystyle 3x\) years.After $\displaystyle 5$ years each of them is $\displaystyle 5$ years older, so their ages then will be \(\displaystyle x+5\) and \(\displaystyle 3x+5\). We are told these add up to $\displaystyle 70$: \[(x+5)+(3x+5)=70 \]Collect the like terms on the left: \[4x+10=70 \]Subtract \(\displaystyle 10\) from both sides (whatever is done to one side must be done to the other, so the two sides stay equal): \[4x=60 \]Divide both sides by \(\displaystyle 4\): \[x=15 \]So Salil is \(\displaystyle 15\) and his mother is \(\displaystyle 3 \times 15 = 45\).Check. In $\displaystyle 5$ years they will be \(\displaystyle 20\) and \(\displaystyle 50\), and \(\displaystyle 20+50=70\), as required. Also \(\displaystyle 45 = 3 \times 15\), so the "three times" condition holds too.Salil is $\displaystyle 15$ years old and his mother is $\displaystyle 45$ years old.

A ratio lets you describe both numbers with a single unknown.The two integers are in the ratio \(\displaystyle 2:5\), which means they can be written as \(\displaystyle 2k\) and \(\displaystyle 5k\) for the same positive number \(\displaystyle k\) — the ratio only fixes their relative sizes, and \(\displaystyle k\) fixes the actual size.Since \(\displaystyle 5k > 2k\), the larger integer is \(\displaystyle 5k\) and the difference is \[5k-2k=3k \]We are told this difference is $\displaystyle 63$: \[3k=63 \quad\Longrightarrow\quad k=21 \]Therefore the integers are \[2k = 2 \times 21 = 42, \qquad 5k = 5 \times 21 = 105 \]Check. \(\displaystyle 105-42=63\), and \(\displaystyle 42:105\) simplifies (dividing both by $\displaystyle 21$) to \(\displaystyle 2:5\). Both are positive integers, as required.The two integers are $\displaystyle 42$ and 105.

Count the coins with one unknown, then add up their value, not their number.The number of two-rupee coins is described in terms of the five-rupee coins, so let the five-rupee coins be the unknown.Let Ruby have \(\displaystyle x\) five-rupee coins. Then she has \(\displaystyle 3x\) two-rupee coins.Their values in rupees are: \[\text{five-rupee coins: } 5 \times x = 5x, \qquad \text{two-rupee coins: } 2 \times 3x = 6x \]The total value is ₹$\displaystyle 88$: \[5x+6x=88 \] \[11x=88 \quad\Longrightarrow\quad x=8 \]So there are \(\displaystyle 8\) five-rupee coins and \(\displaystyle 3 \times 8 = 24\) two-rupee coins.Check. \(\displaystyle 8 \times 5 = 40\) rupees and \(\displaystyle 24 \times 2 = 48\) rupees, and \(\displaystyle 40+48=88\). Also \(\displaystyle 24 = 3 \times 8\), so the "$\displaystyle 3$ times as many" condition holds.Notice that there are more two-rupee coins ($\displaystyle 24$) than five-rupee coins ($\displaystyle 8$), yet the two-rupee coins are worth only a little more in total — counting coins and counting rupees are different things.Ruby has $\displaystyle 24$ two-rupee coins and $\displaystyle 8$ five-rupee coins ($\displaystyle 32$ coins in all).

Let the shorter piece be the unknown, since the longer one is described in terms of it.Let the shorter piece be \(\displaystyle x\) feet long. The longer piece is four times as long, so it is \(\displaystyle 4x\) feet.The two pieces together make up the whole $\displaystyle 300$-foot fence: \[x+4x=300 \] \[5x=300 \quad\Longrightarrow\quad x=60 \]So the shorter piece is \(\displaystyle 60\) feet and the longer piece is \(\displaystyle 4 \times 60 = 240\) feet.Check. \(\displaystyle 60+240=300\) feet, and \(\displaystyle 240 \div 60 = 4\), so the longer piece really is four times the shorter one. The two pieces are indeed of different sizes, as the question says.The two pieces are $\displaystyle 60$ feet and $\displaystyle 240$ feet long.

Write both dimensions in terms of the width, then use the perimeter formula.Let the width of the rectangle be \(\displaystyle w\) cm. The length is "three more than twice its width", so the length is \(\displaystyle 2w+3\) cm.The perimeter of a rectangle is twice the sum of its length and width: \[P = 2(\text{length}+\text{width}) \] \[2\big((2w+3)+w\big)=24 \]Divide both sides by $\displaystyle 2$ first — it keeps the numbers small: \[(2w+3)+w=12 \] \[3w+3=12 \] \[3w=9 \quad\Longrightarrow\quad w=3 \]Then the length is \(\displaystyle 2(3)+3 = 6+3 = 9\) cm.Check. Length \(\displaystyle 9\) is three more than twice the width \(\displaystyle 2 \times 3 = 6\), and the perimeter is \(\displaystyle 2(9+3)=2 \times 12=24\) cm, as required.The rectangle is $\displaystyle 9$ cm long and $\displaystyle 3$ cm wide.

A fixed starting amount plus the same addition every month.She starts with ₹$\displaystyle 500$ already in the account, and ₹$\displaystyle 150$ is added each month. (We assume she spends none of it, and we ignore any interest.)Working month by month, each figure is ₹$\displaystyle 150$ more than the one before it:

End of month	Amount in the account (₹)
$\displaystyle 1$	\(\displaystyle 500+150 = 650\)
$\displaystyle 2$	\(\displaystyle 650+150 = 800\)
$\displaystyle 3$	\(\displaystyle 800+150 = 950\)
$\displaystyle 4$	\(\displaystyle 950+150 = 1100\)
$\displaystyle 5$	\(\displaystyle 1100+150 = 1250\)

So from the second month onwards she has ₹$\displaystyle 800$, ₹$\displaystyle 950$, ₹$\displaystyle 1100$, ₹$\displaystyle 1250$, … — the amount goes up by exactly ₹$\displaystyle 150$ each time.To get the expression, notice that adding ₹$\displaystyle 150$ once a month for \(\displaystyle n\) months adds \(\displaystyle 150 \times n\) rupees in all, on top of the ₹$\displaystyle 500$ she began with: \[A_n = 500+150n \]Check. For \(\displaystyle n=2\), \(\displaystyle A_2 = 500+150(2) = 500+300 = 800\), which matches the table; for \(\displaystyle n=4\), \(\displaystyle A_4 = 500+600 = 1100\), which also matches.The highest power of \(\displaystyle n\) here is \(\displaystyle 1\), so \(\displaystyle A_n\) is a linear expression in \(\displaystyle n\). The constant \(\displaystyle 500\) is her starting balance and the coefficient \(\displaystyle 150\) is the fixed monthly increase.\(\displaystyle A_n = 500+150n\) rupees at the end of the \(\displaystyle n^{\text{th}}\) month; from the second month onwards the balances are ₹$\displaystyle 800$, ₹$\displaystyle 950$, ₹$\displaystyle 1100$, ₹$\displaystyle 1250$, …

A fixed starting number, with the same amount subtracted every hour.The rally begins with $\displaystyle 120$ members, and $\displaystyle 9$ leave in each hour that passes. Because members are lost rather than gained, the fixed number is subtracted each time:

Hours elapsed	Members remaining
$\displaystyle 1$	\(\displaystyle 120-9 = 111\)
$\displaystyle 2$	\(\displaystyle 111-9 = 102\)
$\displaystyle 3$	\(\displaystyle 102-9 = 93\)
$\displaystyle 4$	\(\displaystyle 93-9 = 84\)

After \(\displaystyle n\) hours, \(\displaystyle 9\) members have dropped out \(\displaystyle n\) times over, which is \(\displaystyle 9n\) members in total. So the number remaining is \[M_n = 120-9n \]Check. \(\displaystyle M_1 = 120-9 = 111\) and \(\displaystyle M_3 = 120-27 = 93\), both matching the table.This is a linear expression in \(\displaystyle n\) (the highest power of \(\displaystyle n\) is \(\displaystyle 1\)). The negative coefficient \(\displaystyle -9\) is what makes the count fall instead of rise.One sensible caution: the expression only describes the real situation while there are still members left. \(\displaystyle M_{13} = 120-117 = 3\), but \(\displaystyle M_{14} = 120-126 = -6\), and a rally cannot have \(\displaystyle -6\) members. So the pattern holds for \(\displaystyle n \le 13\) hours, after which everyone has gone.\(\displaystyle M_n = 120-9n\) members remain at the end of the \(\displaystyle n^{\text{th}}\) hour: $\displaystyle 111$ after $\displaystyle 1$ hour, $\displaystyle 102$ after $\displaystyle 2$ hours, $\displaystyle 93$ after $\displaystyle 3$ hours, and so on (valid up to \(\displaystyle n=13\)).

Start from the total and subtract a fixed number of pages each day.The book has $\displaystyle 500$ pages, and $\displaystyle 20$ pages disappear from the "still to read" pile every day. After \(\displaystyle d\) days she has read \(\displaystyle 20 \times d = 20d\) pages, so the number of pages still left is \[P = 500-20d \]After $\displaystyle 15$ days: \[P = 500-20(15) = 500-300 = 200 \]Check the other way round. In $\displaystyle 15$ days she reads \(\displaystyle 20 \times 15 = 300\) pages out of $\displaystyle 500$, and \(\displaystyle 500-300 = 200\) pages remain — the same answer reached by a different route.\(\displaystyle P = 500-20d\) is a linear pattern in \(\displaystyle d\): the highest power of \(\displaystyle d\) is \(\displaystyle 1\), the constant \(\displaystyle 500\) is the length of the book, and the coefficient \(\displaystyle -20\) is the fixed daily reading that makes the number fall.This pattern makes sense only until the book is finished. Setting \(\displaystyle 500-20d=0\) gives \(\displaystyle d=25\), so she finishes on the 25th day; beyond that the expression would give a negative number of pages, which is impossible.$\displaystyle 200$ pages will be left after $\displaystyle 15$ days; the linear pattern is \(\displaystyle P = 500-20d\) (valid for \(\displaystyle 0 \le d \le 25\)).

A fixed starting height plus the same growth every month.The plant is $\displaystyle 1.75$ feet tall to begin with, and each month adds another $\displaystyle 0.5$ feet. Write \(\displaystyle t\) for the number of months and \(\displaystyle h\) for the height in feet.(i) Height after $\displaystyle 7$ months.In $\displaystyle 7$ months the plant gains \(\displaystyle 0.5 \times 7 = 3.5\) feet, on top of the $\displaystyle 1.75$ feet it already had: \[h = 1.75+3.5 = 5.25 \ \text{feet} \] In fractions this is \(\displaystyle \tfrac{7}{4}+\tfrac{7}{2} = \tfrac{7}{4}+\tfrac{14}{4} = \tfrac{21}{4} = 5\tfrac{1}{4}\) feet, which agrees.(ii) Table of heights for \(\displaystyle t = 0\) to \(\displaystyle t = 10\).Each row is obtained from the row above by adding $\displaystyle 0.5$ feet.

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

The value at \(\displaystyle t=7\) is $\displaystyle 5.25$ feet, matching part (i).(iii) The expression, and why the growth is linear.After \(\displaystyle t\) months the plant has gained \(\displaystyle 0.5t\) feet, so \[h = 1.75+0.5t \]Two reasons this is linear growth:First, as an algebraic expression, \(\displaystyle 1.75+0.5t\) is a polynomial of degree $\displaystyle 1$ in \(\displaystyle t\) — the variable \(\displaystyle t\) appears only to the first power, with no \(\displaystyle t^{2}\), no \(\displaystyle t^{3}\) and no \(\displaystyle t\) in a denominator. That is exactly the definition of a linear polynomial.Second, and this is what "linear" means in the growing plant itself, the increase over any one month is always the same: \[h(t+1)-h(t) = \big(1.75+0.5(t+1)\big)-\big(1.75+0.5t\big) = 0.5 \] The \(\displaystyle 1.75\) and the \(\displaystyle 0.5t\) cancel out and only \(\displaystyle 0.5\) is left, whatever \(\displaystyle t\) is. So the plant grows by $\displaystyle 0.5$ feet in the first month, $\displaystyle 0.5$ feet in the fifth month, $\displaystyle 0.5$ feet in the tenth — equal increases in equal intervals of time, which is precisely the constant-rate growth the column of numbers in the table shows.It is worth seeing what would not be linear: if the plant instead grew by $\displaystyle 10$% of its current height each month, the increase would be larger every month (bigger plant, bigger $\displaystyle 10$%), the differences between successive rows would not be constant, and the pattern would not be a linear one.In \(\displaystyle h = 1.75+0.5t\), the constant \(\displaystyle 1.75\) is the height at \(\displaystyle t=0\) (where the plant starts) and the coefficient \(\displaystyle 0.5\) is the fixed monthly rate of growth.(i) $\displaystyle 5.25$ feet. (ii) See the table above: heights $\displaystyle 1.75$, $\displaystyle 2.25$, $\displaystyle 2.75$, …, $\displaystyle 6.75$ feet for \(\displaystyle t = 0\) to \(\displaystyle 10\). (iii) \(\displaystyle h = 1.75+0.5t\) — linear because \(\displaystyle t\) appears only to the first power, so the height rises by the same $\displaystyle 0.5$ feet in every month.

Constant yearly drop — linear decay.The phone loses the same amount, ₹$\displaystyle 800$, in every single year. That constant change per year is exactly what makes the pattern linear.(i) Value after $\displaystyle 3$ years.In $\displaystyle 3$ years the total drop is \(\displaystyle 3 \times 800 = 2400\). \[v = 10000 - 2400 = 7600 \] So the phone is worth ₹$\displaystyle 7,600$ after $\displaystyle 3$ years.(ii) Table of values.After \(\displaystyle t\) years the total drop is \(\displaystyle 800t\), so the value left is \(\displaystyle 10000 - 800t\).

\(\displaystyle t\) (years)	$\displaystyle 0$	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$	$\displaystyle 4$	$\displaystyle 5$	$\displaystyle 6$	$\displaystyle 7$	$\displaystyle 8$
\(\displaystyle v\) (in ₹)	$\displaystyle 10000$	$\displaystyle 9200$	$\displaystyle 8400$	$\displaystyle 7600$	$\displaystyle 6800$	$\displaystyle 6000$	$\displaystyle 5200$	$\displaystyle 4400$	$\displaystyle 3600$

Read the second row across: each entry is exactly $\displaystyle 800$ less than the one before it. The column \(\displaystyle t = 3\) again gives ₹$\displaystyle 7,600$, which checks part (i).(iii) The expression relating \(\displaystyle v\) and \(\displaystyle t\). \[v = 10000 - 800t \]Why this is linear decay: One sensible-model remark: \(\displaystyle v = 0\) when \(\displaystyle 800t = 10000\), that is \(\displaystyle t = 12.5\) years, so the formula only describes the phone up to about $\displaystyle 12$ and a half years — after that the value cannot keep falling below zero.Answer: (i) ₹$\displaystyle 7$,600. (iii) \(\displaystyle v = 10000 - 800t\), a degree-$\displaystyle 1$ polynomial in \(\displaystyle t\) with the negative constant rate \(\displaystyle -800\) per year, hence linear decay.

Constant yearly increase — linear growth.The same number of people, $\displaystyle 50$, is added every year. A fixed change per year makes the pattern linear.(i) Population after $\displaystyle 6$ years.In $\displaystyle 6$ years the total number who move in is \(\displaystyle 6 \times 50 = 300\). \[P = 750 + 300 = 1050 \] So after $\displaystyle 6$ years the village has $\displaystyle 1050$ people.(ii) Table of values.After \(\displaystyle t\) years the increase is \(\displaystyle 50t\), so the population is \(\displaystyle 750 + 50t\).

\(\displaystyle t\) (years)	$\displaystyle 0$	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$	$\displaystyle 4$	$\displaystyle 5$	$\displaystyle 6$	$\displaystyle 7$	$\displaystyle 8$	$\displaystyle 9$	$\displaystyle 10$
\(\displaystyle P\) (people)	$\displaystyle 750$	$\displaystyle 800$	$\displaystyle 850$	$\displaystyle 900$	$\displaystyle 950$	$\displaystyle 1000$	$\displaystyle 1050$	$\displaystyle 1100$	$\displaystyle 1150$	$\displaystyle 1200$	$\displaystyle 1250$

Each entry in the second row is $\displaystyle 50$ more than the previous one, and the column \(\displaystyle t = 6\) reads $\displaystyle 1050$, which checks part (i).(iii) The expression relating \(\displaystyle P\) and \(\displaystyle t\). \[P = 750 + 50t \]Why this is linear growth: (This is a model of migration only. It assumes nobody is born, dies, or leaves — so like all such models it should not be pushed too far into the future.)Answer: (i) $\displaystyle 1050$ people. (iii) \(\displaystyle P = 750 + 50t\), a degree-$\displaystyle 1$ polynomial in \(\displaystyle t\) with the positive constant rate \(\displaystyle +50\) per year, hence linear growth.

Start value minus a constant daily amount.(i) The model.The balance starts at ₹$\displaystyle 600$ and loses ₹$\displaystyle 15$ on each day of use. After \(\displaystyle x\) days the total spent is \(\displaystyle 15x\), so \[b(x) = 600 - 15x \]Why this represents linear decay: (ii) When does the balance run out?The balance runs out when \(\displaystyle b(x) = 0\): \[600 - 15x = 0 \quad \Longrightarrow \quad 15x = 600 \quad \Longrightarrow \quad x = \frac{600}{15} = 40 \] Check: \(\displaystyle 15 \times 40 = 600\), so exactly the whole recharge is used up.So the balance runs out after $\displaystyle 40$ days. (For \(\displaystyle x > 40\) the formula would give a negative balance, which has no meaning here, so the model is valid for \(\displaystyle 0 \le x \le 40\).)(iii) Table of values.

\(\displaystyle x\) (days)	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$	$\displaystyle 4$	$\displaystyle 5$	$\displaystyle 6$	$\displaystyle 7$	$\displaystyle 8$	$\displaystyle 9$	$\displaystyle 10$
\(\displaystyle b(x)\) (in ₹)	$\displaystyle 585$	$\displaystyle 570$	$\displaystyle 555$	$\displaystyle 540$	$\displaystyle 525$	$\displaystyle 510$	$\displaystyle 495$	$\displaystyle 480$	$\displaystyle 465$	$\displaystyle 450$

Each value is $\displaystyle 15$ less than the one before it — the constant step that shows the decay is linear.Answer: (i) \(\displaystyle b(x) = 600 - 15x\); linear decay because it is degree $\displaystyle 1$ in \(\displaystyle x\) with the constant negative rate \(\displaystyle -15\) per day. (ii) After $\displaystyle 40$ days.

Two readings give two equations; subtract to remove the fixed fee.Here \(\displaystyle y = ax + b\), where \(\displaystyle x\) is the number of modules accessed, \(\displaystyle a\) is the cost of one module and \(\displaystyle b\) is the fixed monthly fee.Put the two observations into the relation. Both bills contain the same fixed fee \(\displaystyle b\). So if we subtract equation ($\displaystyle 1$) from equation ($\displaystyle 2$), the \(\displaystyle b\) cancels and only the cost of the extra modules is left: \[(14a + b) - (10a + b) = 500 - 400 \] \[4a = 100 \quad \Longrightarrow \quad a = 25 \]That makes sense on its own: $\displaystyle 4$ extra modules cost ₹$\displaystyle 100$ extra, so one module costs ₹25.Now substitute \(\displaystyle a = 25\) into equation ($\displaystyle 1$) to get \(\displaystyle b\): \[10(25) + b = 400 \quad \Longrightarrow \quad 250 + b = 400 \quad \Longrightarrow \quad b = 150 \]Check in the equation we did not use for the substitution, equation ($\displaystyle 2$): \[14(25) + 150 = 350 + 150 = 500 \quad \checkmark \]So the rule for the bill is \(\displaystyle y = 25x + 150\): ₹$\displaystyle 25$ per module plus a fixed ₹$\displaystyle 150$ a month.Answer: \(\displaystyle a = 25\) and \(\displaystyle b = 150\), so \(\displaystyle y = 25x + 150\).

Two readings give two equations; subtract to remove the fixed fee.In \(\displaystyle y = ax + b\), \(\displaystyle x\) is the number of hours of court use, \(\displaystyle a\) is the charge per hour and \(\displaystyle b\) is the fixed monthly fee. The fixed fee \(\displaystyle b\) is the same in both bills, so subtracting ($\displaystyle 1$) from ($\displaystyle 2$) removes it and leaves only what the extra hours cost: \[(15a + b) - (10a + b) = 1100 - 800 \] \[5a = 300 \quad \Longrightarrow \quad a = 60 \]Sense check: $\displaystyle 5$ extra hours added ₹$\displaystyle 300$, so one hour costs ₹60.Substitute \(\displaystyle a = 60\) in equation ($\displaystyle 1$): \[10(60) + b = 800 \quad \Longrightarrow \quad 600 + b = 800 \quad \Longrightarrow \quad b = 200 \]Check with equation ($\displaystyle 2$): \[15(60) + 200 = 900 + 200 = 1100 \quad \checkmark \]So the bill follows \(\displaystyle y = 60x + 200\): ₹$\displaystyle 60$ per hour on the court plus a fixed ₹$\displaystyle 200$ a month.Answer: \(\displaystyle a = 60\) and \(\displaystyle b = 200\), so \(\displaystyle y = 60x + 200\).

Two known temperature pairs, two unknowns.We are told the relation has the form \[{}^{\circ}\mathrm{C} = a \cdot {}^{\circ}\mathrm{F} + b \] Write \(\displaystyle C\) for the Celsius reading and \(\displaystyle F\) for the Fahrenheit reading, so \(\displaystyle C = aF + b\).Substitute the two known pairs.Melting ice: \(\displaystyle C = 0\) when \(\displaystyle F = 32\): \[0 = 32a + b \qquad \ldots (1) \] Boiling water: \(\displaystyle C = 100\) when \(\displaystyle F = 212\): \[100 = 212a + b \qquad \ldots (2) \]Subtract ($\displaystyle 1$) from ($\displaystyle 2$) so that \(\displaystyle b\) cancels: \[100 - 0 = (212a + b) - (32a + b) \] \[100 = 180a \quad \Longrightarrow \quad a = \frac{100}{180} = \frac{5}{9} \]This number has a meaning: between freezing and boiling there are $\displaystyle 100$ Celsius degrees but $\displaystyle 180$ Fahrenheit degrees, so one Fahrenheit degree is only \(\displaystyle \tfrac{5}{9}\) of a Celsius degree.Find \(\displaystyle b\) from equation ($\displaystyle 1$): \[0 = 32 \cdot \frac{5}{9} + b \quad \Longrightarrow \quad b = -\frac{160}{9} \]Check with equation ($\displaystyle 2$): \[212 \cdot \frac{5}{9} - \frac{160}{9} = \frac{1060 - 160}{9} = \frac{900}{9} = 100 \quad \checkmark \]The relationship. Putting the two values back, \[C = \frac{5}{9}F - \frac{160}{9} = \frac{5}{9}\left(F - 32\right) \] which is the familiar conversion rule: subtract $\displaystyle 32$, then take five-ninths.Answer: \(\displaystyle a = \dfrac{5}{9}\) and \(\displaystyle b = -\dfrac{160}{9}\), so \(\displaystyle {}^{\circ}\mathrm{C} = \dfrac{5}{9}\left({}^{\circ}\mathrm{F} - 32\right)\).

Two points fix a line; then read off what \(\displaystyle a\) and \(\displaystyle b\) do.Every equation here has the form \(\displaystyle y = ax + b\), which is a polynomial of degree $\displaystyle 1$, so each graph is a straight line. That means you only need two points to draw it — plot a third as a check that the three are in a row.Two facts to keep in mind while plotting, and to confirm afterwards from the picture: (i) \(\displaystyle y = 4x,\ y = 2x,\ y = x\)

\(\displaystyle x\)	\(\displaystyle -1\)	$\displaystyle 0$	$\displaystyle 1$
\(\displaystyle y = 4x\)	\(\displaystyle -4\)	$\displaystyle 0$	$\displaystyle 4$
\(\displaystyle y = 2x\)	\(\displaystyle -2\)	$\displaystyle 0$	$\displaystyle 2$
\(\displaystyle y = x\)	\(\displaystyle -1\)	$\displaystyle 0$	$\displaystyle 1$

Plot each row against the \(\displaystyle x\) row and join. Here \(\displaystyle b = 0\) for all three, so all three lines pass through the origin. All three rise to the right because \(\displaystyle a > 0\), and the larger \(\displaystyle a\) is, the steeper the climb: \(\displaystyle y = 4x\) is the steepest and \(\displaystyle y = x\) the gentlest.(ii) \(\displaystyle y = -6x,\ y = -3x,\ y = -x\)

\(\displaystyle x\)	\(\displaystyle -1\)	$\displaystyle 0$	$\displaystyle 1$
\(\displaystyle y = -6x\)	$\displaystyle 6$	$\displaystyle 0$	\(\displaystyle -6\)
\(\displaystyle y = -3x\)	$\displaystyle 3$	$\displaystyle 0$	\(\displaystyle -3\)
\(\displaystyle y = -x\)	$\displaystyle 1$	$\displaystyle 0$	\(\displaystyle -1\)

Again \(\displaystyle b = 0\), so all three pass through the origin. But now \(\displaystyle a < 0\), so all three fall to the right. Steepness still grows with the size of \(\displaystyle a\) ignoring its sign: \(\displaystyle y = -6x\) is the steepest of the three.(iii) \(\displaystyle y = 5x,\ y = -5x\)

\(\displaystyle x\)	\(\displaystyle -1\)	$\displaystyle 0$	$\displaystyle 1$
\(\displaystyle y = 5x\)	\(\displaystyle -5\)	$\displaystyle 0$	$\displaystyle 5$
\(\displaystyle y = -5x\)	$\displaystyle 5$	$\displaystyle 0$	\(\displaystyle -5\)

The two slopes have the same size but opposite signs, so the lines are equally steep but tilted opposite ways, and they cross at the origin. Each is the mirror image of the other in the y-axis (equivalently, in the x-axis). Only the sign of \(\displaystyle a\) has changed, and only the direction of the tilt has changed with it.(iv) \(\displaystyle y = 3x - 1,\ y = 3x,\ y = 3x + 1\)

\(\displaystyle x\)	\(\displaystyle -1\)	$\displaystyle 0$	$\displaystyle 1$
\(\displaystyle y = 3x - 1\)	\(\displaystyle -4\)	\(\displaystyle -1\)	$\displaystyle 2$
\(\displaystyle y = 3x\)	\(\displaystyle -3\)	$\displaystyle 0$	$\displaystyle 3$
\(\displaystyle y = 3x + 1\)	\(\displaystyle -2\)	$\displaystyle 1$	$\displaystyle 4$

Now \(\displaystyle a = 3\) is the same for all three while \(\displaystyle b\) changes. The three lines are parallel — same slope, so same tilt, and they never meet. They cut the y-axis at \(\displaystyle (0, -1)\), \(\displaystyle (0, 0)\) and \(\displaystyle (0, 1)\) respectively. Changing \(\displaystyle b\) simply slides the whole line up or down without turning it.(v) \(\displaystyle y = -2x - 3,\ y = -2x,\ y = 2x + 3\)

\(\displaystyle x\)	\(\displaystyle -1\)	$\displaystyle 0$	$\displaystyle 1$
\(\displaystyle y = -2x - 3\)	\(\displaystyle -1\)	\(\displaystyle -3\)	\(\displaystyle -5\)
\(\displaystyle y = -2x\)	$\displaystyle 2$	$\displaystyle 0$	\(\displaystyle -2\)
\(\displaystyle y = 2x + 3\)	$\displaystyle 1$	$\displaystyle 3$	$\displaystyle 5$

The first two both have \(\displaystyle a = -2\): they are parallel falling lines, one slid down $\displaystyle 3$ units from the other, cutting the y-axis at \(\displaystyle (0, -3)\) and \(\displaystyle (0, 0)\).The third has \(\displaystyle a = +2\), so it is not parallel to them; it rises. Notice that \(\displaystyle 2x + 3 = -(-2x - 3)\), so \(\displaystyle y = 2x + 3\) is the mirror image of \(\displaystyle y = -2x - 3\) in the x-axis, and the two meet exactly on the x-axis: setting \(\displaystyle y = 0\) in either gives \(\displaystyle x = -\tfrac{3}{2}\), so they cross at \(\displaystyle \left(-\tfrac{3}{2},\, 0\right)\).(If your copy of the book prints the third line as \(\displaystyle y = -2x + 3\), then all three have \(\displaystyle a = -2\), all three are parallel, and their y-intercepts \(\displaystyle -3, 0, 3\) illustrate the same point as part (iv).)Answer: \(\displaystyle a\) controls the direction and steepness of the line — positive \(\displaystyle a\) rises, negative \(\displaystyle a\) falls, and a larger \(\displaystyle |a|\) is steeper; two lines with the same \(\displaystyle a\) are parallel. \(\displaystyle b\) controls the height — the line meets the y-axis at \(\displaystyle (0, b)\), so changing \(\displaystyle b\) slides the line up or down without changing its tilt. When \(\displaystyle b = 0\) the line passes through the origin.

Build it term by term from the conditions.Two conditions must hold, and nothing else is fixed: The coefficients of \(\displaystyle x\) and of the constant term are not restricted at all, so we may choose them freely. Taking the \(\displaystyle x^3\) coefficient as $\displaystyle 1$, the \(\displaystyle x\) coefficient as $\displaystyle 2$ and the constant as $\displaystyle 5$ gives \[p(x) = x^3 - 7x^2 + 2x + 5 \]Check it against both conditions: the highest power is \(\displaystyle x^3\) with coefficient \(\displaystyle 1 \ne 0\), so the degree is $\displaystyle 3$; and the coefficient of \(\displaystyle x^2\) is \(\displaystyle -7\). Both conditions hold.This question has many correct answers. Any polynomial of the form \[p(x) = kx^3 - 7x^2 + mx + n, \qquad k \ne 0, \] with \(\displaystyle m\) and \(\displaystyle n\) any numbers, works. For instance \(\displaystyle 4x^3 - 7x^2\), \(\displaystyle -x^3 - 7x^2 + 9\) and \(\displaystyle x^3 - 7x^2 - \tfrac{1}{2}x + \sqrt{3}\) are all valid. Your answer need not match the one above; it only has to satisfy the two conditions.The one thing that would be wrong is taking \(\displaystyle k = 0\): \(\displaystyle -7x^2 + 2x + 5\) has the right \(\displaystyle x^2\) coefficient but degree $\displaystyle 2$, not 3.Answer: \(\displaystyle p(x) = x^{3} - 7x^{2} + 2x + 5\) is one such polynomial; any \(\displaystyle kx^{3} - 7x^{2} + mx + n\) with \(\displaystyle k \neq 0\) is equally correct.

Substitute the given value in place of the variable.The value of a polynomial at a number is found by replacing the variable by that number everywhere and simplifying.(i) \(\displaystyle 5x^{2} - 3x + 7\) at \(\displaystyle x = 1\).Put \(\displaystyle x = 1\) in each term: \[5(1)^{2} - 3(1) + 7 = 5 \times 1 - 3 + 7 = 5 - 3 + 7 = 9 \](A useful shortcut to check with: at \(\displaystyle x = 1\) the value of any polynomial is just the sum of its coefficients, since every power of $\displaystyle 1$ is 1. Here \(\displaystyle 5 + (-3) + 7 = 9\), which agrees.)(ii) \(\displaystyle 4t^{3} - t^{2} + 6\) at \(\displaystyle t = a\).Here the variable is being replaced by another letter, not by a number. That is allowed — substitution does not care whether the thing you put in is a number or a symbol. Replace every \(\displaystyle t\) by \(\displaystyle a\): \[4a^{3} - a^{2} + 6 \] This cannot be simplified further, because \(\displaystyle a\) stands for an unknown number, so \(\displaystyle 4a^3\) and \(\displaystyle a^2\) are not like terms.Answer: (i) \(\displaystyle 9\) (ii) \(\displaystyle 4a^{3} - a^{2} + 6\).

Translate the words into an equation, then undo the operations.Let the unknown number be \(\displaystyle x\). So the equation is \[\frac{5}{2}x + \frac{2}{3} = -\frac{7}{12} \]Undo the addition first. Subtract \(\displaystyle \dfrac{2}{3}\) from both sides: \[\frac{5}{2}x = -\frac{7}{12} - \frac{2}{3} \] The LCM of $\displaystyle 12$ and $\displaystyle 3$ is $\displaystyle 12$, and \(\displaystyle \dfrac{2}{3} = \dfrac{8}{12}\), so \[\frac{5}{2}x = -\frac{7}{12} - \frac{8}{12} = -\frac{15}{12} = -\frac{5}{4} \]Now undo the multiplication. Multiply both sides by \(\displaystyle \dfrac{2}{5}\) (the reciprocal of \(\displaystyle \dfrac{5}{2}\)): \[x = -\frac{5}{4} \times \frac{2}{5} = -\frac{10}{20} = -\frac{1}{2} \]Check by going forwards through the original words: \[\frac{5}{2} \times \left(-\frac{1}{2}\right) + \frac{2}{3} = -\frac{5}{4} + \frac{2}{3} = -\frac{15}{12} + \frac{8}{12} = -\frac{7}{12} \quad \checkmark \]Answer: the number is \(\displaystyle -\dfrac{1}{2}\).

Name the smaller number, express the other in terms of it, then test both readings.Let the smaller number be \(\displaystyle x\). Then the other number is \(\displaystyle 5x\), and since the numbers are positive, \(\displaystyle x > 0\).After $\displaystyle 21$ is added to each, the two new numbers are \[x + 21 \qquad \text{and} \qquad 5x + 21 \]The sentence says one of the new numbers is twice the other, without saying which. Since it does not say, we must try both readings and keep the one that gives positive numbers.Case $\displaystyle 1$: the bigger new number is twice the smaller new number. \[5x + 21 = 2(x + 21) \] \[5x + 21 = 2x + 42 \] Take \(\displaystyle 2x\) to the left and $\displaystyle 21$ to the right: \[3x = 21 \quad \Longrightarrow \quad x = 7 \] Then the numbers are \(\displaystyle x = 7\) and \(\displaystyle 5x = 35\), both positive — acceptable so far.Case $\displaystyle 2$: the smaller new number is twice the bigger new number. \[x + 21 = 2(5x + 21) \] \[x + 21 = 10x + 42 \quad \Longrightarrow \quad -21 = 9x \quad \Longrightarrow \quad x = -\frac{21}{9} = -\frac{7}{3} \] This is negative, so it is rejected: the question says the numbers are positive. (It is also unsurprising — a smaller number cannot be double a bigger one when both are positive.)Check Case $\displaystyle 1$ against the original statement. The numbers are $\displaystyle 7$ and $\displaystyle 35$, and indeed \(\displaystyle 35 = 5 \times 7\). Adding $\displaystyle 21$ to each gives \(\displaystyle 7 + 21 = 28\) and \(\displaystyle 35 + 21 = 56\), and \(\displaystyle 56 = 2 \times 28\). Both conditions hold.Answer: the numbers are $\displaystyle 7$ and 35.

A fixed start plus a constant monthly addition — a linear pattern.You begin with ₹$\displaystyle 800$ and add ₹$\displaystyle 250$ at the end of each month. After \(\displaystyle n\) months you have added \(\displaystyle 250n\), so the amount is \[A = 800 + 250n \](i) After $\displaystyle 6$ months.Here \(\displaystyle n = 6\): \[A = 800 + 250 \times 6 = 800 + 1500 = 2300 \] So you have ₹$\displaystyle 2,300$.(ii) After $\displaystyle 2$ years.First convert to months, because the saving happens monthly: \(\displaystyle 2 \text{ years} = 2 \times 12 = 24\) months. So \(\displaystyle n = 24\): \[A = 800 + 250 \times 24 = 800 + 6000 = 6800 \] So you have ₹$\displaystyle 6,800$.The linear pattern. \[A = 800 + 250n \]

\(\displaystyle n\) (months)	$\displaystyle 0$	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$	$\displaystyle 4$	$\displaystyle 5$	$\displaystyle 6$
\(\displaystyle A\) (in ₹)	$\displaystyle 800$	$\displaystyle 1050$	$\displaystyle 1300$	$\displaystyle 1550$	$\displaystyle 1800$	$\displaystyle 2050$	$\displaystyle 2300$

Each entry is $\displaystyle 250$ more than the one before, and the last column agrees with part (i).This is a linear polynomial in \(\displaystyle n\): the highest power of \(\displaystyle n\) is $\displaystyle 1$, the coefficient of \(\displaystyle n\) is the constant \(\displaystyle +250\) (equal amounts added in equal times, so the graph is a straight line rising steadily), and the constant term $\displaystyle 800$ is the amount at \(\displaystyle n = 0\), before any saving.Answer: (i) ₹$\displaystyle 2,300$ (ii) ₹$\displaystyle 6,800$; the pattern is \(\displaystyle A = 800 + 250n\), where \(\displaystyle n\) is the number of months.

Write a two-digit number in terms of its digits.Let the tens digit be \(\displaystyle a\) and the units digit be \(\displaystyle b\). Then the number itself is worth \[10a + b \] and the number formed by interchanging the digits is worth \[10b + a \]Use the "sum is $\displaystyle 143$" condition. \[(10a + b) + (10b + a) = 143 \] \[11a + 11b = 143 \quad \Longrightarrow \quad 11(a + b) = 143 \quad \Longrightarrow \quad a + b = \frac{143}{11} = 13 \]Notice what happened: reversing a two-digit number and adding always gives \(\displaystyle 11 \times (\text{sum of the digits})\), so this condition tells us the digit sum, nothing more.Use the "digits differ by $\displaystyle 3$" condition.So we need two digits with \[a + b = 13, \qquad |a - b| = 3 \] Adding \(\displaystyle a + b = 13\) and \(\displaystyle a - b = 3\) gives \(\displaystyle 2a = 16\), so \(\displaystyle a = 8\) and then \(\displaystyle b = 5\). Taking \(\displaystyle b - a = 3\) instead simply swaps the two, giving \(\displaystyle a = 5,\ b = 8\).So the digits are $\displaystyle 8$ and $\displaystyle 5$, and the two-digit numbers they form are $\displaystyle 85$ and $\displaystyle 58$. Either can be regarded as "the original" — the other is then the interchanged one.Check. The digits $\displaystyle 8$ and $\displaystyle 5$ differ by \(\displaystyle 8 - 5 = 3\) ✓. And \(\displaystyle 85 + 58 = 143\) ✓. Both are genuine two-digit numbers (neither has $\displaystyle 0$ in the tens place).Answer: the two numbers are $\displaystyle 85$ and 58.

Rewrite each equation as \(\displaystyle y = ax + b\), then read \(\displaystyle a\) and \(\displaystyle b\) straight off.Once an equation is in the form \(\displaystyle y = ax + b\), the slope is \(\displaystyle a\) and the y-intercept is \(\displaystyle b\); the line meets the y-axis at \(\displaystyle (0, b)\), because putting \(\displaystyle x = 0\) leaves \(\displaystyle y = b\).(i) \(\displaystyle y = -3x + 4\) is already in that form. Slope \(\displaystyle = -3\), y-intercept \(\displaystyle = 4\), cuts the y-axis at \(\displaystyle (0, 4)\). Plotting points: \(\displaystyle (0, 4),\ (1, 1),\ (2, -2)\).(ii) \(\displaystyle 2y = 4x + 7\). Divide every term by $\displaystyle 2$: \[y = 2x + \frac{7}{2} \] Slope \(\displaystyle = 2\), y-intercept \(\displaystyle = \dfrac{7}{2}\), cuts the y-axis at \(\displaystyle \left(0, \tfrac{7}{2}\right)\). Plotting points: \(\displaystyle (0, 3.5),\ (1, 5.5),\ (-1, 1.5)\).(iii) \(\displaystyle 5y = 6x - 10\). Divide by $\displaystyle 5$: \[y = \frac{6}{5}x - 2 \] Slope \(\displaystyle = \dfrac{6}{5}\), y-intercept \(\displaystyle = -2\), cuts the y-axis at \(\displaystyle (0, -2)\). Plotting points: \(\displaystyle (0, -2),\ (5, 4),\ (-5, -8)\) — multiples of $\displaystyle 5$ for \(\displaystyle x\) keep \(\displaystyle y\) a whole number.(iv) \(\displaystyle 3y = 6x - 11\). Divide by $\displaystyle 3$: \[y = 2x - \frac{11}{3} \] Slope \(\displaystyle = 2\), y-intercept \(\displaystyle = -\dfrac{11}{3}\), cuts the y-axis at \(\displaystyle \left(0, -\tfrac{11}{3}\right)\). Plotting points: \(\displaystyle \left(0, -\tfrac{11}{3}\right) \approx (0, -3.7)\) and \(\displaystyle \left(3, \tfrac{7}{3}\right) \approx (3, 2.3)\).Draw each line by plotting its two or three points on the same pair of axes and joining them with a ruler.Summary.

Equation	Slope	y-intercept	Point on the y-axis
\(\displaystyle y = -3x + 4\)	\(\displaystyle -3\)	\(\displaystyle 4\)	\(\displaystyle (0,\ 4)\)
\(\displaystyle 2y = 4x + 7\)	\(\displaystyle 2\)	\(\displaystyle \tfrac{7}{2}\)	\(\displaystyle \left(0,\ \tfrac{7}{2}\right)\)
\(\displaystyle 5y = 6x - 10\)	\(\displaystyle \tfrac{6}{5}\)	\(\displaystyle -2\)	\(\displaystyle (0,\ -2)\)
\(\displaystyle 3y = 6x - 11\)	\(\displaystyle 2\)	\(\displaystyle -\tfrac{11}{3}\)	\(\displaystyle \left(0,\ -\tfrac{11}{3}\right)\)

Are any of the lines parallel?Two lines are parallel exactly when they have the same slope but different y-intercepts (same slope and same intercept would make them the very same line).Looking down the slope column: \(\displaystyle -3\), \(\displaystyle 2\), \(\displaystyle \tfrac{6}{5}\), \(\displaystyle 2\). The value $\displaystyle 2$ occurs twice, for (ii) and (iv). Their intercepts \(\displaystyle \tfrac{7}{2}\) and \(\displaystyle -\tfrac{11}{3}\) are different, so they are two distinct lines with the same tilt.Answer: slopes and intercepts are \(\displaystyle -3\) and \(\displaystyle 4\); \(\displaystyle 2\) and \(\displaystyle \tfrac{7}{2}\); \(\displaystyle \tfrac{6}{5}\) and \(\displaystyle -2\); \(\displaystyle 2\) and \(\displaystyle -\tfrac{11}{3}\), so the lines cut the y-axis at \(\displaystyle (0,4)\), \(\displaystyle \left(0,\tfrac{7}{2}\right)\), \(\displaystyle (0,-2)\) and \(\displaystyle \left(0,-\tfrac{11}{3}\right)\). Yes — lines (ii) and (iv) are parallel, both having slope 2.

Substitute into the given rule for (i); reverse the steps for (ii).The rule is \[y = \frac{9}{5}(x - 273) + 32 \] with \(\displaystyle x\) in kelvin and \(\displaystyle y\) in degrees Fahrenheit.(i) Temperature in Fahrenheit when \(\displaystyle x = 313\) K.Substitute \(\displaystyle x = 313\), working inside the bracket first: \[y = \frac{9}{5}(313 - 273) + 32 = \frac{9}{5}(40) + 32 \] \[\frac{9}{5} \times 40 = 9 \times 8 = 72, \qquad \text{so } y = 72 + 32 = 104 \]So the temperature is $\displaystyle 104$ °F.(ii) Temperature in kelvin when \(\displaystyle y = 158\) °F.Now we know \(\displaystyle y\) and want \(\displaystyle x\), so we undo the operations in reverse order. Substitute \(\displaystyle y = 158\): \[158 = \frac{9}{5}(x - 273) + 32 \] Subtract $\displaystyle 32$ from both sides: \[126 = \frac{9}{5}(x - 273) \] Multiply both sides by \(\displaystyle \dfrac{5}{9}\): \[x - 273 = 126 \times \frac{5}{9} = 14 \times 5 = 70 \] Add $\displaystyle 273$ to both sides: \[x = 70 + 273 = 343 \]So the temperature is $\displaystyle 343$ K.Check part (ii) by putting $\displaystyle 343$ back into the original rule: \[\frac{9}{5}(343 - 273) + 32 = \frac{9}{5}(70) + 32 = 126 + 32 = 158 \quad \checkmark \]Answer: (i) $\displaystyle 104$ °F (ii) $\displaystyle 343$ K.

Turn the sentence into a formula, then fix the force to get a line.The statement "work done is the product of the constant force and the distance travelled in the direction of the force" says, in symbols, that for a constant force \(\displaystyle F\), \[w = F \times d \]Taking the constant force as $\displaystyle 3$ units, \(\displaystyle F = 3\), so the linear equation in the two variables \(\displaystyle w\) and \(\displaystyle d\) is \[w = 3d, \qquad \text{or equivalently} \qquad 3d - w = 0 \]This is a linear equation in two variables: each variable appears to the first power only, so its graph is a straight line. There is no constant term, so \(\displaystyle w = 0\) when \(\displaystyle d = 0\) — no distance, no work — and the line passes through the origin. Its slope is $\displaystyle 3$: every extra unit of distance costs $\displaystyle 3$ more units of work.Table of values for the graph.

\(\displaystyle d\) (units of distance)	$\displaystyle 0$	$\displaystyle 1$	$\displaystyle 2$	$\displaystyle 3$
\(\displaystyle w\) (units of work)	$\displaystyle 0$	$\displaystyle 3$	$\displaystyle 6$	$\displaystyle 9$

Drawing it. Take \(\displaystyle d\) along the horizontal axis and \(\displaystyle w\) along the vertical axis. Plot \(\displaystyle (0, 0)\), \(\displaystyle (1, 3)\), \(\displaystyle (2, 6)\) and \(\displaystyle (3, 9)\); they lie in a straight line, so join them with a ruler and extend. (Only \(\displaystyle d \ge 0\) is meaningful here, since a distance travelled cannot be negative.)Work done when the distance is $\displaystyle 2$ units. \[w = 3 \times 2 = 6 \]Verifying it on the graph. Start at \(\displaystyle d = 2\) on the horizontal axis, go straight up until you meet the line, then straight across to the vertical axis: you land on \(\displaystyle w = 6\). The point \(\displaystyle (2, 6)\) is one of the points already plotted from the table, so the graph and the calculation agree.Answer: \(\displaystyle w = 3d\); the work done over a distance of $\displaystyle 2$ units is $\displaystyle 6$ units, shown by the point \(\displaystyle (2, 6)\) on the line.

A linear polynomial has two unknowns; two points on the graph give two equations.A linear polynomial has the form \[p(x) = ax + b \]Saying the graph passes through a point means the coordinates of that point satisfy \(\displaystyle y = p(x)\).(i) Finding \(\displaystyle p(x)\).Through \(\displaystyle (1, 5)\): \(\displaystyle p(1) = 5\), so \[a(1) + b = 5 \quad \Longrightarrow \quad a + b = 5 \qquad \ldots (1) \] Through \(\displaystyle (3, 11)\): \(\displaystyle p(3) = 11\), so \[a(3) + b = 11 \quad \Longrightarrow \quad 3a + b = 11 \qquad \ldots (2) \]Subtract ($\displaystyle 1$) from ($\displaystyle 2$) to eliminate \(\displaystyle b\): \[(3a + b) - (a + b) = 11 - 5 \quad \Longrightarrow \quad 2a = 6 \quad \Longrightarrow \quad a = 3 \] Substitute in ($\displaystyle 1$): \(\displaystyle 3 + b = 5\), so \(\displaystyle b = 2\). Hence \[p(x) = 3x + 2 \]Check both given points: \(\displaystyle p(1) = 3 + 2 = 5\) ✓ and \(\displaystyle p(3) = 9 + 2 = 11\) ✓.(ii) Where the graph cuts the axes.The y-axis is the set of points with \(\displaystyle x = 0\). Put \(\displaystyle x = 0\): \[p(0) = 3(0) + 2 = 2 \] so the graph cuts the y-axis at \(\displaystyle (0,\ 2)\).The x-axis is the set of points with \(\displaystyle y = 0\). Put \(\displaystyle p(x) = 0\): \[3x + 2 = 0 \quad \Longrightarrow \quad 3x = -2 \quad \Longrightarrow \quad x = -\frac{2}{3} \] so the graph cuts the x-axis at \(\displaystyle \left(-\dfrac{2}{3},\ 0\right)\).(iii) Drawing the graph and verifying.

\(\displaystyle x\)	\(\displaystyle -1\)	$\displaystyle 0$	$\displaystyle 1$	$\displaystyle 3$
\(\displaystyle y = 3x + 2\)	\(\displaystyle -1\)	$\displaystyle 2$	$\displaystyle 5$	$\displaystyle 11$

Plot \(\displaystyle (-1, -1)\), \(\displaystyle (0, 2)\), \(\displaystyle (1, 5)\) and \(\displaystyle (3, 11)\) and join them with a ruler — they lie on one straight line, which confirms the polynomial is linear and that it does pass through the two given points.On that drawing, the line meets the vertical axis at height $\displaystyle 2$, matching \(\displaystyle (0, 2)\); and it crosses the horizontal axis a little to the left of the origin, between \(\displaystyle x = -1\) and \(\displaystyle x = 0\), at about \(\displaystyle -0.67\), which matches \(\displaystyle -\tfrac{2}{3}\).Answer: (i) \(\displaystyle p(x) = 3x + 2\). (ii) It cuts the y-axis at \(\displaystyle (0,\ 2)\) and the x-axis at \(\displaystyle \left(-\dfrac{2}{3},\ 0\right)\).

Turn each of the three conditions into an equation in \(\displaystyle a, b, c, d\), then solve.We are given \(\displaystyle p(x) = ax + b\) and \(\displaystyle q(x) = cx + d\).Condition (i): \(\displaystyle p(0) = 5\). \[p(0) = a(0) + b = b, \qquad \text{so } b = 5 \]Condition (iii): \(\displaystyle p(x) + q(x) = 6x + 4\) for all real \(\displaystyle x\).Add the two polynomials by collecting like terms: \[p(x) + q(x) = (ax + b) + (cx + d) = (a + c)x + (b + d) \] Two polynomials that are equal for all \(\displaystyle x\) must have matching coefficients — the \(\displaystyle x\)-terms must agree and the constant terms must agree. Comparing with \(\displaystyle 6x + 4\): \[a + c = 6 \qquad \ldots (1) \] \[b + d = 4 \] Since \(\displaystyle b = 5\), the second of these gives \[d = 4 - 5 = -1 \]Condition (ii): \(\displaystyle p(x) - q(x)\) cuts the x-axis at \(\displaystyle (3, 0)\).Subtract: \[p(x) - q(x) = (a - c)x + (b - d) \] and \(\displaystyle b - d = 5 - (-1) = 6\), so \(\displaystyle p(x) - q(x) = (a - c)x + 6\).Cutting the x-axis at \(\displaystyle (3, 0)\) means the value of this polynomial is $\displaystyle 0$ when \(\displaystyle x = 3\): \[(a - c)(3) + 6 = 0 \quad \Longrightarrow \quad 3(a - c) = -6 \quad \Longrightarrow \quad a - c = -2 \qquad \ldots (2) \]Solve ($\displaystyle 1$) and ($\displaystyle 2$) together.Add them: \(\displaystyle (a + c) + (a - c) = 6 + (-2)\), so \(\displaystyle 2a = 4\) and \(\displaystyle a = 2\). Then from ($\displaystyle 1$), \(\displaystyle c = 6 - 2 = 4\).So \[p(x) = 2x + 5, \qquad q(x) = 4x - 1 \]Check all three conditions. Answer: \(\displaystyle p(x) = 2x + 5\) and \(\displaystyle q(x) = 4x - 1\).

Growing by the same amount each time.Read the building rule carefully first, because the whole question follows from it. Stage $\displaystyle 1$ is a single hexagon. A hexagon has $\displaystyle 6$ sides, so stage $\displaystyle 1$ uses \(\displaystyle 6\) matchsticks.At every later stage one more hexagon is joined on at the end, and the new hexagon shares a side with the hexagon already there. That shared side is a matchstick that has already been counted once. So the new hexagon does not cost $\displaystyle 6$ fresh matchsticks — it costs only its other \[6-1=5 \] sides. This is the key idea: every stage after the first needs exactly $\displaystyle 5$ more matchsticks than the stage before it.So \[\text{Stage }1=6,\qquad \text{Stage }2=6+5=11,\qquad \text{Stage }3=11+5=16 . \](i) The next two stages.To draw stage $\displaystyle 4$, copy the stage-$\displaystyle 3$ strip of three hexagons and attach a fourth hexagon at the free end so that it touches the third hexagon along exactly one matchstick. Stage $\displaystyle 5$ is the same strip with a fifth hexagon attached in the same way. (Keep the hexagons in a straight row, each one sharing exactly one side with its neighbour.)Adding $\displaystyle 5$ each time, \[\text{Stage }4=16+5=21,\qquad \text{Stage }5=21+5=26 . \]So stage $\displaystyle 4$ needs \(\displaystyle 21\) matchsticks and stage $\displaystyle 5$ needs \(\displaystyle 26\) matchsticks.(ii) The completed table.

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

(iii) The rule.Start at $\displaystyle 6$ matchsticks and add $\displaystyle 5$ once for every stage after the first. Going from stage $\displaystyle 1$ up to stage \(\displaystyle n\) we add $\displaystyle 5$ exactly \(\displaystyle n-1\) times, so the number of matchsticks is \[M(n)=6+5(n-1)=6+5n-5=5n+1 . \]Here is a second way to see the same rule, which explains where the \(\displaystyle +1\) comes from. Stand one single matchstick upright at the left end of the strip. Now build the hexagons one after another from left to right: each hexagon closes itself off using $\displaystyle 5$ new matchsticks, because its left-hand side is already in place. With \(\displaystyle n\) hexagons that is \(\displaystyle 5n\) matchsticks, plus the one we started with: \[M(n)=5n+1 . \]Checking against the counts we already know: \(\displaystyle M(1)=6\), \(\displaystyle M(2)=11\), \(\displaystyle M(3)=16\), \(\displaystyle M(4)=21\), \(\displaystyle M(5)=26\). All five agree, so the rule is right.Notice that \(\displaystyle M(n)=5n+1\) is a linear polynomial in \(\displaystyle n\) — that is exactly what a pattern with a constant increase of $\displaystyle 5$ per step looks like, and the $\displaystyle 5$ is the coefficient of \(\displaystyle n\).(iv) The 15th stage. \[M(15)=5\times 15+1=75+1=76 . \](v) Can $\displaystyle 200$ matchsticks form a stage?A stage uses $\displaystyle 200$ matchsticks only if \(\displaystyle M(n)=200\) for some whole number \(\displaystyle n\) (there is no such thing as stage \(\displaystyle 39.8\)). So solve \[5n+1=200 \;\Longrightarrow\; 5n=199 \;\Longrightarrow\; n=\frac{199}{5}=39.8 , \] which is not a whole number. Hence no stage of this pattern uses $\displaystyle 200$ matchsticks.A neater way to say the same thing: \(\displaystyle 5n+1\) always leaves remainder \(\displaystyle 1\) on division by \(\displaystyle 5\) (the counts run \(\displaystyle 6,11,16,21,26,\dots\), always ending in $\displaystyle 1$ or $\displaystyle 6$), whereas $\displaystyle 200$ is exactly divisible by 5. The pattern jumps straight past it: stage $\displaystyle 39$ uses \(\displaystyle 5\times 39+1=196\) matchsticks and stage $\displaystyle 40$ uses \(\displaystyle 5\times 40+1=201\).Stage $\displaystyle 4$ needs $\displaystyle 21$ matchsticks and stage $\displaystyle 5$ needs $\displaystyle 26$; the rule is \(\displaystyle M(n)=5n+1\); the 15th stage needs $\displaystyle 76$ matchsticks; and no, $\displaystyle 200$ matchsticks cannot form a stage, because \(\displaystyle 5n+1=200\) gives \(\displaystyle n=39.8\), which is not a whole number.

Two points fix a line; parallel lines share the same coefficient of \(\displaystyle x\).Step $\displaystyle 1$: turn "the graph passes through" into equations.Saying the graph of \(\displaystyle p\) passes through \(\displaystyle (2,3)\) means that when \(\displaystyle x=2\) the value of the polynomial is \(\displaystyle 3\), i.e. \(\displaystyle p(2)=3\). Similarly \(\displaystyle p(6)=11\). Substituting into \(\displaystyle p(x)=ax+b\): \[2a+b=3 \qquad \text{and} \qquad 6a+b=11 . \]Step $\displaystyle 2$: solve for \(\displaystyle a\) and \(\displaystyle b\).Both equations contain \(\displaystyle +b\), so subtracting the first from the second removes \(\displaystyle b\): \[(6a+b)-(2a+b)=11-3 \;\Longrightarrow\; 4a=8 \;\Longrightarrow\; a=2 . \] Putting \(\displaystyle a=2\) back into \(\displaystyle 2a+b=3\): \[4+b=3 \;\Longrightarrow\; b=-1 . \] So \[p(x)=2x-1 . \] Check both given points: \(\displaystyle p(2)=4-1=3\) and \(\displaystyle p(6)=12-1=11\). Both are correct.Step $\displaystyle 3$: use "parallel" to get \(\displaystyle c\).For a linear polynomial \(\displaystyle cx+d\), the coefficient \(\displaystyle c\) is what controls the steepness and direction of the line: increasing \(\displaystyle x\) by $\displaystyle 1$ changes the value by \(\displaystyle c\). Two lines are parallel exactly when they rise at the same rate, so they must have the same coefficient of \(\displaystyle x\). (If the coefficients were different, one line would gain on the other and they would have to cross.) Hence \[c=a=2 . \]Step $\displaystyle 4$: use the point \(\displaystyle (4,-1)\) to get \(\displaystyle d\).\(\displaystyle q(4)=-1\), and \(\displaystyle q(x)=2x+d\), so \[2(4)+d=-1 \;\Longrightarrow\; 8+d=-1 \;\Longrightarrow\; d=-9 . \] So \[q(x)=2x-9 . \] Check: \(\displaystyle q(4)=8-9=-1\), which is correct. Also \(\displaystyle d=-9\) is different from \(\displaystyle b=-1\), so \(\displaystyle p\) and \(\displaystyle q\) really are two different parallel lines and not the same line drawn twice.Step $\displaystyle 5$: where each line meets the \(\displaystyle x\)-axis.Every point of the \(\displaystyle x\)-axis has \(\displaystyle y\)-coordinate \(\displaystyle 0\). So the line meets the \(\displaystyle x\)-axis where the polynomial takes the value \(\displaystyle 0\) — that is, at its zero.For \(\displaystyle p\): \[2x-1=0 \;\Longrightarrow\; 2x=1 \;\Longrightarrow\; x=\tfrac12 , \] so the point is \(\displaystyle \left(\tfrac12,\,0\right)\).For \(\displaystyle q\): \[2x-9=0 \;\Longrightarrow\; 2x=9 \;\Longrightarrow\; x=\tfrac92 , \] so the point is \(\displaystyle \left(\tfrac92,\,0\right)\).These are two different points, as they must be: parallel lines can never meet the \(\displaystyle x\)-axis at the same place, otherwise they would meet each other there.\(\displaystyle p(x)=2x-1\) and \(\displaystyle q(x)=2x-9\). The graph of \(\displaystyle p(x)\) meets the \(\displaystyle x\)-axis at \(\displaystyle \left(\tfrac12,\,0\right)\) and the graph of \(\displaystyle q(x)\) meets it at \(\displaystyle \left(\tfrac92,\,0\right)\).

Take out the common factor \(\displaystyle a\).The two coefficients in \(\displaystyle f(x)=ax+a\) are the same number, so \(\displaystyle a\) can be factored out: \[f(x)=ax+a=a(x+1). \] Every function in this family is therefore just the single expression \(\displaystyle x+1\) multiplied by a positive number. That one observation explains everything they share.1. They all have the same zero, so all their graphs pass through the point \(\displaystyle (-1,0)\).Put \(\displaystyle x=-1\): \[f(-1)=a(-1+1)=a\times 0=0 . \] The value of \(\displaystyle a\) never enters, because it is multiplied by \(\displaystyle 0\). So \(\displaystyle x=-1\) is a zero of every function of this form, and every one of these lines cuts the \(\displaystyle x\)-axis at exactly the same point \(\displaystyle (-1,0)\). Picture them as a fan of lines all pivoting about that one point.For example, \(\displaystyle f(x)=x+1\) (with \(\displaystyle a=1\)), \(\displaystyle f(x)=3x+3\) (with \(\displaystyle a=3\)) and \(\displaystyle f(x)=\tfrac12 x+\tfrac12\) (with \(\displaystyle a=\tfrac12\)) are three different lines, but each one gives \(\displaystyle 0\) at \(\displaystyle x=-1\).2. They are all increasing (rising) lines.The coefficient of \(\displaystyle x\) is \(\displaystyle a\), and we are told \(\displaystyle a>0\). A positive coefficient of \(\displaystyle x\) means the value goes up by \(\displaystyle a\) each time \(\displaystyle x\) goes up by $\displaystyle 1$, so every graph in this family slopes upward from left to right.3. The value at \(\displaystyle x=0\) equals the coefficient of \(\displaystyle x\).\(\displaystyle f(0)=a\), and the coefficient of \(\displaystyle x\) is also \(\displaystyle a\). So each of these graphs crosses the \(\displaystyle y\)-axis at \(\displaystyle (0,a)\), at exactly the height given by its own steepness — a steeper line in this family is also the one that meets the \(\displaystyle y\)-axis higher up. This crossing is always above the origin, since \(\displaystyle a>0\).4. They all take positive values to the right of \(\displaystyle -1\) and negative values to the left.Since \(\displaystyle a>0\), the sign of \(\displaystyle a(x+1)\) is the sign of \(\displaystyle x+1\): so \(\displaystyle f(x)>0\) when \(\displaystyle x>-1\) and \(\displaystyle f(x)<0\) when \(\displaystyle x<-1\), for every member of the family. In particular \(\displaystyle f(x)>0\) for all \(\displaystyle x>0\), so none of these graphs ever enters the fourth quadrant.The single most important shared feature is the first one: the common zero.Every function \(\displaystyle f(x)=ax+a=a(x+1)\) with \(\displaystyle a>0\) has \(\displaystyle -1\) as its zero, so all of their graphs are straight lines passing through the one common point \(\displaystyle (-1,0)\); and because \(\displaystyle a>0\), all of them are increasing lines that cross the \(\displaystyle y\)-axis at \(\displaystyle (0,a)\) — the height there being equal to the line's own coefficient of \(\displaystyle x\).