Introduction to Linear Polynomials — CBSE Class 9 Maths Textbook Solutions

Terms: $200l$200l, $160w$160w, $50lw$50lw. Variables: $l$l and $w$w. Coefficients: coefficient of $l$l is $200$200, coefficient of $w$w is $160$160, and coefficient of $lw$lw is $50$50. There is no constant (variable-free) term in this expression.

Example 1's expression, $4x+5y+3$4x+5y+3, has a genuine constant term ($3$3), and every term ($4x$4x, $5y$5y) has degree $1$1 in a single variable. Example 2's expression, $200l+160w+50lw$200l+160w+50lw, has no constant term, and it includes the term $50lw$50lw, which is the product of both variables — degree $1$1 in $l$l and degree $1$1 in $w$w, so degree $2$2 overall. So, unlike Example 1, Example 2 is not simply a sum of separate one-variable, degree-1 terms.

Terms: $10x$10x and $-x^2$-x². Variable: $x$x only. Coefficients: coefficient of $x^2$x² is $-1$-1, coefficient of $x$x is $10$10. There is no constant term.

Similarity: both are built from numeric coefficients multiplying powers of variable(s). Difference: Example 1's expression ($4x+5y+3$4x+5y+3) has two variables, each appearing only to the first power, so it is linear. Example 3's expression ($10x-x^2$10x-x²) has only one variable but includes a degree-2 term ($x^2$x²), making it a quadratic, not linear, polynomial.

(i) Highest power of $x$x is $2$2, so degree $= 2$= 2.
(ii) Highest power of $y$y is $3$3, so degree $= 3$= 3.
(iii) $-9=-9x^0$-9=-9x⁰, so degree $=0$=0 (a constant polynomial).
(iv) Highest power of $z$z is $1$1, so degree $=1$=1 (a linear polynomial).

Any valid example works, e.g.:
Degree $1$1: $2x+3$2x+3.
Degree $2$2: $x^2-4x+1$x²-4x+1.
Degree $3$3: $x^3+2x^2-x+5$x³+2x²-x+5.

In $x^4-3x^3+6x^2-2x+7$x⁴-3x³+6x²-2x+7: the coefficient of $x^3$x³ is $-3$-3, and the coefficient of $x^2$x² is $6$6.

In $4z^3+5z^2-11$4z³+5z²-11, there is no term containing $z^1$z¹ at all, so the coefficient of $z$z is $0$0.

In $9x^3+5x^2-8x-10$9x³+5x²-8x-10, the constant (variable-free) term is $-10$-10.

Using perimeter $=4\times\text{side}$=4× side:

side $1$1 cm $\to$→ perimeter $4(1)=4$4(1)=4 cm
side $1.5$1.5 cm $\to$→ perimeter $4(1.5)=6$4(1.5)=6 cm
side $2$2 cm $\to$→ perimeter $4(2)=8$4(2)=8 cm
side $2.5$2.5 cm $\to$→ perimeter $4(2.5)=10$4(2.5)=10 cm
side $3$3 cm $\to$→ perimeter $4(3)=12$4(3)=12 cm

As the side increases by a constant $0.5$0.5 cm each time, the perimeter increases by a constant $4\times0.5=2$4×0.5=2 cm each time ($4,6,8,10,12$4,6,8,10,12 — consecutive difference $2$2), confirming the linear relationship perimeter $=4\times\text{side}$=4× side.

Using $200+50m=750$200+50m=750: $50m=550\Rightarrow m=11$50m=550⇒ m=11. So the player played $11$11 matches.

Yes — for the expression $10x-x^2$10x-x², each input value of $x$x (the rectangle's length) produces one output value (the area of the rectangle), just like the input–output machine shown for $2x+3$2x+3.

At $x=6$x=6: output $=10(6)-6^2=60-36=24$=10(6)-6²=60-36=24. So the area is $24\text{ cm}^2$24 cm² when $x=6$x=6 cm.

(i) $5(0)-3=-3$5(0)-3=-3.
(ii) $5(-1)-3=-5-3=-8$5(-1)-3=-5-3=-8.
(iii) $5(2)-3=10-3=7$5(2)-3=10-3=7.

(i) $7(0)^2-4(0)+6=6$7(0)²-4(0)+6=6.
(ii) $7(-3)^2-4(-3)+6=7(9)+12+6=63+12+6=81$7(-3)²-4(-3)+6=7(9)+12+6=63+12+6=81.
(iii) $7(4)^2-4(4)+6=7(16)-16+6=112-16+6=102$7(4)²-4(4)+6=7(16)-16+6=112-16+6=102.

Let Salil's present age be $x$x; his mother's present age is $3x$3x. After $5$5 years: $(x+5)+(3x+5)=70\Rightarrow4x+10=70\Rightarrow4x=60\Rightarrow x=15$(x+5)+(3x+5)=70⇒4x+10=70⇒4x=60⇒ x=15.

So Salil is $15$15 years old and his mother is $3(15)=45$3(15)=45 years old.

Let the integers be $2k$2k and $5k$5k. Then $5k-2k=63\Rightarrow3k=63\Rightarrow k=21$5k-2k=63⇒3k=63⇒ k=21.

The integers are $2(21)=42$2(21)=42 and $5(21)=105$5(21)=105.

Let the number of five-rupee coins be $x$x, so the number of two-rupee coins is $3x$3x. Total value: $5x+2(3x)=88\Rightarrow5x+6x=88\Rightarrow11x=88\Rightarrow x=8$5x+2(3x)=88⇒5x+6x=88⇒11x=88⇒ x=8.

So she has $8$8 five-rupee coins and $3(8)=24$3(8)=24 two-rupee coins — a total of $8+24=32$8+24=32 coins.

Let the shorter piece be $x$x ft; the longer piece is $4x$4x ft. $x+4x=300\Rightarrow5x=300\Rightarrow x=60$x+4x=300⇒5x=300⇒ x=60.

Shorter piece $=60$=60 ft, longer piece $=240$=240 ft.

Let width $=w$=w; length $=2w+3$=2w+3. Perimeter: $2(l+w)=24\Rightarrow l+w=12\Rightarrow(2w+3)+w=12\Rightarrow3w+3=12\Rightarrow3w=9\Rightarrow w=3$2(l+w)=24⇒ l+w=12⇒(2w+3)+w=12⇒3w+3=12⇒3w=9⇒ w=3.

Length $=2(3)+3=9$=2(3)+3=9. So width $=3$=3 cm and length $=9$=9 cm. (Check: perimeter $=2(9+3)=24$=2(9+3)=24 ✓.)

The pattern of square counts is $1,3,5,7,9,\ldots$1,3,5,7,9,… — each stage adds $2$2 more squares than the previous one. Continuing from Stage $5=9$5=9: Stage $6=11$6=11, Stage $7=13$7=13.

Full sequence for Stages $1$1–$7$7: $1,\,3,\,5,\,7,\,9,\,11,\,13$1, 3, 5, 7, 9, 11, 13.

Using tiles $=2n-1$=2n-1:

Stage $15$15: $2(15)-1=29$2(15)-1=29 tiles.
Stage $26$26: $2(26)-1=51$2(26)-1=51 tiles.

For $21$21 tiles: $2n-1=21\Rightarrow2n=22\Rightarrow n=11$2n-1=21⇒2n=22⇒ n=11 (Stage $11$11).
For $47$47 tiles: $2n-1=47\Rightarrow2n=48\Rightarrow n=24$2n-1=47⇒2n=48⇒ n=24 (Stage $24$24).

Using amount left $=100-5n$=100-5n:

On day $15$15: $100-5(15)=100-75=₹25$100-5(15)=100-75=₹25 left.

Entire amount spent means the amount left is $0$0: $100-5n=0\Rightarrow n=20$100-5n=0⇒ n=20. So it takes $20$20 days to spend the entire ₹100.

Using fare $=15n-5$=15n-5 (for $n\ge2$n≥2 km): $15n-5=130\Rightarrow15n=135\Rightarrow n=9$15n-5=130⇒15n=135⇒ n=9.

So the fare is ₹130 for a $9$9 km journey. (Check: fare $=25+15(9-2)=25+105=130$=25+15(9-2)=25+105=130 ✓.)

At the end of month $1$1: $500+150=₹650$500+150=₹650. At the end of month $2$2: $650+150=₹800$650+150=₹800. At the end of month $3$3: $800+150=₹950$800+150=₹950. This continues, increasing by a constant ₹150 each month.

Linear expression for the amount at the end of the $n$nth month: $A(n)=500+150n$A(n)=500+150n.

After $1$1 hour: $120-9=111$120-9=111. After $2$2 hours: $111-9=102$111-9=102. After $3$3 hours: $102-9=93$102-9=93. This continues, decreasing by a constant $9$9 members each hour.

Linear expression: $M(n)=120-9n$M(n)=120-9n.

Area $=13\times\text{breadth}$=13× breadth.

(i) $13\times12=156\text{ cm}^2$13×12=156 cm².
(ii) $13\times10=130\text{ cm}^2$13×10=130 cm².
(iii) $13\times8=104\text{ cm}^2$13×8=104 cm².

Linear pattern: $\text{Area}(b)=13b$Area(b)=13b. Check: as breadth drops by $2$2 (from $12$12 to $10$10, or $10$10 to $8$8), area drops by a constant $26=13\times2$26=13×2 ($156\to130\to104$156→130→104), confirming a linear pattern with slope $13$13.

Volume $=7\times11\times\text{height}=77\times\text{height}$=7×11× height=77× height.

(i) $77\times5=385\text{ cm}^3$77×5=385 cm³.
(ii) $77\times9=693\text{ cm}^3$77×9=693 cm³.
(iii) $77\times13=1001\text{ cm}^3$77×13=1001 cm³.

Linear pattern: $V(h)=77h$V(h)=77h (since length $\times$× breadth $=77$=77 is fixed, volume scales linearly with height, increasing by $77\times4=308$77×4=308 for every $4$4 cm rise in height, e.g. $385\to693$385→693).

Pages left after $15$15 days $=500-20(15)=500-300=200$=500-20(15)=500-300=200 pages.

Linear pattern: pages left after $d$d days, $P(d)=500-20d$P(d)=500-20d (linear decay, decreasing by a constant $20$20 pages per day).

Using $C(d)=100+60d$C(d)=100+60d:

$C(15)=100+60(15)=100+900=₹1000$C(15)=100+60(15)=100+900=₹1000.

For $C=700$C=700: $100+60d=700\Rightarrow60d=600\Rightarrow d=10$100+60d=700⇒60d=600⇒ d=10 km.

Using $h(t)=3-0.5t$h(t)=3-0.5t: $h(5)=3-0.5(5)=3-2.5=0.5$h(5)=3-0.5(5)=3-2.5=0.5 m.

(i) $h(7)=1.75+0.5(7)=1.75+3.5=5.25$h(7)=1.75+0.5(7)=1.75+3.5=5.25 ft.

(ii)

(iii) $h(t)=1.75+0.5t$h(t)=1.75+0.5t. This is linear growth because $h$h increases by a constant amount ($0.5$0.5 ft) for every equal increase (one month) in $t$t, and the coefficient of $t$t is positive.

(i) $v(3)=10000-800(3)=10000-2400=₹7600$v(3)=10000-800(3)=10000-2400=₹7600.

(ii)

(iii) $v(t)=10000-800t$v(t)=10000-800t. This is linear decay because the value decreases by a constant ₹800 for every additional year, and the coefficient of $t$t is negative.

(i) $P(6)=750+50(6)=750+300=1050$P(6)=750+50(6)=750+300=1050.

(ii)

(iii) $P(t)=750+50t$P(t)=750+50t. This is linear growth because the population increases by a constant $50$50 people every year, and the coefficient of $t$t is positive.

(i) $b(x)=600-15x$b(x)=600-15x. This is linear decay because the balance decreases by a fixed ₹15 for every additional day, i.e. a constant negative rate of change.

(ii) Balance runs out when $b(x)=0$b(x)=0: $600-15x=0\Rightarrow15x=600\Rightarrow x=40$600-15x=0⇒15x=600⇒ x=40 days.

(iii)

$150$150 is the fixed monthly fee — the amount billed even when $x=0$x=0 GB are used (the $y$y-intercept). $20$20 is the extra cost charged per GB of data used — the rate at which the bill increases for every additional GB (the slope).

$400=10a+b$400=10a+b and $500=14a+b$500=14a+b. Subtracting: $500-400=14a-10a\Rightarrow100=4a\Rightarrow a=25$500-400=14a-10a⇒100=4a⇒ a=25.

Then $b=400-10(25)=400-250=150$b=400-10(25)=400-250=150.

So $a=25$a=25 (₹25 per module) and $b=150$b=150 (₹150 fixed fee).

$800=10a+b$800=10a+b and $1100=15a+b$1100=15a+b. Subtracting: $1100-800=15a-10a\Rightarrow300=5a\Rightarrow a=60$1100-800=15a-10a⇒300=5a⇒ a=60.

Then $b=800-10(60)=800-600=200$b=800-10(60)=800-600=200.

So $a=60$a=60 (₹60 per hour) and $b=200$b=200 (₹200 fixed fee).

$0=32a+b$0=32a+b and $100=212a+b$100=212a+b. Subtracting: $100-0=212a-32a\Rightarrow100=180a\Rightarrow a=\dfrac{100}{180}=\dfrac{5}{9}$100-0=212a-32a⇒100=180a⇒ a=100/180=5/9.

Then $b=0-32\left(\dfrac{5}{9}\right)=-\dfrac{160}{9}$b=0-32(5/9)=-160/9.

So $a=\dfrac59$a=5/9 and $b=-\dfrac{160}{9}$b=-160/9 — this matches the familiar conversion $°C=\dfrac59(°F-32)$°C=5/9(°F-32), since $\dfrac59°F-\dfrac{160}{9}=\dfrac59(°F-32)$5/9°F-160/9=5/9(°F-32). (Check: at $°F=32$°F=32: $\frac59(32)-\frac{160}{9}=\frac{160}{9}-\frac{160}{9}=0$5/9(32)-160/9=160/9-160/9=0 ✓; at $°F=212$°F=212: $\frac59(212)-\frac{160}{9}=\frac{1060}{9}-\frac{160}{9}=\frac{900}{9}=100$5/9(212)-160/9=1060/9-160/9=900/9=100 ✓.)

Using $y=2x+1$y=2x+1:

$x=1\Rightarrow y=3$x=1⇒ y=3 (given); $x=2\Rightarrow y=2(2)+1=5$x=2⇒ y=2(2)+1=5; $x=5\Rightarrow y=2(5)+1=11$x=5⇒ y=2(5)+1=11; $x=7\Rightarrow y=15$x=7⇒ y=15 (given); $x=9\Rightarrow y=2(9)+1=19$x=9⇒ y=2(9)+1=19; $x=12\Rightarrow y=2(12)+1=25$x=12⇒ y=2(12)+1=25; $x=20\Rightarrow y=2(20)+1=41$x=20⇒ y=2(20)+1=41.

Completed table:

Both lines share the same $y$y-intercept, $(0,1)$(0,1) (since $b=1$b=1 for both), so they cross the $y$y-axis at the same point. But their slopes are opposite: $y=3x+1$y=3x+1 has slope $+3$+3 (rises left to right — linear growth), while $y=-3x+1$y=-3x+1 has slope $-3$-3 (falls left to right — linear decay). The two lines cross only at $(0,1)$(0,1) and are mirror images of one another in the vertical line through that point.

Yes — when the slope $a$a is kept fixed and only the $y$y-intercept $b$b changes, all the resulting lines are parallel to each other (same steepness and direction) but cross the $y$y-axis at different points $(0,b)$(0,b). Increasing $b$b shifts the line straight up; decreasing $b$b shifts it straight down — the slope itself never changes.

All three lines pass through the origin (since $b=0$b=0 for all). Their slopes are $4$4, $2$2, $1$1 respectively — $y=4x$y=4x is the steepest, $y=x$y=x is inclined equally to both axes (at $45°$45°), and $y=2x$y=2x lies between them. As $a$a increases (with $b$b fixed at $0$0), the line becomes steeper.

All three lines pass through the origin. Their slopes are $-6$-6, $-3$-3, $-1$-1 — all negative, so all fall left to right (linear decay through the origin). $y=-6x$y=-6x is the steepest downward slope, $y=-x$y=-x is the shallowest of the three.

Both pass through the origin and are equally steep ($|a|=5$|a|=5) but in opposite directions: $y=5x$y=5x rises left to right, $y=-5x$y=-5x falls left to right. They are mirror images of each other (reflections in the $x$x-axis, and also in the $y$y-axis) and meet only at the origin.

All three share the same slope $a=3$a=3, so they are parallel to one another, but have different $y$y-intercepts: $-1$-1, $0$0, $+1$+1. So $y=3x-1$y=3x-1 crosses the $y$y-axis below the origin, $y=3x$y=3x passes through the origin, and $y=3x+1$y=3x+1 crosses above it — changing only $b$b shifts a parallel line straight up or down.

The first two lines, $y=-2x-3$y=-2x-3 and $y=-2x$y=-2x, share slope $-2$-2, so they are parallel to each other (shifted vertically by their differing $y$y-intercepts, $-3$-3 and $0$0). The third, $y=2x+3$y=2x+3, has the opposite slope $+2$+2, so it is not parallel to the first two — it rises where they fall; it merely happens to have a $y$y-intercept ($+3$+3) that is the negative of the first line's ($-3$-3).

Any valid answer works, e.g. $x^3-7x^2+2x+5$x³-7x²+2x+5 (degree $3$3, and the coefficient of $x^2$x² is $-7$-7).

(i) $5(1)^2-3(1)+7=5-3+7=9$5(1)²-3(1)+7=5-3+7=9.

(ii) Substituting $t=a$t=a: $4a^3-a^2+6$4a³-a²+6.

Let the number be $x$x. Then $5x+2=17\Rightarrow5x=15\Rightarrow x=3$5x+2=17⇒5x=15⇒ x=3.

The number is $3$3.

Let the smaller number be $x$x; the other is $5x$5x. Adding $21$21 to both: $x+21$x+21 and $5x+21$5x+21. Since $5x+21$5x+21 is naturally the larger, it equals twice the smaller new number: $5x+21=2(x+21)\Rightarrow5x+21=2x+42\Rightarrow3x=21\Rightarrow x=7$5x+21=2(x+21)⇒5x+21=2x+42⇒3x=21⇒ x=7.

So the numbers are $7$7 and $5(7)=35$5(7)=35. (Check: new numbers are $28$28 and $56$56, and $56=2\times28$56=2×28 ✓.)

Using $A(n)=800+250n$A(n)=800+250n ($n$n = number of months):

(i) After $6$6 months: $800+250(6)=800+1500=₹2300$800+250(6)=800+1500=₹2300.

(ii) $2$2 years $=24$=24 months: $800+250(24)=800+6000=₹6800$800+250(24)=800+6000=₹6800.

Linear pattern: $A(n)=800+250n$A(n)=800+250n.

Let the tens digit be $x$x and the units digit be $y$y. Original number $=10x+y$=10x+y; reversed number $=10y+x$=10y+x. Given: $(10x+y)+(10y+x)=143\Rightarrow11x+11y=143\Rightarrow x+y=13$(10x+y)+(10y+x)=143⇒11x+11y=143⇒ x+y=13, and $|x-y|=3$|x-y|=3.

Case $x-y=3$x-y=3: with $x+y=13$x+y=13, adding gives $2x=16\Rightarrow x=8,\;y=5$2x=16⇒ x=8, y=5 — number $=85$=85 (reversed $=58$=58); check $85+58=143$85+58=143 ✓ and digits differ by $3$3 ✓.

Case $y-x=3$y-x=3: gives $x=5,\;y=8$x=5, y=8 — number $=58$=58 (reversed $=85$=85) — the same pair of numbers.

So the two numbers are 58 and 85.

Rewrite each in the form $y=ax+b$y=ax+b:

(i) $y=-3x+4$y=-3x+4: slope $=-3$=-3, $y$y-intercept $=4$=4, cuts the $y$y-axis at $(0,4)$(0,4).

(ii) $2y=4x+7\Rightarrow y=2x+3.5$2y=4x+7⇒ y=2x+3.5: slope $=2$=2, $y$y-intercept $=3.5$=3.5, cuts the $y$y-axis at $(0,3.5)$(0,3.5).

(iii) $5y=6x-10\Rightarrow y=1.2x-2$5y=6x-10⇒ y=1.2x-2: slope $=1.2$=1.2 (i.e. $\tfrac65$6/5), $y$y-intercept $=-2$=-2, cuts the $y$y-axis at $(0,-2)$(0,-2).

(iv) $3y=6x-11\Rightarrow y=2x-\tfrac{11}{3}$3y=6x-11⇒ y=2x-11/3: slope $=2$=2, $y$y-intercept $=-\tfrac{11}{3}\approx-3.67$=-11/3≈-3.67, cuts the $y$y-axis at $\left(0,-\tfrac{11}{3}\right)$(0,-11/3).

To graph each, plot the $y$y-intercept and use the slope (rise/run) to find a second point, then draw the line through them.

Parallel lines: (ii) and (iv) both have slope $2$2, so lines (ii) and (iv) are parallel to each other.

(i) $y=\dfrac95(313-273)+32=\dfrac95(40)+32=72+32=104°F$y=9/5(313-273)+32=9/5(40)+32=72+32=104°F.

(ii) $158=\dfrac95(x-273)+32\Rightarrow126=\dfrac95(x-273)\Rightarrow x-273=126\times\dfrac59=70\Rightarrow x=343\text{ K}$158=9/5(x-273)+32⇒126=9/5(x-273)⇒ x-273=126× 5/9=70⇒ x=343 K.

Work $=$= force $\times$× distance, so with force $=3$=3 (constant): $w=3d$w=3d. This is a linear equation with slope $3$3 and $y$y-intercept $0$0 — its graph is a straight line through the origin passing through points like $(0,0)$(0,0), $(1,3)$(1,3), $(2,6)$(2,6).

When $d=2$d=2: $w=3(2)=6$w=3(2)=6 units. On the graph, the point $(2,6)$(2,6) lies exactly on the line $w=3d$w=3d, confirming the answer.

(i) Let $p(x)=ax+b$p(x)=ax+b. From $(1,5)$(1,5): $5=a+b$5=a+b. From $(3,11)$(3,11): $11=3a+b$11=3a+b. Subtracting: $6=2a\Rightarrow a=3$6=2a⇒ a=3; then $b=5-3=2$b=5-3=2. So $p(x)=3x+2$p(x)=3x+2.

(ii) $y$y-axis: $x=0\Rightarrow p(0)=2$x=0⇒ p(0)=2, point $(0,2)$(0,2). $x$x-axis: $p(x)=0\Rightarrow3x+2=0\Rightarrow x=-\tfrac23$p(x)=0⇒3x+2=0⇒ x=-2/3, point $\left(-\tfrac23,0\right)$(-2/3,0).

(iii) Plot $(0,2)$(0,2) and $(1,5)$(1,5) (or $(3,11)$(3,11)) and draw the line through them; it should also pass through $\left(-\tfrac23,0\right)$(-2/3,0). Check: $p(1)=3(1)+2=5$p(1)=3(1)+2=5 ✓, $p(3)=3(3)+2=11$p(3)=3(3)+2=11 ✓ — matching the given points.

(i) $p(0)=b=5$p(0)=b=5.

(iii) $p(x)+q(x)=(a+c)x+(b+d)=6x+4\Rightarrow a+c=6$p(x)+q(x)=(a+c)x+(b+d)=6x+4⇒ a+c=6 and $b+d=4$b+d=4. Since $b=5$b=5, $d=4-5=-1$d=4-5=-1.

(ii) $p(x)-q(x)=(a-c)x+(b-d)$p(x)-q(x)=(a-c)x+(b-d), and this is $0$0 at $x=3$x=3: $(a-c)(3)+(b-d)=0$(a-c)(3)+(b-d)=0. Here $b-d=5-(-1)=6$b-d=5-(-1)=6, so $3(a-c)+6=0\Rightarrow a-c=-2$3(a-c)+6=0⇒ a-c=-2.

Now solve $a+c=6$a+c=6 and $a-c=-2$a-c=-2: adding gives $2a=4\Rightarrow a=2$2a=4⇒ a=2, so $c=4$c=4.

So $p(x)=2x+5$p(x)=2x+5 and $q(x)=4x-1$q(x)=4x-1.

Check: $p(0)=5$p(0)=5 ✓; $p(x)-q(x)=(2x+5)-(4x-1)=-2x+6$p(x)-q(x)=(2x+5)-(4x-1)=-2x+6, which is $0$0 at $x=3$x=3 ✓; $p(x)+q(x)=(2x+5)+(4x-1)=6x+4$p(x)+q(x)=(2x+5)+(4x-1)=6x+4 ✓.

Figure: Stage 1 is a single hexagon ($6$6 matchsticks). Stage 2 attaches a second hexagon sharing one edge with the first (adding $6-1=5$6-1=5 new matchsticks). Stage 3 attaches a third hexagon sharing an edge with the second (another $+5$+5), and so on — the hexagons form a chain, each new one welded onto the free end of the last.

(i) Stage 4 $=16+5=21$=16+5=21 matchsticks. Stage 5 $=21+5=26$=21+5=26 matchsticks (continue the chain by attaching one more hexagon at each stage, sharing an edge with the previous one).

(ii)

(iii) The first stage uses $6$6 matchsticks, and every later stage adds $5$5: matchsticks$(n) = 6+5(n-1)=5n+1$(n) = 6+5(n-1)=5n+1.

(iv) Stage $15$15: $5(15)+1=76$5(15)+1=76 matchsticks.

(v) Solve $5n+1=200\Rightarrow5n=199\Rightarrow n=39.8$5n+1=200⇒5n=199⇒ n=39.8 — not a whole number. Since $n$n must be a positive integer, no, $200$200 matchsticks cannot form any complete stage of this pattern.

Finding p(x): From $(2,3)$(2,3): $3=2a+b$3=2a+b. From $(6,11)$(6,11): $11=6a+b$11=6a+b. Subtracting: $8=4a\Rightarrow a=2$8=4a⇒ a=2; then $b=3-2(2)=-1$b=3-2(2)=-1. So $p(x)=2x-1$p(x)=2x-1.

Finding q(x): Since $q$q is parallel to $p$p, $c=a=2$c=a=2. Using $(4,-1)$(4,-1): $-1=2(4)+d=8+d\Rightarrow d=-9$-1=2(4)+d=8+d⇒ d=-9. So $q(x)=2x-9$q(x)=2x-9.

Where each line meets the x-axis:

$p$p: $0=2x-1\Rightarrow x=0.5$0=2x-1⇒ x=0.5, point $(0.5,\,0)$(0.5, 0).

$q$q: $0=2x-9\Rightarrow x=4.5$0=2x-9⇒ x=4.5, point $(4.5,\,0)$(4.5, 0).

Factor: $f(x)=ax+a=a(x+1)$f(x)=ax+a=a(x+1). So $f(-1)=a(-1+1)=0$f(-1)=a(-1+1)=0 for every positive value of $a$a — meaning every such line passes through the same point, $(-1,\,0)$(-1, 0), on the $x$x-axis, regardless of $a$a. Also, since $a>0$a>0, every such line has a positive slope (linear growth), and its $y$y-intercept ($b=a$b=a) always equals its slope.