Two Variables, One Line (Do Variables, Ek Line): Class 9 Maths Chapter 13 Hinglish mein

Ganita Manjari Part 2 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 13 Oct 2026 · Updated 13 Oct 2026

85 min
2 simulations
44 practice questions
23 MCQs

Is chapter mein aap kya seekhenge

Do variables wali linear equation ko standard form ax + by + c = 0 mein likho aur uske coefficients aur constant term ke naam batao.
Check karo ki koi ordered pair solution hai ya nahi, linear equation ke solutions nikaalo, aur samjhao ki solutions infinitely many kyun hote hain.
Linear equation ka graph uske do points se banao aur graph se aur solutions padho.
Do points se line ka slope nikaalo, aur y = mx + d se slope aur y-intercept padho.
Kisi situation se linear equations ka pair banao aur use substitution, elimination aur graphs banakar solve karo.
Ratios a1/a2, b1/b2 aur c1/c2 se tay karo ki pair ka ek solution hai, koi solution nahi hai ya infinitely many solutions hain.

NCERT Mathematics Class 9, Chapter 13, pp. 85–122 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

Tum $y = 2 x + 1$ jaise rules ko pehle hi straight lines ke roop mein bana chuke ho (Part 1, Chapter 2, Introduction to Linear Polynomials). Yahan hum is baat ko ulta karte hain. Hum ek aisi equation se shuru karte hain jo do unknown numbers ko aapas mein jodti hai, aur poochhte hain ki numbers ke kaunse pairs usmein fit hote hain. Answers hamesha ek straight line par line up hote hain. Phir hum aisi do equations ko saath rakhte hain aur woh pair dhoondhte hain, agar koi hai, jo dono mein fit ho.

13.1 Do Variables wali Linear Equations

School canteen mein ek samosa ₹15 ka hai aur idli ki ek plate ₹20 ki. Ishaan apne dosto ke liye dono mein se kuch order karta hai, aur bill ₹110 aata hai. Baad mein use yaad nahi rehta ki usne kaun si cheez kitni order ki thi. Kya hum phir bhi jo pata hai woh likh sakte hain?

Do amounts unknown hain, isliye har ek ko apna letter milta hai. Maan lo $x$ samosas ki sankhya hai aur $y$ idli plates ki sankhya. Samosas ki cost $15 x$ rupees hai aur plates ki cost $20 y$ rupees. Dono milkar bill banate hain:

15 x + 20 y = 110

Left side dono tarah ke khaane ki cost jodti hai; right side woh hai jo Ishaan ne sach mein pay kiya. Aisi equation mein do letters hote hain, har ek sirf first power mein, aur kabhi aapas mein multiply nahi hote. Ise linear equation in two variables (दो चरों वाला रैखिक समीकरण) kehte hain.

3/3 $x = 2, y = 4$ aur $x = 6, y = 1$ dono se ₹110 ka bill banta hai.

Fig. 13.1Ek bill, do alag orders jo usmein fit hote hain.NCERT p. 85

Upar do price tags: samosa 15 rupees ka ek aur idli plate 20 rupees ki ek, neeche unknown counts x samosas aur y plates likhe hain. Doosre step mein beech mein ek bill box aata hai jismein likha hai 15x plus 20y equals 110. Teesre step mein neeche do trays aati hain: left tray mein 2 samosas aur 4 idli plates hain, label x equals 2, y equals 4 aur 30 plus 80 equals 110; right tray mein 6 samosas aur 1 plate hai, label x equals 6, y equals 1 aur 90 plus 20 equals 110.

Fig. 13.1 mein trays dekho: sirf bill se pata nahi chalta ki Ishaan ne kya order kiya, kyunki do alag orders usmein fit hote hain. Yeh baat yaad rakho. Is chapter ka bada hissa isi baare mein hai ki do variables wali ek equation humein kya bata sakti hai, aur kya nahi.

13.1.1 Do Variables wali Linear Equations ka Standard Form

Do variables wali kisi bhi linear equation ko aise rearrange kiya ja sakta hai ki har term ek side par ho aur doosri side par 0. Is saaf layout se equations ko compare karna aasaan ho jaata hai.

Definition

Do variables wali linear equation

Aisi equation jise $a x + b y + c = 0$ mein rearrange kiya ja sake, kuch real numbers $a,$ $b,$ $c,$ ke liye, jahan $a$ aur $b$ mein se kam se kam ek non-zero ho.

Example $15 x + 20 y - 110 = 0,$ jahan $a = 15,$ $b = 20,$ $c = - 110.$

Standard form

a x + b y + c = 0

(13.1)

jahan

$a$: $x$ ka coefficient
$b$: $y$ ka coefficient
$c$: constant term ( $a$ aur $b$ dono 0 nahi)

Hum $a x + b y + c = 0$ ko standard form (मानक रूप) kehte hain. Jo number $x$ ko multiply karta hai, woh coefficient of $x,$ kehlata hai, aur isi tarah $y,$ ke liye bhi, jabki $c$ constant term hai. Ishaan ke bill ka standard form hai $15 x + 20 y - 110 = 0.$

Practice mein do baatein kaam aati hain. Pehli, coefficients fractions ya decimals bhi ho sakte hain. Jaise, $\frac{2}{3} x - \frac{1}{4} y + \frac{5}{6} = 0$ pehle se standard form mein hai. Har term ko 12 se multiply karne par, jo 3, 4 aur 6 ka LCM hai, zyada saaf $8 x - 3 y + 10 = 0,$ milta hai aur solutions nahi badalte.

Doosri, koi ek letter gayab bhi ho sakta hai. Equation $9 - 3 x = 0$ ko $- 3 x + 0 \cdot y + 9 = 0,$ likha ja sakta hai, isliye yeh $b = 0.$ ke saath do variables wali linear equation gini jaati hai. Isi tarah, $2 y = 7$ asal mein $0 \cdot x + 2 y - 7 = 0$ hai, jahan $a = 0.$ Sirf dono letters ka gayab hona allowed nahi hai, kyunki tab equation in dono ke baare mein kuch nahi kehti: $x$ ya $y .$

Example 13.1NCERT Example 1 · p. 86

Equations ko standard form mein likhna

Har equation ko $a x + b y + c = 0$ form mein likho aur batao $a,$ $b$ aur $c .$

$3 x - 4 y = 7.5$
$\frac{y}{3} + 2 = \frac{5 x}{2}$
$6 = 0.4 y - 2.5 x$
$2 x = - 9$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Inmein se kaunsi do variables wali linear equation nahi hai?

2 x - 7 = 3 y

x = 5

x^{2} + y = 4

\frac{1}{2} x + y = 0

Answer dekho

Answer (C) $x^{2} + y = 4$

x^{2}

second power hai, isliye

x^{2} + y = 4

linear nahi hai.

x = 5

mein lagta hai ki sirf ek letter hai, lekin yeh

x + 0 \cdot y - 5 = 0,

hai, jo allowed hai.

Practice · 3 questions

Practice 13.1: answers ke saath

Short answerEasyNCERT Exercise Set 13.1, Q1
Do variables wali ek linear equation likho jismein $a = 0,$ $b = 5$ aur $c = - \frac{3}{4} .$ Phir use whole-number coefficients ke saath dobara likho.
Solution dikhao
Answer $0 \cdot x + 5 y - \frac{3}{4} = 0,$ ya $20 y - 3 = 0.$
Substitute karne par $0 \cdot x + 5 y - \frac{3}{4} = 0,$ milta hai, yaani $5 y - \frac{3}{4} = 0.$ Har term ko 4 se multiply karne par milta hai $20 y - 3 = 0.$ Dono har us point ko describe karte hain jahan $y = 0.15,$ chahe $x$ kuch bhi ho.
2
Short answerEasyNCERT Exercise Set 13.1, Q2
Har equation ko positive first coefficient ke saath standard form mein likho. Phir uske do coefficients aur constant term batao.
1. $y + 4 = 3 x$
  Solution dikhao
  $3 x - y - 4 = 0 ⁣ :$ $x$ ka coefficient 3 hai, $y$ ka $- 1,$ constant $- 4.$
2. $5 y = 2 x$
  Solution dikhao
  $2 x - 5 y = 0 ⁣ :$ coefficients 2 aur $- 5,$ constant 0.
3. $x = - 6$
  Solution dikhao
  $x + 0 \cdot y + 6 = 0 ⁣ :$ coefficients 1 aur 0, constant 6.
4. $4 y = 9$
  Solution dikhao
  $0 \cdot x + 4 y - 9 = 0 ⁣ :$ coefficients 0 aur 4, constant $- 9.$
3
Short answerEasyNCERT Exercise Set 13.1, Q3
Tay karo ki kaunsi equation har situation ko describe karti hai.
1. Chawal ka ek bag cheeni ke bag se teen guna bhaari hai. Chawal ka bag $r$ kg aur cheeni ka bag $s$ kg ho, to Asha likhti hai $r = 3 s$ aur Dev likhta hai $s = 3 r .$ Kaun sahi hai?
  Asha, $r = 3 s$ Dev, $s = 3 r$ Dono, kyunki dono ek hi baat kehte hain
  Answer dekho
  Answer (A) Asha, $r = 3 s$
  Chawal ka bag zyada bhaari hai, isliye $r$ ko 3 times $s .$ hona chahiye. Test karo: agar $s = 2,$ to chawal 6 kg ka hai, aur yeh sirf $r = 3 s$ se milta hai.
2. Ek kabaddi match mein do raiders ne milkar 23 points banaye. Ek ne $x$ points banaye aur doosre ne $y .$ Kya $x + y = 23$ sahi equation hai? Kya $x - y = 23$ sahi hai?
  Solution dikhao
  Answer $x + y = 23$ sahi hai; $x - y = 23$ nahi.
  'Milkar' ka matlab hai ki dono scores jode jaate hain, isliye $x + y = 23$ sahi hai. Equation $x - y = 23$ yeh kahegi ki ek raider ne doosre se 23 points zyada banaye, jo ek alag baat hai.

13.2 Do Variables wali Linear Equation ka Solution

Tum yeh pehle se solve kar sakte ho: $2 x + 5 = 17 ⁣ :$ iska bilkul ek answer hai, $x = 6.$ Do letters ke saath answer ka idea badalna padta hai, kyunki ek akele number ko test nahi kiya ja sakta, jaise is equation mein: $3 x + 4 y = 24.$ Humein har letter ke liye ek value chahiye.

13.2.1 Solution ka Algebraic Matlab

Do letters wali equation ke liye solution (हल) mein har letter ki ek value chahiye: ek $x$ -value aur ek $y$ -value jo saath milkar dono sides ko barabar kar dein. Lo $3 x + 4 y = 24.$ Jab $x = 4$ aur $y = 3,$ to left side hai $3 (4) + 4 (3) = 12 + 12 = 24.$ To $x = 4,$ $y = 3$ ek solution hai.

Is solution ko hum $(4, 3),$ likhte hain, pehle $x$ aur phir $y$ . Order matter karta hai, isiliye $(4, 3)$ ko ordered pair (क्रमित युग्म) kehte hain. Numbers ki jagah badlo aur $(3, 4) ⁣ :$ test karo: $3 (3) + 4 (4) = 9 + 16 = 25,$ 24 nahi. To $(3, 4)$ solution nahi hai, bhale hi usmein wahi do numbers hain.

Kitne solutions hain? $x$ ki koi bhi value chuno aur rakh do; tumhare paas ek one-variable equation bachti hai, is ke liye: $y .$ $x = 0$ se $4 y = 24,$ milta hai, isliye $(0, 6)$ ek solution hai. $x = 1$ se $4 y = 21,$ milta hai, isliye $(1, \frac{21}{4})$ bhi ek hai. $x = - 4$ se $4 y = 36,$ milta hai, isliye $(- 4, 9)$ bhi chalta hai. $x$ ki har value bilkul ek $y,$ deti hai, aur chunne ke liye values ka koi ant nahi. Isliye equation ke infinitely many solutions hain.

13.2.2 Solutions ko Geometrically Dekhna

Ab $3 x + 4 y = 24$ ke kuch solutions ko grid par points ki tarah plot karo. Pehle $y$ ko subject banao. $4 y = - 3 x + 24$ se humein milta hai

y = - \frac{3}{4} x + 6

Iska form $y = m x + d,$ hai, jise tum Part 1 mein is roop mein mil chuke ho: $y = a x + b .$ Wahan tumne dekha tha ki aisa rule hamesha straight line deta hai. To $3 x + 4 y = 24$ ka har solution ek hi line par ek point hai, aur us line ka har point ek solution hai. Yeh line equation ka graph (आलेख) hai: uske saare infinitely many solutions ki ek saath ek picture.

4/4 $(3, 4)$ se $3 (3) + 4 (4) = 25,$ milta hai, isliye yeh line par nahi hai.

Fig. 13.2

3 x + 4 y = 24

ka har solution ek hi straight line par hota hai.NCERT Fig. 13.1 · p. 89

Ek grid jo across minus 5 se 9 tak aur upar minus 2 se 10 tak hai, right mein working likhi hai. Step 1 mein (0, 6) aur (8, 0) plot hote hain, x equals 0 aur x equals 8 se. Step 2 mein (4, 3) aur (minus 4, 9) judte hain. Step 3 mein chaaron points se hokar ek straight line banti hai, aur note likha hai ki sab ek hi line par hain. Step 4 mein point (3, 4) ko line se zara hatkar dashed ring se mark kiya jaata hai aur note hai ki yeh 24 nahi, 25 deta hai, isliye solution nahi hai.

SimulationKya point line par hai?

Ek coordinate grid jo dono axes par minus 6 se 10 tak hai. Teen fixed points, C (minus 4, 9), A (0, 6) aur B (8, 0), line 3x plus 4y equals 24 par hain aur measured segments se jude hain, isliye line C se B tak bani hai. Chautha point P (6, 6) se shuru hota hai. Tum P ko drag kar sakte ho; yeh half units par snap hota hai aur apne coordinates dikhata hai, taaki tum test kar sako ki woh equation satisfy karta hai ya nahi.

Fig. 13.3

C

A

hote hue

B

tak ke segments is line ka hissa hain:

3 x + 4 y = 24.

P

ko drag karo aur uske coordinates test karo.

Try karo

$P$ ko line par kisi aise point par drag karo jiske coordinates whole numbers hon, aur jo $A,$ $B$ aur $C .$ na ho. Check karo ki wahan $3 x + 4 y = 24$ hai.
$P$ ko $(6, 1.5),$ par rakho, phir is point par: $(6, 2) .$ Kaunsa line par hai? Dono ko equation mein test karo.
$P$ ko is point par rakho: $(3, 4) .$ Yeh line ke kareeb dikhta hai: wahan $3 x + 4 y$ kya deta hai?

Kyunki do points ek line fix kar dete hain, use banane ke liye koi bhi do solutions kaafi hain. Sabse jaldi milne wale do aksar axis crossings hote hain: $y$ -axis ke har point par $x = 0,$ aur $x$ -axis ke har point par $y = 0.$

Example 13.2NCERT Example 2 · p. 89

Do aasaan points se graph

$x + 2 y = 5$ ke do solutions nikaalo aur uska graph banao. Graph se chaar aur solutions dhoondho.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

4/4Line par pade grid corners aur solutions hain.

Fig. 13.4Do axis crossings is line ko fix kar dete hain:

x + 2 y = 5;

phir aur solutions padhe ja sakte hain.NCERT Fig. 13.2 · p. 90

Ek grid jo across minus 3 se 8 tak aur upar minus 2 se 4 tak hai. Step 1 mein y-axis par point (0, 2.5) ko ring kiya gaya hai, jo x equals 0 rakhkar mila. Step 2 mein x-axis par (5, 0) ko ring kiya gaya hai, jo y equals 0 rakhkar mila. Step 3 mein dono points ko straight line se joda jaata hai. Step 4 mein chaar aur points mark hote hain jahan line grid corners se guzarti hai: (minus 1, 3), (1, 2), (3, 1) aur (7, minus 1).

Example 13.3NCERT Example 3 · p. 90

Do lines, do cases

Har equation ke do solutions nikaalo aur uska graph banao: (i) $5 x + 2 y = 10$ (ii) $3 x - 4 y = 0.$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 13.4NCERT Example 4 · p. 92

Missing coefficients nikaalna

Pair $(2, - 5)$ in dono equations ka solution hai: $p x - 2 y = 16$ aur $3 x + q y = - 14.$ Nikaalo $p$ aur $q .$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

13.3 Do Variables wali Linear Equation ka Slope-Intercept Form

13.3.1 Slope ko samajhna

Railway station par wheelchair ramp dheere-dheere upar chadhta hai, jabki playground ki slide tezi se chadhti hai. Dono upar jaate hain; fark yeh hai ki har kadam aage badhne par har ek kitna upar uthta hai. Graph par lines bhi isi tarah alag hoti hain, aur maths is baat ko ek number se pakadta hai: line ka slope (ढाल), jise gradient bhi kehte hain.

Graph ko left se right padho, jaise tum ek sentence padhte ho. Agar chalte-chalte line upar chadhti hai, to uska slope positive hai. Agar neeche girti hai, to uska slope negative hai. Chadhaai ya girawat jitni steep hogi, slope ka size utna bada hoga.

Slope naapne ke liye line par do points chuno, $A (x_{1}, y_{1})$ aur $B (x_{2}, y_{2}),$ aur $A$ se $B$ tak do moves mein chalo: pehle across, phir upar ya neeche. Across wale move $x_{2} - x_{1},$ ko run kehte hain, aur upar-ya-neeche wale move $y_{2} - y_{1},$ ko rise kehte hain. Dono ke saath sign hota hai: left ki taraf ya neeche ki taraf move negative gina jaata hai. Slope rise ko run se divide karke milta hai.

Slope = rise ÷ run

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

(13.2)

jahan

$y_{2} - y_{1}$: pehle point se doosre point tak rise
$x_{2} - x_{1}$: run, jo 0 nahi hona chahiye

4/4 $A$ se is point tak chalo, $C (9, 8) ⁣ :$ rise 6 aur run 8, aur phir se $\frac{6}{8} = \frac{3}{4}$

Fig. 13.5Line par kinhi bhi do points ke liye rise-to-run ratio same rehta hai.NCERT Fig. 13.7 · p. 96

First quadrant ka ek grid jo across 0 se 10 tak aur upar 0 se 9 tak hai, aur usmein A (1, 2) aur B (5, 5) se guzarti ek upar jaati line hai. Step 2 mein A se right ki taraf 4 units ka arrow banta hai, label run equals 4. Step 3 mein B tak 3 units upar ka arrow banta hai, label rise equals 3, aur side panel mein rise over run equals 3 over 4 likha hai. Step 4 mein C (9, 8) mark hota hai aur line ke upar A se C tak ek bada dashed triangle banta hai, rise 6 aur run 8 ke saath, aur panel mein 6 over 8 equals 3 over 4 dikhta hai.

Slope same aaya, chahe $A$ se $B$ jao ya $A$ se $C .$ Yeh koi sanyog nahi hai. Straight line par har rise-and-run triangle ki shape same hoti hai, bas chhota ya bada hota hai, isliye ratio kabhi nahi badalta. Tum kaunse do points chunte ho, isse koi fark nahi padta.

SimulationDo points ke beech rise aur run

Ek coordinate grid jo dono axes par minus 2 se 10 tak hai, aur usmein do draggable points hain, A (1, 2) par aur B (5, 5) par, jo ek measured segment se jude hain. Segment ke right-angled legs bane hain aur unpar delta x aur delta y likha hai, yaani run aur rise, bina signs ke sizes. Drag karte waqt points whole units par snap hote hain aur apne coordinates dikhate hain.

Fig. 13.6Legs

Δ x

(run) aur

Δ y

(rise) in do points ke beech bane hain:

A

aur

B;

ye sirf sizes dikhate hain, isliye signs khud tay karo.

Try karo

$Δ x$ aur $Δ y$ padho, in points ke liye: $A (1, 2)$ aur $B (5, 5) .$ Slope $Δ y \div Δ x$ kya hai?
$B$ ko $(9, 8),$ tak drag karo, usi line par aur aage. Run aur rise badalte hain: kya slope badalta hai?
$B$ ko yahan drag karo: $(5, - 1) .$ Ab rise upar hai ya neeche, aur slope ka sign kya hai? Phir $B$ ko theek upar rakho is point ke: $A,$ yaani $(1, 7) ⁣ :$ slope kyun nahi nikaala ja sakta?

Example 13.5NCERT Example 5 · p. 97

Teen points se slope

Points $P (- 2, - 1),$ $Q (0, 3)$ aur $R (3, 9)$ ek hi line par hain. Uska slope nikaalo, aur points ke ek doosre pair se check karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 13.6NCERT Example 6 · p. 97

Neeche girti line

Us line ka slope nikaalo jo in points se guzarti hai: $A (- 3, 4)$ aur $B (3, 1) .$ Kya isse fark padta hai ki tum kaunsa point pehle lete ho?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

13.3.2 Slope-Intercept Form

Part 1 mein tum rule $y = a x + b,$ se mile the, jahan $a$ steepness tay karta tha aur $b$ woh height thi jahan line $y$ -axis ko cross karti thi. Yeh chapter, tumhari textbook ki tarah, isi rule ko in letters ke saath likhta hai: $m$ aur $d .$

Slope-intercept form

y = m x + d

(13.3)

jahan

$m$: line ka slope
$d$: $y$ -intercept, isliye $(0, d)$ graph par hai

Yahan $m$ slope hai aur $d$ $y$ -intercept (अंतःखंड) hai. Kyunki dono seedhe equation se padhe ja sakte hain, $y = m x + d$ ko slope-intercept form kehte hain. Lekin $m$ theek rise-over-run slope hi kyun ho? Iska reason yeh hai.

Derivation

m

slope kyun hai

Maan lo $A (x_{1}, y_{1})$ aur $B (x_{2}, y_{2})$ line $y = m x + d,$ ke koi bhi do points hain, jahan $x_{1} \neq x_{2} .$ Dono points equation satisfy karte hain, isliye $y_{1} = m x_{1} + d$ aur $y_{2} = m x_{2} + d .$

Doosri mein se pehli ghatao. Dono $d$ 's cancel ho jaate hain, aur bachta hai $y_{2} - y_{1} = m x_{2} - m x_{1},$ aur common factor $m$ bahar nikaalne par milta hai $y_{2} - y_{1} = m (x_{2} - x_{1}) .$ Shabdon mein:

rise = m \times run

Run $x_{2} - x_{1}$ zero nahi hai, isliye hum isse divide kar sakte hain:

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

$d$ 's cancel ho gaye, aur $A$ aur $B$ line ke koi bhi do points the. Isliye points ka har pair same rise-over-run ratio deta hai, aur woh ratio hai $m .$

Koi bhi equation $a x + b y + c = 0$ jismein $b \neq 0$ ho, is form mein likhi ja sakti hai. $a x$ aur $c$ ko right-hand side par le jao, phir is se divide karo: $b ⁣ :$

Standard form se

y = - \frac{a}{b} x - \frac{c}{b}

(13.4)

jahan

$m$: $- \frac{a}{b},$ slope
$d$: $- \frac{c}{b},$ $y$ -intercept

ExampleNCERT p. 99

Slope aur intercept padhna

$4 x - 2 y = 6$ ko slope-intercept form mein likho. $m$ aur $d$ tumhe kya batate hain? Values ki ek chhoti table se line banao.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

$m$ ka sign lines ko alag-alag kinds mein baant deta hai, jaisa Table 13.1 aur Fig. 13.7 dikhate hain. $y = - 2$ jaisi horizontal line kabhi upar nahi uthti, isliye $m = 0.$ $x = - 3$ jaisi vertical line sabse alag hai: uske kinhi bhi do points ke beech run 0 hai, aur hum 0 se divide nahi kar sakte. Isliye vertical line ka koi slope nahi hota (uska slope undefined hai), aur yahi ek aisi line hai jise $y = m x + d$ describe nahi kar sakta.

Table 13.1Chaar tarah ki lines

Slope	Line	Example
$m > 0$	right ki taraf upar jaati hai	$y = \frac{1}{2} x + 1$
$m < 0$	right ki taraf neeche girti hai	$y = - 2 x + 3$
$m = 0$	horizontal	$y = - 2$
koi nahi	vertical	$x = - 3$

5/5Chaaron tarah ki lines ek saath.

Fig. 13.7Upar jaati, neeche girti, flat aur khadi lines.NCERT Figs. 13.12–13.13 · p. 100

Ek grid jo dono axes par minus 5 se 5 tak hai, right mein legend hai. Step 1 mein solid line y equals one half x plus 1 dikhti hai, jo upar jaati hai: m greater than 0. Step 2 mein dashed line y equals minus 2x plus 3 dikhti hai, jo neeche girti hai: m less than 0. Step 3 mein dotted horizontal line y equals minus 2 dikhti hai: m equals 0. Step 4 mein dash-dot vertical line x equals minus 3 dikhti hai, jiska koi slope nahi hai. Step 5 mein chaaron lines poore legend ke saath ek saath dikhti hain.

Quick check

4 mein se 1

Us line ka slope nikaalo jo in points se guzarti hai:

(- 1, 4)

aur

(3, - 2) .

Answer dekho

Answer -1.5

Rise

= - 2 - 4 = - 6

aur run

= 3 - (- 1) = 4,

isliye

m = \frac{- 6}{4} = - 1.5.

Minus sign batata hai ki line neeche girti hai.

Practice · 4 questions

Practice 13.3: answers ke saath

1
Short answerEasyNCERT Exercise Set 13.3, Q1
Ek skate park ke teen ramps table mein describe kiye gaye hain. Unhe steepness ke order mein lagao, sabse kam steep se sabse steep tak.
Ramp Rise Run
P 1.2 m 2 m
Q 1.5 m 3 m
R 2 m 2.5 m
1. Ramp P
2. Ramp R
3. Ramp Q
Answer dekho
Answer
Ramp Q
Ramp P
Ramp R
Slopes: Q $= \frac{1.5}{3} = 0.5,$ P $= \frac{1.2}{2} = 0.6,$ R $= \frac{2}{2.5} = 0.8.$ Q, P se zyada upar uthta hai lekin lambe run par, isliye woh kam steep hai.
Short answerEasyNCERT Exercise Set 13.3, Q2
Ek tin ki chhat 6 m ke horizontal run par 1.8 m upar uthti hai. Chhat ka slope kya hai?
Answer dekho
Answer 0.3
$\frac{1.8}{6} = 0.3 ⁣ :$ chhat across har metre par 0.3 m upar uthti hai.
3
Short answerMediumNCERT Exercise Set 13.3, Q3
Points ke har pair se guzarti line ka slope nikaalo.
1. $(- 3, 2)$ aur $(4, 2)$
  Answer dekho
  Answer 0
  Rise $= 0,$ run $= 7,$ isliye $m = 0 ⁣ :$ ek horizontal line.
2. $(- 2, 5)$ aur $(2, - 3)$
  Answer dekho
  Answer -2
  Rise $= - 8,$ run $= 4,$ isliye $m = - 2.$
3. $(- 4, - 2)$ aur $(4, 2)$
  Answer dekho
  Answer 0.5
  Rise $= 4,$ run $= 8,$ isliye $m = \frac{1}{2} .$
4. $(- 1, - 3)$ aur $(- 1, 4)$
  07undefined
  Answer dekho
  Answer (C) undefined
  Run hai $- 1 - (- 1) = 0.$ Line vertical hai aur uska slope undefined hai, 0 nahi.
Short answerMediumNCERT Exercise Set 13.3, Q4
Ek school 4 seedhiyon ke paas wheelchair ramp bana raha hai, har seedhi 15 cm oonchi hai. Guideline hai ki har 12 cm horizontal run par 1 cm rise ho. Ramp ka run kitna lamba hona chahiye, metres mein?
m
Answer dekho
Answer 7.2 m
Total rise $= 4 \times 15 = 60$ cm. Run rise ka 12 guna hai: $60 \times 12 = 720$ cm $= 7.2$ m.

Ramp	Rise	Run
P	1.2 m	2 m
Q	1.5 m	3 m
R	2 m	2.5 m

13.4 Do Variables wali Linear Equations ka Pair

Do letters wali ek akeli equation ke anginat solutions hote hain, jaisa Ishaan ke bill ne dikhaya. Ek pakka answer paane ke liye humein ek doosri, alag jaankari chahiye. Yeh rahi ek number puzzle: Mere mann mein do numbers hain. Unka sum 31 hai aur unka difference 7 hai.

Tum guess and check kar sakte ho, jaise Table 13.2 mein. Wahan har guess ka sum pehle se sahi hai, aur hum use tab tak adjust karte hain jab tak difference bhi sahi na ho jaaye.

Table 13.2Number puzzle ke liye guess and check

Pehla number	Doosra number	Sum	Difference
16	15	31	1
20	11	31	9
19	12	31	7

Chhote whole numbers ke liye guess karna chal jaata hai, lekin jaldi hi slow ho jaata hai. Algebra zyada tez hai. Numbers ko $x$ aur $y,$ kehne par dono facts ban jaate hain $x + y = 31$ aur $x - y = 7.$ Unhi do unknowns ke baare mein likhe do facts milkar pair of linear equations in two variables (रैखिक समीकरण युग्म) banate hain. Koi ordered pair is pair ko tabhi solve karta hai jab woh dono equations ko ek saath true banaye; sirf ek mein fit hona kaafi nahi hai.

3/3Lines yahan cross karti hain: $(19, 12),$ yahi ek pair hai jo dono mein fit hota hai.

Fig. 13.8Har clue ke akele answers ki poori line hoti hai; sirf crossing point dono mein fit hota hai.NCERT p. 104

Ek grid jo dono axes par 0 se 32 tak hai, har 5 par gridlines hain. Step 1 mein solid line x plus y equals 31 banti hai aur (10, 21) aur (25, 6) mark hote hain, sum 31 wale do pairs. Step 2 mein dashed line x minus y equals 7 judti hai, saath mein (10, 3) aur (31, 24), difference 7 wale do pairs. Step 3 mein crossing point (19, 12) ko ring kiya jaata hai, yahi ek pair hai jo dono clues mein fit hota hai.

Rozana ki problems se bhi aise pairs bilkul natural tarike se bante hain. Ek planetarium mein 2 adults aur 3 bachchon ki family tickets ke ₹410 deti hai, aur 1 adult aur 4 bachchon ki doosri family ₹380 deti hai. Agar adult ticket ₹ $x$ ka hai aur bachche ka ticket ₹ $y,$ to $2 x + 3 y = 410$ aur $x + 4 y = 380.$ Is pair ko solve karne par dono prices mil jaate hain; yeh tum Section 13.5 ke quick check mein karoge.

Saare chhe numbers ki jagah letters rakhne par har aisa pair neeche wali shape ka hota hai. Har equation mein dono coefficients mein se kam se kam ek non-zero hona chahiye.

Linear equations ka ek pair

\begin{aligned} a_{1} x + b_{1} y + c_{1} & = 0 \\ a_{2} x + b_{2} y + c_{2} & = 0 \end{aligned}

(13.5)

Quick check

3 mein se 1

Ek pen ₹

p

ka hai aur ek pencil ki price hai ₹

q .

Teen pens aur do pencils ₹47 ke hain, aur ek pen pencil se ₹4 mehnga hai. Kaunsa pair ise describe karta hai?

3 p + 2 q = 47

aur

p = q + 4

3 p + 2 q = 47

aur

q = p + 4

2 p + 3 q = 47

aur

p = q + 4

3 p + 2 q = 47

aur

p + q = 4

Answer dekho

Answer (A) $3 p + 2 q = 47$ aur $p = q + 4$

Pen mehnga item hai, isliye

p = q + 4,

aur pehla fact deta hai

3 p + 2 q = 47.

q = p + 4

likhne se pencil mehngi ho jaati. (Solve karne par

q = 7,

p = 11

milta hai.)

Practice · 3 questions

Practice 13.4: answers ke saath

Har situation ke liye diye gaye letters use karo aur dono facts ko do equations mein badlo. Unhe solve karna baad mein aayega.

Short answerEasyNCERT Exercise Set 13.4, Q1
Ek bakery cakes ₹ $x$ har ek aur cookies ke boxes ₹ $y$ har ek ke hisaab se bechti hai. 3 cakes aur 5 boxes ka order ₹1,400 ka hai, aur 2 cakes aur 1 box ka order ₹700 ka.
Solution dikhao
Answer $3 x + 5 y = 1400$ aur $2 x + y = 700.$
Pehla order deta hai $3 x + 5 y = 1400$ aur doosra deta hai $2 x + y = 700.$ (Baad mein solve karne par $x = 300$ aur $y = 100$ milta hai.)
Short answerEasyNCERT Exercise Set 13.4, Q2
Ek auto-rickshaw ek fixed amount ₹ $x$ aur har kilometre ke ₹ $y$ leta hai. Neha ki 4 km ki ride ₹86 ki padi aur Arjun ki 9 km ki ride ₹161 ki.
Solution dikhao
Answer $x + 4 y = 86$ aur $x + 9 y = 161.$
Har kiraya fixed part plus distance times rate hai: $x + 4 y = 86$ aur $x + 9 y = 161.$ (Inse milta hai $y = 15$ aur $x = 26$ .)
Short answerMediumNCERT Exercise Set 13.4, Q3
School play ke tickets adults ke liye ₹80 aur students ke liye ₹50 ke hain. Kul 320 tickets bike aur ₹19,600 ikattha hue. Maan lo $x$ adult tickets aur $y$ student tickets bike.
Solution dikhao
Answer $x + y = 320$ aur $80 x + 50 y = 19600.$
Tickets ginne par: $x + y = 320.$ Paise ginne par: $80 x + 50 y = 19600.$ (Solve karne par 120 adult aur 200 student tickets milte hain.)

13.5 Linear Equations ke Pair ke Solutions Nikaalna

Pair ke liye har method ek hi trick par tika hai: do letters wali do equations ko ek letter wali ek equation mein badal do, jise tum pehle se solve karna jaante ho. Ek letter pata chalte hi, use wapas rakhne par doosra mil jaata hai. Is section ke do methods ek letter ko gayab karne ke do tarike hain.

13.5.1 Linear Equations ke Pair ko Solve karne ke Algebraic Methods

Kya tumne shortcut pakda? Board 1 ko aadha karne par milta hai 'kite + top = ₹130'. Board 2 mein theek 'kite + top' hai, isliye hum uski jagah 130 rakh sakte hain: tab 2 drums ki cost $210 - 130 = 80$ rupees hai, aur ek drum ₹40 ka hai. Ab Board 3 se 2 kites = ₹140 milta hai, isliye ek kite ₹70 ki hai aur ek top ₹60 ka. Kisi cheez ki jagah uske barabar jaani hui amount rakhna neeche ke dono methods ka dil hai: substitution method aur elimination method.

Substitution method

Substitution method (प्रतिस्थापन विधि) mein hum ek equation se ek letter ko doosre letter ke terms mein likhte hain, aur us expression ko doosri equation mein rakh dete hain. Steps ye hain:

Woh equation chuno jismein kisi letter ko alag karna sabse aasaan ho, aur us letter ko subject banao.
Is expression ko doosri equation mein substitute karo aur ek letter wali equation solve karo.
Value wapas rakhkar doosra letter nikaalo.
Pair ko dono original equations mein check karo.

Example 13.7NCERT Example 7 · p. 106

Fraction answer wala substitution

Is pair ko substitution se solve karo: equation (1) hai $4 x + 9 y = 5$ aur equation (2) hai $x - 2 y = 3.$

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 13.8NCERT Example 8 · p. 107

Straight line par angles

Do angles ek linear pair banate hain, isliye milkar woh banate hain $180^{\circ} .$ Bada angle chhote angle ke double se $30^{\circ}$ kam hai. Dono angles nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Elimination method

Elimination method (विलोपन विधि) mein hum ek letter ke coefficients ko size mein barabar banate hain, aur phir equations ko jodte ya ghatate hain taaki woh letter cancel ho jaaye. Jab kisi letter ka coefficient 1 nahi hota, tab yeh aksar mehnat bachata hai.

Ek ya dono equations ko suitable numbers se multiply karo taaki kisi ek letter ke coefficients dono mein same size ke ho jaayein.
Agar un coefficients ka sign same hai to equations ko subtract karo; agar signs opposite hain to add karo. Ek letter eliminate ho jaata hai.
Bache hue letter ko solve karo, phir wapas substitute karke doosra nikaalo.
Dono original equations mein check karo.

4/4(1) se milta hai $15 + 2 y = 19,$ isliye $y = 2;$ (2) mein check: $25 - 8 = 17.$

Fig. 13.9

y

ko eliminate karna, in se:

3 x + 2 y = 19

aur

5 x - 4 y = 17.

NCERT p. 108

Equations ek ke neeche ek likhi hain, left mein tags hain. Step 1 mein (1) 3x plus 2y equals 19 aur (2) 5x minus 4y equals 17 dikhte hain. Step 2 mein (3) judta hai, equation (1) times 2, yaani 6x plus 4y equals 38, jismein terms minus 4y aur plus 4y highlight kiye gaye hain aur note hai ki y-parts ko match karo. Step 3 mein ek rule (line) khinchti hai aur (2) aur (3) jodkar 11x equals 55 milta hai, isliye x equals 5. Step 4 mein x equals 5 ko (1) mein rakha jaata hai, jisse 15 plus 2y equals 19 milta hai, isliye y equals 2, aur ise (2) mein check kiya jaata hai: 25 minus 8 equals 17.

Example 13.9NCERT Example 9 · p. 108

Income aur kharche

Do dosto ki monthly incomes ka ratio $5 : 3,$ hai, aur unke monthly kharchon ka ratio hai $2 : 1.$ Dono mein se har ek mahine mein ₹2,500 bachata hai. Unki monthly incomes nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 13.10NCERT Example 10 · p. 109

Ek number aur uska ulta

Ek two-digit number ko uske digits ulte karke bane number mein joda jaata hai, aur total 121 aata hai. Dono digits mein 3 ka fark hai. Number kya ho sakta hai? Har possibility nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Jab elimination dono letters hata de

Ab tak har pair ka bilkul ek solution tha. Ab dekho $2 x + 3 y = 7$ (1) aur $6 x + 9 y = 21$ (2). $x,$ ko eliminate karne ke liye (1) ko 3 se multiply karo. Yeh ban jaati hai $6 x + 9 y = 21,$ jo bilkul equation (2) hai, isliye subtract karne par bachta hai $0 = 0.$ Dono letters gayab ho gaye, aur jo bacha woh hamesha true hai.

Dono equations asal mein ek hi equation hai jo do baar likhi gayi hai, isliye $2 x + 3 y = 7$ ka har solution is equation ko bhi solve karta hai: $6 x + 9 y = 21.$ Pair ke infinitely many solutions hain. Test yeh hai ki kya ek equation doosri ki scaled copy hai: kya koi number $k \neq 0$ hai jiske saath $a_{1} = k a_{2},$ $b_{1} = k b_{2}$ aur $c_{1} = k c_{2}$ ? Agar $a_{2},$ $b_{2}$ aur $c_{2}$ sab non-zero hain, to yeh kehne jaisa hai ki teen ratios barabar hain:

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}

Ab sirf ek number badlo: $2 x + 3 y = 7$ (1) aur $6 x + 9 y = 20$ (2). (1) ko 3 se multiply karne par milta hai $6 x + 9 y = 21.$ Ab dono equations daawa karti hain ki bilkul wahi expression, $6 x + 9 y,$ 21 ke barabar bhi hai aur 20 ke barabar bhi. Yeh impossible hai (subtract karne par false statement $0 = 1$ milta hai), isliye koi bhi $(x, y)$ dono ko satisfy nahi karta, aur pair ka koi solution nahi hai. Yahan $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{1}{3},$ lekin $\frac{c_{1}}{c_{2}} = \frac{7}{20} .$

Teeno outcomes ko saath rakhne par kisi bhi pair ke liye (jahan $a_{2},$ $b_{2},$ $c_{2}$ non-zero hon) ek quick test milta hai, solve karna shuru karne se pehle hi:

Table 13.3Linear equations ke pair ke teen possible outcomes

Solutions	Ratios	Elimination ka ant
bilkul ek	$a_{1} / a_{2} \neq b_{1} / b_{2}$	ek letter bachta hai
infinitely many	$a_{1} / a_{2} = b_{1} / b_{2} = c_{1} / c_{2}$	$0 = 0$
koi nahi	$a_{1} / a_{2} = b_{1} / b_{2} \neq c_{1} / c_{2}$	ek false statement, jaise $0 = 1$

History seNCERT Thoda sa History · p. 112

Ek bahut purana method

Ishaan ka bill poochhta hai ki kaunse whole numbers ek aisi equation mein fit hote hain: $a x + b y = c .$ Āryabhaṭa (499 CE) ne theek isi ke liye ek step-by-step method diya, kuṭṭaka yaani 'pulveriser'. Lagbhag 130 saal baad, Brahmagupta (628 CE) ne kai unknowns ko ek saath handle kiya, har ek ko apna colour name dekar, kuch waise hi jaise aaj hum use karte hain $x$ aur $y .$

Log usse bahut pehle se equations ko pairs mein solve kar rahe the. Lagbhag 4,000 saal pehle Babylonian scribes clay tablets par do unknowns wali problems dete the. Lagbhag 2,000 saal purani Chinese classic Nine Chapters on the Mathematical Art mein inke liye fangcheng naam ka poora chapter hai. Iska procedure, jo board par bamboo counting rods se kiya jaata tha, is section ka elimination method hi hai: ek equation ke multiples doosri mein se tab tak ghatao jab tak koi unknown gayab na ho jaaye.

Quick check

5 mein se 1

2 x + 3 y = 11

aur

5 x - 2 y = 18,

y

ko eliminate karne ke liye sabse achha move kaunsa hai?

Pehli ko 2 se aur doosri ko 3 se multiply karo, phir add karoPehli ko 2 se aur doosri ko 3 se multiply karo, phir subtract karoPehli ko 5 se aur doosri ko 2 se multiply karo, phir add karoDono equations ko jaisi hain waise hi add karo

Answer dekho

Answer (A) Pehli ko 2 se aur doosri ko 3 se multiply karo, phir add karo

Isse

y

-terms ban jaate hain

+ 6 y

aur

- 6 y,

aur add karne se hat jaate hain:

19 x = 76,

isliye

x = 4

aur

y = 1.

Subtract karne par bachta

12 y .

5 aur 2 se multiply karne par

x

-terms match hote hain,

y

-terms nahi.

13.6 Do Variables wali Linear Equations ke Pair ko Solve karne ka Graphical Method

Pair ki har equation ka graph ek straight line hota hai, aur us line ke points theek uske solutions hote hain. Isliye pair ka solution dono lines par hona chahiye. To agar hum dono lines ko same axes par banayein, to solutions bas woh points hain jo dono lines mein common hain.

Try karke dekhoNCERT p. 113

Chaar pairs banao

Tumhe chahiye

graph paper
ek ruler
alag-alag colours ki do pencils

Pehle guess karo

Banane se pehle: do straight lines mein kitne points common ho sakte hain?

hamesha bilkul ekek, koi nahi ya infinitely manyhamesha kam se kam ekbilkul do

Har pair ke liye ek chhoti table banao jismein har line ke kam se kam do points hon. Dono lines ko same axes par banao, har ek ke liye alag colour use karke, aur note karo ki woh kaise baithti hain.

$x + y = 5$ aur $2 x - y = 4$
$x - 2 y = 2$ aur $3 x - 6 y = 6$
$2 x + y = 4$ aur $4 x + 2 y = 2$
$x + 3 y = 3$ aur $x - y = - 1$

Tumhari chaaron drawings woh sirf teen tarike dikhati hain jinse do lines ek plane mein baith sakti hain, jo Fig. 13.10 mein dikhaye gaye hain:

Intersecting lines (प्रतिच्छेदी रेखाएँ) bilkul ek point par milti hain, isliye pair ka bilkul ek solution hota hai: woh point.
Parallel lines (समांतर रेखाएँ) kabhi nahi milti, isliye pair ka koi solution nahi hota.
Coincident lines (संपाती रेखाएँ) ek doosre ke upar padti hain, isliye har point common hota hai aur pair ke infinitely many solutions hote hain.

1/3 $x + y = 5$ aur $2 x - y = 4$ is point par milti hain: $(3, 2) ⁣ :$ ek solution.

Fig. 13.10Intersecting, parallel aur coincident lines.NCERT p. 113

Ek grid jo across minus 1 se 6 tak aur upar minus 2 se 5 tak hai, right mein dono equations aur ek verdict hai. Step 1 mein x plus y equals 5 aur dashed line 2x minus y equals 4 ringed point (3, 2) par cross karti dikhti hain: woh ek baar milti hain, isliye ek solution hai. Step 2 mein 2x plus y equals 4 aur dashed 4x plus 2y equals 2 dikhti hain, do parallel lines jo kabhi nahi milti: koi solution nahi. Step 3 mein x minus 2y equals 2 ek chaudi halki band ki tarah dikhti hai aur dashed 3x minus 6y equals 6 uske saath-saath chalti hai: ek hi line, isliye infinitely many solutions.

Example 13.11NCERT Example 11 · p. 113

Drawing se solve karna

Pair $x + 2 y = 8$ aur $3 x - 2 y = 0$ ko graphically solve karo, aur batao ki iska solution kis kind ka hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3/3Woh yahan cross karti hain: $(2, 3) .$

Fig. 13.11Example 13.11 ki dono lines yahan milti hain:

(2, 3) .

NCERT Fig. 13.14 · p. 114

Ek grid jo across minus 1 se 9 tak aur upar minus 1 se 7 tak hai. Step 1 mein solid line x plus 2y equals 8 banti hai jo (0, 4) aur (8, 0) se guzarti hai. Step 2 mein dashed line 3x minus 2y equals 0 judti hai jo origin aur (4, 6) se guzarti hai. Step 3 mein crossing point (2, 3) ko ring kiya jaata hai aur note hai ki lines wahan milti hain.

Example 13.12NCERT Example 12 · p. 114

Pehle predict karo, phir banao

Tay karo ki lines $x - 2 y + 2 = 0$ aur $2 x - 4 y - 8 = 0$ intersect karti hain, parallel hain ya coincide karti hain. Phir unhe banakar check karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

3/3Dono 2 across, 1 upar jaati hain, aur vertically 3 units door rehti hain.

Fig. 13.12Same slopes aur alag intercepts: Example 13.12 ki lines kabhi nahi milti.NCERT Fig. 13.15 · p. 115

Ek grid jo across minus 3 se 7 tak aur upar minus 3 se 5 tak hai. Step 1 mein solid line x minus 2y plus 2 equals 0 banti hai jo (minus 2, 0), (0, 1) aur (4, 3) se guzarti hai. Step 2 mein dashed line 2x minus 4y minus 8 equals 0 judti hai jo (0, minus 2), (4, 0) aur (6, 1) se guzarti hai. Step 3 mein har line par 2 across aur 1 upar ka step banta hai aur ek double arrow unke beech 3 units ka constant vertical gap dikhata hai, saath mein note hai ki woh kabhi nahi milti.

PropertySame slopes, parallel lines

Same slope wali do alag lines kabhi nahi milti: woh parallel hoti hain.

Example 13.13NCERT Example 13 · p. 115

Ek kharidari jo kuch naya nahi jodti

Meera 3 pencils aur 2 erasers ₹24 mein khareedti hai. Uska bhai usi tarah ki 6 pencils aur 4 erasers ₹48 mein khareedta hai. Equations ka pair likho aur uske graph ko describe karo. Kya tum prices nikaal sakte ho?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Neeche diye explorer mein jo pair chaho try karo. Har equation ke numbers badlo aur dono lines, unke ratios, unke slopes aur solutions ki sankhya dekho. Phir Make it par switch karo aur khud intersecting, parallel aur coincident pairs banao.

InteractiveDo lines ka explorer

Do linear equations a x plus b y plus c equals 0 form mein, har ek mein a, b aur c ke liye minus 9 se 9 tak plus aur minus buttons, jo minus 8 se 8 tak ke grid par ek solid aur ek dashed line ki tarah bani hain. Ek ring crossing point ko uske coordinates ke saath mark karti hai. Ek status box batata hai ki lines intersect karti hain (exact solution ke saath), parallel hain (koi solution nahi) ya coincide karti hain (infinitely many), aur tables ratios a1 over a2, b1 over b2, c1 over c2 aur har line ka slope dikhati hain. Preset buttons ek intersecting, ek parallel aur ek coincident pair load karte hain. Make it mode line 1 ko fix karta hai aur tumse line 2 badalne ko kehta hai taaki lines parallel ho jaayein, ek hi line ban jaayein, ya diye gaye point par milein, hints aur score ke saath.

Fig. 13.13Har line ke liye

a,

b

aur

c

badlo aur dekho ki lines intersect karti hain, parallel hain ya coincide karti hain.

Quick check

4 mein se 1

Linear equations ke ek pair ke graphs parallel lines hain. Pair ke kitne solutions hain?

bilkul ekkoi nahiinfinitely manydo

Answer dekho

Answer (B) koi nahi

Solutions common points hote hain, aur parallel lines ka koi common point nahi hota. Infinitely many solutions coincident lines se aate hain, parallel lines se nahi.

Practice · 11 questions

Practice 13.5: answers ke saath

1
Short answerMediumNCERT Exercise Set 13.5, Q1 (i)–(iv)
Har problem ke liye do equations likho, phir unhe solve karo.
1. Do integers ka sum $- 4$ hai aur unka difference 18 hai. Unhe nikaalo.
  Solution dikhao
  Answer7 aur $- 11.$
  $x + y = - 4$ aur $x - y = 18.$ Jodne par: $2 x = 14,$ isliye $x = 7$ aur $y = - 11.$ Check: $7 - (- 11) = 18.$
2. Do numbers mein 45 ka fark hai, aur unmein se ek doosre ka chaar guna hai. Unhe nikaalo.
  Solution dikhao
  Answer60 aur 15.
  $x - y = 45$ aur $x = 4 y .$ Substitute karne par: $3 y = 45,$ isliye $y = 15$ aur $x = 60.$
3. Ek hockey club 4 sticks aur 10 balls ₹5,100 mein khareedta hai, aur baad mein 3 sticks aur 4 balls ₹3,300 mein. Ek stick aur ek ball ka price nikaalo.
  Solution dikhao
  AnswerStick ₹900, ball ₹150.
  $4 s + 10 b = 5100$ aur $3 s + 4 b = 3300.$ 3 aur 4 se multiply karo: $12 s + 30 b = 15300$ aur $12 s + 16 b = 13200.$ Subtract karne par $14 b = 2100,$ isliye $b = 150,$ aur phir $3 s = 3300 - 600,$ isliye $s = 900.$
4. Ek courier fixed fee aur per kilogram ek rate leta hai. 3 kg ke parcel ki cost ₹110 hai aur 8 kg ke parcel ki ₹210. Fee aur rate nikaalo. 12 kg ka parcel kitne ka padega?
  Solution dikhao
  AnswerFee ₹50, ₹20 per kg; 12 kg ka parcel ₹290 ka.
  $f + 3 r = 110$ aur $f + 8 r = 210.$ Subtract karne par $5 r = 100,$ isliye $r = 20$ aur $f = 50.$ 12 kg ke liye: $50 + 12 \times 20 = 290.$
2
Short answerHardNCERT Exercise Set 13.5, Q1 (v)–(vii)
Har problem ke liye do equations likho, phir unhe solve karo.
1. Ek fraction ke numerator aur denominator dono mein 1 jodne par woh $\frac{1}{2}$ ban jaata hai, aur dono mein 4 jodne par $\frac{3}{5}$ ban jaata hai. Fraction nikaalo.
  Solution dikhao
  Answer $\frac{5}{11}$
  $2 (x + 1) = y + 1$ se milta hai $2 x - y = - 1;$ $5 (x + 4) = 3 (y + 4)$ se milta hai $5 x - 3 y = - 8.$ Pehli se $y = 2 x + 1;$ phir $5 x - 6 x - 3 = - 8,$ isliye $x = 5,$ $y = 11.$ Check: $\frac{6}{12} = \frac{1}{2}$ aur $\frac{9}{15} = \frac{3}{5} .$
2. Agar ek fraction ke numerator se 3 ghataya jaaye, to woh ban jaata hai $\frac{1}{3} .$ Agar iski jagah denominator mein 4 joda jaaye, to woh ban jaata hai $\frac{1}{2} .$ Fraction nikaalo.
  Solution dikhao
  Answer $\frac{5}{6}$
  $3 (x - 3) = y$ aur $2 x = y + 4.$ Substitute karo $y = 3 x - 9 ⁣ :$ $2 x = 3 x - 5,$ isliye $x = 5$ aur $y = 6.$ Check: $\frac{2}{6} = \frac{1}{3}$ aur $\frac{5}{10} = \frac{1}{2} .$
3. Chhe saal pehle Kavya apne cousin se chaar guna badi thi. Chaar saal baad woh usse do guna badi hogi. Unki abhi ki ages nikaalo.
  Solution dikhao
  AnswerKavya 26, cousin 11.
  $x - 6 = 4 (y - 6)$ se milta hai $x - 4 y = - 18;$ $x + 4 = 2 (y + 4)$ se milta hai $x - 2 y = 4.$ Subtract karne par $2 y = 22,$ isliye $y = 11$ aur $x = 26.$ Check: $20 = 4 \times 5$ aur $30 = 2 \times 15.$
3
Short answerHardNCERT Exercise Set 13.5, Q1 (viii)–(x)
Har problem ke liye do equations likho, phir unhe solve karo.
1. Ek two-digit number ke digits ka sum 12 hai, aur number ka 7 guna, digits ulte karke bane number ke 4 gune ke barabar hai. Number nikaalo.
  Solution dikhao
  Answer48
  $x + y = 12$ aur $7 (10 x + y) = 4 (10 y + x),$ jo simplify hokar banta hai $66 x = 33 y,$ isliye $y = 2 x .$ Phir $3 x = 12,$ $x = 4,$ $y = 8.$ Check: $7 \times 48 = 336 = 4 \times 84.$
2. Rohit bank se ₹4,100 nikaalta hai, sirf ₹200 aur ₹500 ke notes mein, aur use kul 13 notes milte hain. Use har tarah ke kitne notes milte hain?
  Solution dikhao
  Answer₹200 ke 8 notes aur ₹500 ke 5 notes.
  $x + y = 13$ aur $200 x + 500 y = 4100.$ Substitute karo $x = 13 - y ⁣ :$ $2600 + 300 y = 4100,$ isliye $y = 5$ aur $x = 8.$
3. Ek cycle shop pehle 2 ghanton ke liye ek fixed amount leti hai aur har extra ghante ke liye ek tay amount. 6 ghante ki hire ₹140 ki padti hai aur 9 ghante ki hire ₹200 ki. Fixed amount aur hourly charge nikaalo.
  Solution dikhao
  Answer₹60 fixed, har extra ghante ke ₹20.
  6 ghante ki hire mein 4 extra ghante hain aur 9 ghante ki hire mein 7: $x + 4 y = 140$ aur $x + 7 y = 200.$ Subtract karne par $3 y = 60,$ isliye $y = 20$ aur $x = 60.$
DiagramMediumNCERT Exercise Set 13.5, Q2
12 runners ki ek relay squad mein ladkiyon se 2 ladke zyada hain. Equations ka ek pair banao aur graphs banakar ladkiyon aur ladkon ki sankhya nikaalo.
Solution dikhao
Answer5 ladkiyan aur 7 ladke.
$x + y = 12$ in points se guzarti hai: $(0, 12),$ $(6, 6)$ aur $(12, 0) .$ $y = x + 2$ in points se guzarti hai: $(0, 2),$ $(4, 6)$ aur $(5, 7) .$ Lines yahan cross karti hain: $(5, 7),$ isliye 5 ladkiyan aur 7 ladke hain. Check: $5 + 7 = 12$ aur $7 - 5 = 2.$
5
Short answerMediumNCERT Exercise Set 13.5, Q3
Bina banaye, $\frac{a_{1}}{a_{2}},$ $\frac{b_{1}}{b_{2}}$ aur $\frac{c_{1}}{c_{2}}$ compare karke lines ke har pair ko intersecting, parallel ya coincident mein classify karo.
1. $3 x + 5 y - 7 = 0$ aur $4 x - 2 y + 1 = 0$
  intersect karti hainparallel haincoincide karti hain
  Answer dekho
  Answer (A) intersect karti hain
  $\frac{3}{4} \neq \frac{5}{- 2},$ isliye lines intersect karti hain.
2. $4 x - 6 y + 10 = 0$ aur $6 x - 9 y + 15 = 0$
  intersect karti hainparallel haincoincide karti hain
  Answer dekho
  Answer (C) coincide karti hain
  $\frac{4}{6} = \frac{- 6}{- 9} = \frac{10}{15} = \frac{2}{3},$ isliye lines coincide karti hain.
3. $5 x + 2 y - 3 = 0$ aur $10 x + 4 y + 7 = 0$
  intersect karti hainparallel haincoincide karti hain
  Answer dekho
  Answer (B) parallel hain
  $\frac{5}{10} = \frac{2}{4} = \frac{1}{2},$ lekin $\frac{- 3}{7} \neq \frac{1}{2},$ isliye lines parallel hain.
6
DiagramMediumNCERT Exercise Set 13.5, Q4
Pehle tay karo ki har pair solve ho bhi sakta hai ya nahi. Phir graphs banao aur jo bhi solutions hon unhe padho.
1. $2 x - y = 3$ aur $4 x - 2 y = 6$
  Solution dikhao
  AnswerInfinitely many: is line ka har point $y = 2 x - 3.$
  Teeno ratios $\frac{1}{2},$ ke barabar hain, isliye lines coincide karti hain. $y = 2 x - 3,$ par har point, jaise $(0, - 3),$ $(1, - 1)$ ya $(3, 3),$ ek solution hai.
2. $x + 2 y = 6$ aur $3 x + 6 y = 10$
  Solution dikhao
  AnswerKoi solution nahi.
  $\frac{1}{3} = \frac{2}{6}$ lekin $\frac{6}{10} \neq \frac{1}{3},$ isliye lines parallel hain aur kabhi nahi milti.
3. $x - y = 1$ aur $2 x + y = 8$
  Solution dikhao
  Answer $(3, 2)$
  $\frac{1}{2} \neq \frac{- 1}{1},$ isliye lines intersect karti hain. $x - y = 1$ in points se guzarti hai: $(1, 0)$ aur $(4, 3);$ $2 x + y = 8$ in points se guzarti hai: $(4, 0)$ aur $(2, 4) .$ Woh yahan cross karti hain: $(3, 2) ⁣ :$ $3 - 2 = 1$ aur $6 + 2 = 8.$
4. $3 x + y - 2 = 0$ aur $6 x + 2 y - 9 = 0$
  Solution dikhao
  AnswerKoi solution nahi.
  $\frac{3}{6} = \frac{1}{2}$ lekin $\frac{- 2}{- 9} = \frac{2}{9} \neq \frac{1}{2},$ isliye lines parallel hain.
Short answerMediumNCERT Exercise Set 13.5, Q5
Ek rectangular plot ka perimeter 58 m hai, aur uski length uski width ke double se 1 m kam hai. Uski length aur width nikaalo.
Solution dikhao
AnswerLength 19 m, width 10 m.
$l + w = 29$ aur $l = 2 w - 1.$ Substitute karne par: $3 w - 1 = 29,$ isliye $w = 10$ aur $l = 19.$ Check: $2 (19 + 10) = 58.$
8
Short answerMediumNCERT Exercise Set 13.5, Q6
Is line se shuru karo: $3 x - 4 y + 5 = 0.$ Ek partner equation banao jiska graph, is line ke saath milkar, deta hai
1. intersecting lines
  Solution dikhao
  AnswerJaise $x + y - 2 = 0.$
  Koi bhi equation jismein $\frac{3}{a_{2}} \neq \frac{- 4}{b_{2}}$ ho, chalegi, jaise $x + y - 2 = 0,$ kyunki $\frac{3}{1} \neq \frac{- 4}{1} .$
2. parallel lines
  Solution dikhao
  AnswerJaise $6 x - 8 y + 1 = 0.$
  $x$ - aur $y$ -coefficients ko double karo lekin constant ko nahi: $\frac{3}{6} = \frac{- 4}{- 8} = \frac{1}{2},$ jabki $\frac{5}{1} \neq \frac{1}{2} .$
3. coincident lines
  Solution dikhao
  AnswerJaise $6 x - 8 y + 10 = 0.$
  Poori equation ko kisi bhi non-zero number se multiply karo, yahan 2 se, taaki teeno ratios ho jaayein $\frac{1}{2} .$
Short answerHardNCERT Exercise Set 13.5, Q7
Mahāvīrācārya ki Gaṇitasārasaṅgraha (9th century CE) mein phalon ke price wali puzzles hain jinmein do kharidariyon ke numbers aapas mein badal jaate hain. Usi style mein ek nayi puzzle yeh rahi: 8 guavas aur 5 pomegranates 131 coins ke hain, jabki 5 guavas aur 8 pomegranates 116 coins ke. Substitution ya usual elimination steps use kiye bina har phal ka price nikaalo.
Solution dikhao
AnswerEk guava 12 coins ka hai aur ek pomegranate 7 coins ka.
Maan lo ek guava $x$ coins ka hai aur ek pomegranate $y ⁣ :$ $8 x + 5 y = 131$ aur $5 x + 8 y = 116.$ Add karne par $13 x + 13 y = 247,$ isliye $x + y = 19.$ Subtract karne par $3 x - 3 y = 15,$ isliye $x - y = 5.$ Phir $2 x = 24,$ isliye $x = 12$ aur $y = 7.$ Numbers ki adla-badli ki wajah se equations ka sum aur difference in dono ke saaf multiples ban kar aate hain: $x + y$ aur $x - y .$
Short answerMediumNCERT Exercise Set 13.5, Q8
6 rulers aur 4 glue sticks ₹168 ke hain, jabki 3 rulers aur 5 glue sticks ₹138 ke. Ek ruler aur ek glue stick ki cost nikaalo.
Solution dikhao
AnswerRuler ₹16 ka, glue stick ₹18 ki.
$6 r + 4 g = 168$ aur $3 r + 5 g = 138.$ Doosri ko double karne par: $6 r + 10 g = 276.$ Pehli ko subtract karne par: $6 g = 108,$ isliye $g = 18,$ aur $6 r = 168 - 72 = 96,$ isliye $r = 16.$
Short answerMediumNCERT Exercise Set 13.5, Q9
Ek flower pot shelf ke upar rakha hai, aur pot ka top floor se 130 cm upar hai. Jab pot ko shelf ke paas floor par rakh diya jaata hai, to shelf ka top pot ke top se 50 cm ooncha hota hai. Shelf aur pot kitne oonche hain?
Solution dikhao
AnswerShelf 90 cm, pot 40 cm.
$s + p = 130$ aur $s - p = 50.$ Jodne par: $2 s = 180,$ isliye $s = 90$ aur $p = 40.$

Practice · 16 questions

Practice questions answers ke saath (NCERT “Chapter ke end ki exercises”)

1
Short answerMediumNCERT End-of-chapter Exercises, Q1
Ek phone khaali battery se ek jaisi speed se charge hota hai. $x$ minutes baad uska battery level $y$ per cent hai, aur $y = 1.5 x$ ka graph in points se guzarta hai: $(0, 0)$ aur $(4, 6) .$
1. Line ka slope kya hai, aur yahan uska kya matlab hai? $y$ -intercept kya hai, aur uska kya matlab hai?
  Solution dikhao
  AnswerSlope 1.5; intercept 0.
  Slope $= \frac{6 - 0}{4 - 0} = 1.5 ⁣ :$ battery har minute 1.5 per cent badhti hai. $y$ -intercept 0 hai: $x = 0$ par battery khaali hai.
2. Battery ko 90 per cent tak pahunchne mein kitna time lagta hai?
  min
  Answer dekho
  Answer 60 min
  $1.5 x = 90$ se milta hai $x = 60$ minutes.
3. Kya point $(10, 16)$ is line par hai?
  haannahi
  Answer dekho
  Answer (B) nahi
  $x = 10$ par line deti hai $y = 1.5 \times 10 = 15,$ 16 nahi.
4. Yeh line, line $x = 6$ se kahan milti hai?
  Solution dikhao
  Answer $(6, 9)$
  Line $x = 6$ ke har point par hota hai $x = 6;$ $y = 1.5 x$ par isse milta hai $y = 9.$ To woh yahan milti hain: $(6, 9) .$
2
Short answerMediumNCERT End-of-chapter Exercises, Q2
Fahrenheit aur Celsius mein temperatures is linear equation se jude hain: $F = \frac{9}{5} C + 32.$
1. $C$ across aur $F$ upar rakhkar graph banane ke liye do points nikaalo. Uska slope aur intercept kya hai?
  Solution dikhao
  AnswerJaise $(0, 32)$ aur $(100, 212);$ slope $\frac{9}{5},$ intercept 32.
  $C = 0$ se $F = 32$ milta hai aur $C = 100$ se $F = 212.$ Slope $\frac{9}{5}$ batata hai ki $F$ 9 degrees badhta hai jab bhi 5 degrees badhta hai $C;$ intercept batata hai ki paani 32 °F par jamta hai.
2. Ek suhaana din 25 °C ka hai. Fahrenheit mein yeh kitna hai?
  °F
  Answer dekho
  Answer 77 °F
  $\frac{9}{5} \times 25 + 32 = 45 + 32 = 77.$
3. Ek bachche ka temperature 104 °F hai. Celsius mein yeh kitna hai?
  °C
  Answer dekho
  Answer 40 °C
  $104 - 32 = 72$ aur $72 \times \frac{5}{9} = 40.$
4. Kis temperature par dono scales same number dikhate hain?
  Answer dekho
  Answer -40
  $F = C ⁣ :$ rakho: $C = \frac{9}{5} C + 32$ se milta hai $- \frac{4}{5} C = 32,$ isliye $C = - 40.$
DiagramEasyNCERT End-of-chapter Exercises, Q3
Graphs banakar pair $x + y = 7$ aur $2 x - y = 2$ solve karo.
Solution dikhao
Answer $(3, 4)$
$x + y = 7$ in points se guzarti hai: $(0, 7)$ aur $(7, 0) .$ $2 x - y = 2$ in points se guzarti hai: $(0, - 2)$ aur $(1, 0) .$ Lines yahan cross karti hain: $(3, 4) .$ Check: $3 + 4 = 7$ aur $6 - 4 = 2.$
Short answerMediumNCERT End-of-chapter Exercises, Q4
Line $l$ in points se guzarti hai: $(0, 6)$ aur $(6, 0) .$ Line $n$ in points se guzarti hai: $(0, - 2)$ aur $(4, 6) .$ Woh point nikaalo jahan dono milti hain.
Solution dikhao
Answer $(\frac{8}{3}, \frac{10}{3})$
Line $l$ ka slope $- 1$ aur intercept 6 hai: $x + y = 6.$ Line $n$ ka slope $\frac{8}{4} = 2$ aur intercept hai $- 2 ⁣ :$ $y = 2 x - 2.$ Substitute karne par: $x + 2 x - 2 = 6,$ isliye $x = \frac{8}{3}$ aur $y = \frac{10}{3} .$ Graph mein crossing grid lines ke beech dikhti hai, isiliye algebra ki zaroorat padti hai.
Short answerMediumNCERT End-of-chapter Exercises, Q5
Line $y = m x + d$ kahan cross karti hai $x$ -axis ko? Kab aisa koi point nahi hota?
Solution dikhao
Answer $(- \frac{d}{m}, 0)$ par, jab $m \neq 0.$
$y = 0 ⁣ :$ rakho: $m x + d = 0,$ isliye $x = - \frac{d}{m},$ basharte $m \neq 0.$ Agar $m = 0$ to line $y = d$ flat hai: jab $d \neq 0,$ to woh $x$ -axis se kabhi nahi milti, aur woh khud $x$ -axis hoti hai jab $d = 0.$
Short answerMediumNCERT End-of-chapter Exercises, Q6
Cab app X ₹40 plus ₹12 per km leta hai. Cab app Y ₹70 plus ₹9 per km leta hai. Kitni distance ke liye dono ki cost same hoti hai? Kab kaunsa sasta padta hai?
Solution dikhao
Answer10 km par barabar; 10 km se kam par X sasta hai, zyada par Y.
$40 + 12 d = 70 + 9 d$ se milta hai $3 d = 30,$ isliye $d = 10$ km, jahan dono ki cost ₹160 hai. X ka fixed charge chhota hai, isliye chhoti rides ke liye woh sasta hai ( $d < 10$ ). Y ka per km rate chhota hai, isliye lambi rides mein woh jeet jaata hai ( $d > 10$ ).
7
Short answerEasyNCERT End-of-chapter Exercises, Q7
Kitni lines hain
1. jinka slope $- 2$ hai?
  Solution dikhao
  AnswerInfinitely many.
  $y = - 2 x + d$ slope $- 2$ wali line hai, chahe kuch bhi ho $d ⁣ :$ parallel lines ki poori family.
2. jinka slope $- 2$ hai aur jo $(1, 3)$ se guzarti hain?
  Solution dikhao
  AnswerBilkul ek: $y = - 2 x + 5.$
  $(1, 3)$ ko is mein rakho: $y = - 2 x + d ⁣ :$ $3 = - 2 + d,$ isliye $d = 5.$ $d$ ki sirf ek value chalti hai.
Short answerMediumNCERT End-of-chapter Exercises, Q8
Pair $6 x + k y - 3 = 0$ aur $3 x - 2 y + 5 = 0$ ka bilkul ek solution hota hai, lagbhag har value ke liye, is ki: $k .$ $k$ ki kaunsi value se bachna chahiye?
Solution dikhao
Answer $k = - 4$ (baaki har value bilkul ek solution deti hai).
$\frac{6}{3} = 2,$ isliye humein chahiye $\frac{k}{- 2} \neq 2,$ yaani $k \neq - 4.$ (Jab $k = - 4$ ho, to lines parallel hoti hain, kyunki $\frac{- 3}{5} \neq 2$ .)
Short answerHardNCERT End-of-chapter Exercises, Q9
$a$ aur $b$ aise chuno ki $(2 a - b) x + (a + b) y = 4 a + 1$ aur $2 x + 3 y = 7$ ek hi line ko describe karein.
Solution dikhao
Answer $a = 5,$ $b = 4.$
Humein chahiye $\frac{2 a - b}{2} = \frac{a + b}{3} = \frac{4 a + 1}{7} .$ Pehla aur teesra dete hain $6 a - 7 b = 2;$ doosra aur teesra dete hain $- 5 a + 7 b = 3.$ Jodne par $a = 5,$ aur phir $b = 4.$ Check: pehli equation ban jaati hai $6 x + 9 y = 21,$ jo is equation ka 3 guna hai: $2 x + 3 y = 7.$
Short answerHardNCERT End-of-chapter Exercises, Q10
$k,$ ki ek value ke liye lines $2 x + 3 y = 4$ aur $(k - 1) x + (k + 2) y = 2 k - 2$ bilkul ek doosre ke upar padti hain. Woh value nikaalo.
Answer dekho
Answer 7
$3 (k - 1) = 2 (k + 2)$ se milta hai $k = 7.$ Tab doosri equation hai $6 x + 9 y = 12,$ jo pehli ka bilkul 3 guna hai, isliye lines coincide karti hain.
11
Short answerHardNCERT End-of-chapter Exercises, Q11
Do delivery drones grid wale ek field ke upar straight lines mein udte hain.
1. Drone A $(0, 0)$ se shuru hota hai aur har 2 units right par 3 units upar jaata hai. Drone B $(8, 0)$ se shuru hota hai aur har 1 unit right par 3 units upar jaata hai. Kya unke paths cross karte hain? Kahan?
  Solution dikhao
  AnswerHaan, yahan: $(16, 24) .$
  A: $y = \frac{3}{2} x$ jab $x \geq 0.$ B: $y = 3 (x - 8)$ jab $x \geq 8.$ Dono ko barabar rakhne par: $1.5 x = 3 x - 24,$ isliye $x = 16$ aur $y = 24.$ Dono drones $x = 16,$ se guzarte hain, isliye paths sach mein wahan cross karte hain.
2. Drone C $(2, 0)$ se shuru hota hai aur har 1 unit right par 1 unit upar jaata hai. Drone D $(6, 0)$ se shuru hota hai aur har 2 units right par 1 unit neeche jaata hai. Kya unke paths cross karte hain?
  Solution dikhao
  AnswerNahi.
  Poori lines $y = x - 2$ aur $y = - \frac{1}{2} (x - 6)$ yahan milti hain: $(\frac{10}{3}, \frac{4}{3}) .$ Lekin D sirf $x = 6,$ se right ki taraf chalta hai, aur $\frac{10}{3} < 6,$ isliye woh point D ke start ke peeche hai. Paths kabhi cross nahi karte.
Short answerHardNCERT End-of-chapter Exercises, Q12
Ek pal mein ghadi 4 bajkar $p$ minute dikhati hai. Bees minute baad woh 5 bajkar $q$ minute dikhati hai. Agar $p$ , $q,$ ka paanch guna hai, to doosri reading ka time kya tha?
Solution dikhao
Answer5 bajkar 10 minute (5:10).
$p + 20 = 60 + q$ aur $p = 5 q .$ Substitute karne par: $4 q = 40,$ isliye $q = 10$ aur $p = 50.$ Readings thi 4:50 aur 5:10.
Short answerMediumNCERT End-of-chapter Exercises, Q13
Ek two-digit number ke digits ka sum 13 hai. Digits ki jagah badalne par original se 27 kam number milta hai. Number nikaalo.
Answer dekho
Answer 85
$x + y = 13$ aur $9 (x - y) = 27,$ isliye $x - y = 3.$ Phir $x = 8,$ $y = 5 ⁣ :$ number 85 hai, aur $85 - 58 = 27.$
Short answerHardNCERT End-of-chapter Exercises, Q14
Ek cyclic quadrilateral $P Q R S,$ mein $∠ P = (2 x + y)^{\circ},$ $∠ Q = (3 y + 25)^{\circ},$ $∠ R = (5 x + y + 10)^{\circ}$ aur $∠ S = (4 x + y + 15)^{\circ} .$ Chaaron angles nikaalo.
Solution dikhao
Answer $∠ P = 55^{\circ},$ $∠ Q = 70^{\circ},$ $∠ R = 125^{\circ},$ $∠ S = 110^{\circ} .$
$∠ P + ∠ R = 180$ se milta hai $7 x + 2 y = 170.$ $∠ Q + ∠ S = 180$ se milta hai $4 x + 4 y = 140,$ isliye $y = 35 - x .$ Phir $7 x + 70 - 2 x = 170,$ isliye $x = 20$ aur $y = 15.$ Angles hain 55°, 70°, 125° aur 110°, jinka sum 360° hai.
Short answerHardNCERT End-of-chapter Exercises, Q15
Ek bus ek fixed distance steady speed se tay karti hai. Agar woh 10 km/h tez chalti, to trip mein 2 ghante kam lagte; agar woh 10 km/h dheere chalti, to 3 ghante zyada lagte. Speed, time aur distance nikaalo.
Solution dikhao
Answer12 ghante ke liye 50 km/h: 600 km.
$(v + 10) (t - 2) = v t$ simplify hokar banta hai $- v + 5 t = 10.$ $(v - 10) (t + 3) = v t$ simplify hokar banta hai $3 v - 10 t = 30.$ Pehli se $v = 5 t - 10;$ phir $15 t - 30 - 10 t = 30,$ isliye $t = 12$ aur $v = 50.$ Distance $= 600$ km. Check: 60 km/h par 10 h lagte hain aur 40 km/h par 15 h.
Short answerHardNCERT End-of-chapter Exercises, Q16
Ek pita ki age unke do bachchon ki ages ke sum ki teen guna hai. 5 saal baad woh unki ages ke sum ke do gune honge. Unki abhi ki age kya hai?
saal
Answer dekho
Answer 45 saal
Lo pita $F$ aur bachchon ka sum $S ⁣ :$ $F = 3 S$ aur $F + 5 = 2 (S + 10) .$ Phir $3 S + 5 = 2 S + 20,$ isliye $S = 15$ aur $F = 45.$

Recap

Summary aur key points

Do variables wali linear equation ko $a x + b y + c = 0,$ likha ja sakta hai, jahan $a$ aur $b$ dono zero nahi hote (Section 13.1.1).
Solution ek ordered pair $(x, y)$ hai jo equation ko true banata hai; order matter karta hai (Section 13.2.1).
$3 x + 4 y = 24$ jaisi ek akeli equation ko infinitely many ordered pairs satisfy karte hain, kyunki $x$ ki har value ek $y$ deti hai (Section 13.2.1).
Iska graph ek straight line hai jo theek iske solutions se bani hai, na zyada na kam (Section 13.2.2).
Axis crossings, jo pehle $x = 0$ aur phir $y = 0,$ rakhkar milti hain, line banane ka aasaan tarika hain (Example 13.2).
Slope $= \frac{rise}{run} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}},$ line ke kinhi bhi do points ke liye same (Section 13.3.1).
$y = m x + d,$ mein $m$ slope hai aur $d$ $y$ -intercept, isliye graph $y$ -axis se is point par milta hai: $(0, d);$ standard form se, $m = - \frac{a}{b}$ (Section 13.3.2).
Positive slope upar jaata hai, negative slope neeche girta hai, $y = k$ ka slope 0 hai, aur vertical line $x = k$ ka koi slope nahi hota (Table 13.1).
Linear equations ka pair us ordered pair se solve hota hai jo dono equations ko satisfy kare (Section 13.4).
Substitution aur elimination dono pair ko ek letter wali ek equation mein badal dete hain (Section 13.5.1).
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} ⁣ :$ ek solution; $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} ⁣ :$ infinitely many; $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} ⁣ :$ koi nahi (Table 13.3).
Graph mein, intersecting lines ek solution deti hain, coincident lines infinitely many, aur parallel lines koi nahi (Section 13.6).

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise Set 13.1, Q1	Practice 13.1 · Q1
NCERT Exercise Set 13.1, Q2	Practice 13.1 · Q2
NCERT Exercise Set 13.1, Q3	Practice 13.1 · Q3
NCERT Exercise Set 13.2, Q1	Practice 13.2 · Q1
NCERT Exercise Set 13.2, Q2	Practice 13.2 · Q2
NCERT Exercise Set 13.2, Q3	Practice 13.2 · Q3
NCERT Exercise Set 13.2, Q4	Practice 13.2 · Q4
NCERT Exercise Set 13.2, Q5	Practice 13.2 · Q5
NCERT Exercise Set 13.2, Q6	Practice 13.2 · Q6
NCERT Exercise Set 13.2, Q7	Practice 13.2 · Q7
NCERT Exercise Set 13.3, Q1	Practice 13.3 · Q1
NCERT Exercise Set 13.3, Q2	Practice 13.3 · Q2
NCERT Exercise Set 13.3, Q3	Practice 13.3 · Q3
NCERT Exercise Set 13.3, Q4	Practice 13.3 · Q4
NCERT Exercise Set 13.4, Q1	Practice 13.4 · Q1
NCERT Exercise Set 13.4, Q2	Practice 13.4 · Q2
NCERT Exercise Set 13.4, Q3	Practice 13.4 · Q3
NCERT Exercise Set 13.5, Q1 (i)–(iv)	Practice 13.5 · Q1
NCERT Exercise Set 13.5, Q1 (v)–(vii)	Practice 13.5 · Q2
NCERT Exercise Set 13.5, Q1 (viii)–(x)	Practice 13.5 · Q3
NCERT Exercise Set 13.5, Q2	Practice 13.5 · Q4
NCERT Exercise Set 13.5, Q3	Practice 13.5 · Q5
NCERT Exercise Set 13.5, Q4	Practice 13.5 · Q6
NCERT Exercise Set 13.5, Q5	Practice 13.5 · Q7
NCERT Exercise Set 13.5, Q6	Practice 13.5 · Q8
NCERT Exercise Set 13.5, Q7	Practice 13.5 · Q9
NCERT Exercise Set 13.5, Q8	Practice 13.5 · Q10
NCERT Exercise Set 13.5, Q9	Practice 13.5 · Q11
NCERT End-of-chapter Exercises, Q1	Chapter ke end ki exercises · Q1
NCERT End-of-chapter Exercises, Q2	Chapter ke end ki exercises · Q2
NCERT End-of-chapter Exercises, Q3	Chapter ke end ki exercises · Q3
NCERT End-of-chapter Exercises, Q4	Chapter ke end ki exercises · Q4
NCERT End-of-chapter Exercises, Q5	Chapter ke end ki exercises · Q5
NCERT End-of-chapter Exercises, Q6	Chapter ke end ki exercises · Q6
NCERT End-of-chapter Exercises, Q7	Chapter ke end ki exercises · Q7
NCERT End-of-chapter Exercises, Q8	Chapter ke end ki exercises · Q8
NCERT End-of-chapter Exercises, Q9	Chapter ke end ki exercises · Q9
NCERT End-of-chapter Exercises, Q10	Chapter ke end ki exercises · Q10
NCERT End-of-chapter Exercises, Q11	Chapter ke end ki exercises · Q11
NCERT End-of-chapter Exercises, Q12	Chapter ke end ki exercises · Q12
NCERT End-of-chapter Exercises, Q13	Chapter ke end ki exercises · Q13
NCERT End-of-chapter Exercises, Q14	Chapter ke end ki exercises · Q14
NCERT End-of-chapter Exercises, Q15	Chapter ke end ki exercises · Q15
NCERT End-of-chapter Exercises, Q16	Chapter ke end ki exercises · Q16

Quick check: MCQs

13.1.1 Do Variables wali Linear Equations ka Standard Form3 questions
13.2.2 Solutions ko Geometrically Dekhna4 questions
13.3.2 Slope-Intercept Form4 questions
13.4 Do Variables wali Linear Equations ka Pair3 questions
Jab elimination dono letters hata de5 questions
13.6 Do Variables wali Linear Equations ke Pair ko Solve karne ka Graphical Method4 questions

Glossary · 15

Key terms (English + Hindi)

coefficientगुणांकgunaank: Woh number jo kisi variable ko multiply karta hai, jaise $3 x$ mein 3.Chapter mein pehli baar yahan aaya →
coincident linesसंपाती रेखाएँsampaati rekhaen: Do equations jinke graphs ek hi line hain, isliye har point common hota hai.Chapter mein pehli baar yahan aaya →

elimination methodविलोपन विधिvilopan vidhi: Pair ko aise solve karna: ek variable ke coefficients match karo aur add ya subtract karo taaki woh cancel ho jaaye.Chapter mein pehli baar yahan aaya →

graph of an equationआलेखaalekh: Un saare points ka set jinke coordinates equation satisfy karte hain; linear equation ke liye yeh ek straight line hai.Chapter mein pehli baar yahan aaya →

intersecting linesप्रतिच्छेदी रेखाएँpratichchhedi rekhaen: Lines jo bilkul ek point par milti hain.Chapter mein pehli baar yahan aaya →

linear equation in two variablesदो चरों वाला रैखिक समीकरणdo charon vaala rekhik samikaran: Aisi equation jise $a x + b y + c = 0$ likha ja sake, jahan $a$ aur $b$ dono zero na hon.Chapter mein pehli baar yahan aaya →

ordered pairक्रमित युग्मkramit yugm: $(x, y)$ ki tarah likhe do numbers, jahan order matter karta hai: $(4, 3)$ aur $(3, 4)$ alag hain.Chapter mein pehli baar yahan aaya →

pair of linear equationsरैखिक समीकरण युग्मrekhik samikaran yugm: Same do variables wali do linear equations, jinhe saath mein dekha jaata hai.Chapter mein pehli baar yahan aaya →
parallel linesसमांतर रेखाएँsamaantar rekhaen: Ek plane ki lines jo kabhi nahi milti; unke slopes barabar hote hain.Chapter mein pehli baar yahan aaya →

slopeढालdhaal: Line ki steepness: line ke kinhi bhi do points ke beech rise divided by run.Chapter mein pehli baar yahan aaya →
slope-intercept formढाल-अंतःखंड रूपdhaal-antahkhand roop: Form $y = m x + d$ , jo slope $m$ aur $y$ -intercept $d$ seedhe dikhata hai.Chapter mein pehli baar yahan aaya →
solutionहलhal: $x$ aur $y$ ki values jo saath milkar equation ki dono sides ko barabar kar dein.Chapter mein pehli baar yahan aaya →
standard formमानक रूपmaanak roop: Layout $a x + b y + c = 0$ , jismein har term ek side par aur doosri side par 0 ho.Chapter mein pehli baar yahan aaya →
substitution methodप्रतिस्थापन विधिpratisthaapan vidhi: Pair ko aise solve karna: ek variable ko doosre ke terms mein likho aur use doosri equation mein rakh do.Chapter mein pehli baar yahan aaya →

y-intercepty-अंतःखंडy-antahkhand: Woh value $d$ jahan line $y$ -axis se milti hai, point $(0, d)$ par.Chapter mein pehli baar yahan aaya →

Two Variables, One Line (Do Variables, Ek Line): Class 9 Maths Chapter 13 Hinglish mein

Is chapter mein aap kya seekhenge

13.1 Do Variables wali Linear Equations

13.1.1 Do Variables wali Linear Equations ka Standard Form

Practice 13.1: answers ke saath

13.2 Do Variables wali Linear Equation ka Solution

13.2.1 Solution ka Algebraic Matlab

13.2.2 Solutions ko Geometrically Dekhna

Practice 13.2: answers ke saath

13.3 Do Variables wali Linear Equation ka Slope-Intercept Form

13.3.1 Slope ko samajhna

13.3.2 Slope-Intercept Form

Practice 13.3: answers ke saath

13.4 Do Variables wali Linear Equations ka Pair

Practice 13.4: answers ke saath

13.5 Linear Equations ke Pair ke Solutions Nikaalna

13.5.1 Linear Equations ke Pair ko Solve karne ke Algebraic Methods

Substitution method

Elimination method

Jab elimination dono letters hata de

13.6 Do Variables wali Linear Equations ke Pair ko Solve karne ka Graphical Method

Practice 13.5: answers ke saath

Practice questions answers ke saath (NCERT “Chapter ke end ki exercises”)

Summary aur key points

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye