Pair of Linear Equations in Two Variables: Class 10 Maths Chapter 3 Hinglish mein

Mathematics · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

60 min
1 simulation
30 practice questions
14 MCQs

Is chapter mein aap kya seekhenge

Kisi rozana ki situation ko model karne ke liye do variables wali linear equations ka ek pair likho.
Dono lines draw karke aur unka common point padh kar linear equations ka pair solve karo.
Graph se aur a1/a2, b1/b2 aur c1/c2 compare karke pair ko consistent, inconsistent ya dependent mein classify karo.
Substitution method se pair solve karo, aur bina variable wale true ya false statement ko infinitely many ya no solutions ki tarah samjho.
Elimination method se pair solve karo, aise multipliers chun kar jo ek variable ke coefficients ko barabar kar dein.
Prices, ages, digits, fractions aur fees wale word problems solve karo, aur har answer ko original conditions par check karo.

NCERT Mathematics Class 10, Chapter 3, pp. 24–37 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

3.1 Introduction

Har July, Van Mahotsav week mein, tumhare school ka eco-club ek plant stall lagata hai. Neem saplings ₹15 each aur tulsi pots ₹10 each mein bikte hain. Stall band hone tak club 30 plants bech chuka hai aur cash box mein ₹380 hain. Kisi ne tally nahi rakhi. Kitne neem saplings bike?

Tum guess se shuru kar sakte ho. Maan lo 10 neem saplings bike. Tab 20 tulsi pots bhi bike, aur cash $150 + 200 = 350$ rupees hota, jo ₹30 kam hai. Ek tulsi pot ki jagah ek neem sapling rakhne se total 30 plants hi rehta hai par ₹5 badh jaata hai. To chhe swaps se gap bhar jaata hai: 16 neem aur 14 tulsi. Fig. 3.1 mein saare guesses saath-saath dikhaye gaye hain.

3/3Solah neem aur chaudah tulsi se theek ₹380 aate hain.

Fig. 3.1Guess karo, check karo aur adjust karo jab tak cash match na ho jaaye.

Collect hue rupees ka ek bar chart, 380 par ek dashed target line ke saath. Step 1 mein 10 neem aur 20 tulsi ka ek bar 350 tak pahunchta hai, us par 30 rupees kam likha hai. Step 2 mein 20 neem aur 10 tulsi ka bar judta hai jo 400 tak jaata hai, us par 20 rupees zyada likha hai. Step 3 mein 16 neem aur 14 tulsi ka ek green bar judta hai jo theek target line tak pahunchta hai, us par spot on likha hai.

Guessing isliye kaam kar gayi kyunki numbers chhote aur aasaan the. Agar stall ne 347 plants beche hote to yeh slow hota, aur ₹12.50 aur ₹8.75 jaise prices ke saath to bilkul bekaar. Algebra ek aisa method deta hai jo hamesha kaam karta hai. Maan lo $x$ neem saplings ki sankhya hai aur $y$ tulsi pots ki sankhya. Stall ke do facts do equations ban jaate hain:

\begin{aligned} x + y & = 30 \\ 15 x + 10 y & = 380 \end{aligned}

Har equation ke akele mein bahut saare solutions hain. Jaise $x = 10$ , $y = 20$ pehli equation mein fit hota hai par doosri mein nahi. Humein woh values chahiye jo dono mein ek saath fit hon. Same do unknowns mein do linear equations, saath mein lekar, ek pair of linear equations in two variables (दो चरों वाले रैखिक समीकरण युग्म) banati hain.

Class IX se yaad karo ki ek linear equation in two variables ko $a x + b y + c = 0$ ki tarah likha ja sakta hai. Yahan $a$ , $b$ aur $c$ real numbers hain, aur $a$ aur $b$ dono zero nahi hain. Iske infinitely many solutions hote hain, aur jab hum unhe plot karte hain to woh sab ek straight line par aate hain. Woh line equation ka graph hai: us par har point ek solution hai, aur har solution us par ek point hai.

Definition

Pair of linear equations in two variables

Same do variables, jaise $x$ aur $y$ , mein do linear equations jinhe saath mein dekha jaata hai. Solution of the pair values ka woh pair hai jo dono equations ko ek hi time par satisfy karta hai.

Example $x + y = 30$ aur $15 x + 10 y = 380$ ka solution $x = 16$ , $y = 14$ hai.

$x$ aur $y$ mein linear equations ke kisi bhi pair ko neeche diye general form mein likha ja sakta hai. Chhe numbers iske coefficients hain; chhota 1 ya 2 batata hai ki number kis equation ka hai.

Pair ka general form

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

(3.1)

jahan

$a_{1}, b_{1}, c_{1}$: real numbers, jahan $a_{1}$ aur $b_{1}$ dono zero nahi hain
$a_{2}, b_{2}, c_{2}$: real numbers, jahan $a_{2}$ aur $b_{2}$ dono zero nahi hain

Eco-club pair ke liye $a_{1} = 1$ , $b_{1} = 1$ , $c_{1} = - 30$ aur $a_{2} = 15$ , $b_{2} = 10$ , $c_{2} = - 380$ . Signs dhyaan se dekho: 30 ko left side le jaane par woh $- 30$ ban jaata hai. Is chapter mein tum solution nikaalne ke teen tareeke dekhoge: graphs draw karna (Section 3.2), substitution (Section 3.3.1) aur elimination (Section 3.3.2).

3.2 Graphical Method of Solution of a Pair of Linear Equations

Pair ki har equation ka graph ek straight line hai, isliye ek pair same plane mein do lines deta hai. Pair ka solution dono lines par hona chahiye. To pair ke solutions theek woh points hain jo dono lines mein common hain.

Fig. 3.2 mein eco-club equations draw ki gayi hain. Solid line 30 plants bechne ka har tareeka collect karti hai. Dashed line ₹380 lene ka har tareeka collect karti hai. Sirf ek point dono lines par hai, $(16, 14)$ , jo wahi answer hai jo humne guess karke nikaala tha.

3/3Sirf $(16, 14)$ dono lines par hai: 16 neem saplings aur 14 tulsi pots.

Fig. 3.2Pair ka solution woh point hai jo uski dono lines mein common hai.

Ek grid jismein neem saplings x across hain, 0 se 32 tak, aur tulsi pots y upar ki taraf, 0 se 40 tak. Step 1 solid blue line x + y = 30 draw karta hai, (10, 20) aur (20, 10) par dots ke saath, jise 30 plants kaha gaya hai. Step 2 dashed orange line 15x + 10y = 380 jodta hai, (4, 32) aur (20, 8) par dots ke saath, jise kul 380 rupees kaha gaya hai. Step 3 crossing point (16, 14) ko ring karta hai aur batata hai ki wahan dono facts sach hain.

SimulationCommon solution dhoondho

Dono axes par minus 2 se 8 tak ek coordinate grid. Fixed points A (0, 4), B (4, 2) aur C (6, 1) equation x + 2y = 8 ke solutions hain, aur fixed points P (0, minus 1), Q (1, 1) aur R (3, 5) equation 2x minus y = 1 ke solutions hain. Point S (6, 6) se shuru hota hai aur use whole-number positions par drag kiya ja sakta hai, saath mein uske coordinates dikhte hain, taaki tum test kar sako ki woh kis set of points ki line mein hai.

Fig. 3.3

A

B

C

equation

x + 2 y = 8

ko solve karte hain aur

P

Q

R

equation

2 x - y = 1

ko.

S

ko drag karo.

Try karo

$S$ ko aise drag karo ki woh $A$ , $B$ aur $C$ ki line mein aa jaaye. Kya uske coordinates $x + 2 y = 8$ ko satisfy karte hain?
Woh ek jagah dhoondho jahan $S$ , $A, B, C$ aur $P, Q, R$ dono ki line mein aata hai. Use dono equations mein test karo.
$S$ ko $(4, 7)$ par rakho. Yeh kaun si equation satisfy karta hai, aur points ki kaun si row ki line mein hai?

Same plane mein bani do lines sirf teen tarah se behave kar sakti hain, jo Fig. 3.4 mein dikhaya gaya hai. Har tareeka batata hai ki pair ke kitne solutions hain.

Woh ek point par intersect karti hain. Pair ka exactly ek solution hota hai, jise unique solution kehte hain.
Woh parallel hain. Lines kabhi nahi milti, isliye pair ka koi solution nahi hota.
Woh coincide karti hain. Ek line bilkul doosri ke upar padi hai. Uska har point common hai, isliye pair ke infinitely many solutions hote hain.

1/4Lines ka har pair in teen cases mein se kisi ek mein aata hai.

Fig. 3.4Plane mein do lines ek baar milti hain, kabhi nahi milti, ya har jagah milti hain.

Teen chhote graphs saath-saath, har ek mein ek solid line 1 aur ek dashed line 2. Figure step by step dikhata hai: opening view mein teenon panels ek saath dikhte hain. Intersecting step left panel ko highlight karta hai, jahan lines ek ringed point par cross karti hain: ek solution, consistent. Parallel step middle panel ko highlight karta hai, jahan lines bina mile saath-saath chalti hain: koi solution nahi, inconsistent. Coincident step right panel ko highlight karta hai, jahan dashed line solid line ke theek upar padi hai: infinitely many solutions, dependent.

Inhi cases se pairs ke naam milte hain. Ek pair consistent (संगत) hai agar uska kam se kam ek solution ho, aur inconsistent (असंगत) hai agar koi solution na ho. Jab ek equation bas doosri ka multiple hoti hai, to dono same line draw karti hain, aur pair ko dependent (आश्रित) kehte hain. Kyunki iske solutions hain (infinitely many), dependent pair bhi consistent gina jaata hai.

Definition

Consistent pair

Linear equations ka aisa pair jiska kam se kam ek solution ho: exactly ek (intersecting lines) ya infinitely many (coincident lines).

Definition

Inconsistent pair

Linear equations ka aisa pair jiska koi solution nahi. Iske graphs parallel lines hote hain.

Definition

Dependent pair

Aisa pair jismein ek equation doosri ka multiple hai, isliye dono same line draw karti hain aur infinitely many solutions share karti hain. Solutions hone ki wajah se yeh consistent gina jaata hai.

Coefficients se case predict karna

Do lines draw karne mein time lagta hai. Kya hum sirf numbers dekh kar case bata sakte hain? Dono equations ko general form (3.1) mein likho aur teen ratios compare karo: $\frac{a_{1}}{a_{2}}$ , $\frac{b_{1}}{b_{2}}$ aur $\frac{c_{1}}{c_{2}}$ . Table 3.1 yeh un teen pairs ke liye karta hai jinke graphs hum pehle se jaante hain:

Pair A: $2 x - y - 1 = 0$ aur $x + 2 y - 8 = 0$ , jo $(2, 3)$ par milti hain
Pair B: $3 x - 2 y + 4 = 0$ aur $6 x - 4 y + 8 = 0$
Pair C: $x - 3 y + 3 = 0$ aur $2 x - 6 y - 6 = 0$

Table 3.1Teen pairs ke coefficient ratios compare karna

	$\frac{a_{1}}{a_{2}}$	$\frac{b_{1}}{b_{2}}$	$\frac{c_{1}}{c_{2}}$	Lines	Solutions
A	$2$	$- \frac{1}{2}$	$\frac{1}{8}$	cross karti hain	ek
B	$\frac{1}{2}$	$\frac{1}{2}$	$\frac{1}{2}$	ek hi line	infinitely many
C	$\frac{1}{2}$	$\frac{1}{2}$	$- \frac{1}{2}$	parallel	koi nahi

Columns ko upar se neeche padho. Jab lines cross karti hain, to pehle do ratios alag hote hain. Jab lines parallel hoti hain ya coincide karti hain, to pehle do ratios barabar hote hain, aur teesra ratio faisla karta hai: coincident lines ke liye barabar, parallel lines ke liye alag.

Compare

Ratio test

Intersecting lines

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
exactly ek solution
consistent

Coincident lines

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
infinitely many solutions
dependent (consistent)

Parallel lines

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
koi solution nahi
inconsistent

PropertyRatio test kaam kyun karta hai

General form (3.1) wale pair ke liye, ratio test ki teen conditions exactly batati hain ki lines kab intersect karti hain, coincide karti hain ya parallel hoti hain. Har condition ulti taraf bhi kaam karti hai: intersecting lines mein hamesha $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ hota hai, aur isi tarah baaki bhi.

Maan lo $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}$ , aur is common value ko $k$ kaho. Tab $a_{1} = k a_{2}$ aur $b_{1} = k b_{2}$ , isliye pehli equation $k (a_{2} x + b_{2} y) + c_{1} = 0$ ban jaati hai.

Agar $c_{1} = k c_{2}$ bhi ho, to pehli equation bas doosri ka $k$ guna hai. Dono ke solutions same hain, isliye woh same line draw karti hain.

Agar $c_{1} \neq k c_{2}$ , to common point ke liye $a_{2} x + b_{2} y = - c_{2}$ aur saath mein $k (a_{2} x + b_{2} y) = - c_{1}$ chahiye. Isse $k c_{2} = c_{1}$ banna padega, jo galat hai. To koi common point nahi hai: lines parallel hain.

Agar $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ , to $a_{1} b_{2} - a_{2} b_{1} \neq 0$ . Is case mein Section 3.3.2 ka elimination method hamesha $x$ ki exactly ek value aur $y$ ki ek value deta hai, isliye lines ek baar cross karti hain.

Agar doosri equation ka koi coefficient 0 hai, to neeche 0 wala ratio nahi ban sakta. Tab $a_{1} b_{2}$ ko $a_{2} b_{1}$ se compare karo (ratios ko cross-multiply karke), ya bas lines draw kar lo. Neeche wali lab mein teenon cases try karo.

InteractiveTwo-line lab

Dono axes par minus 8 se 8 tak ek interactive graph jismein do lines hain, ek solid blue line 1 aur ek dashed orange line 2, dono a x plus b y plus c equals 0 form mein likhi hui. Plus aur minus buttons chhe whole-number coefficients mein se har ek ko minus 9 se 9 tak badalte hain. Ek table ratios a1 over a2, b1 over b2 aur c1 over c2 ko reduced fractions mein dikhati hai, aur ek status box batata hai ki lines intersect karti hain, parallel hain ya coincide karti hain, pair consistent hai, inconsistent hai ya dependent, aur agar exact solution hai to woh deta hai, jo graph par bhi ring kiya jaata hai. Buttons milne wala, parallel aur same-line example load karte hain. Make it mode line 1 ko fix kar deta hai aur tumse line 2 aise badalne ko kehta hai ki lines parallel ho jaayein, coincide karein, ya ek marked point par milein, hints aur score ke saath.

Fig. 3.5Chhe coefficients badlo aur lines, ratios

\frac{a_{1}}{a_{2}}

\frac{b_{1}}{b_{2}}

\frac{c_{1}}{c_{2}}

aur solutions ki sankhya dekho.

Pair ko graph se solve karna

Pair ko graphs se solve karne ke liye har equation ke do ya teen solutions nikaalo, unhe plot karo, dono lines kheencho aur common point padh lo. $x = 0$ ya $y = 0$ rakhne se aksar aasaan points mil jaate hain. Har line par teesra point galtiyon ke khilaaf ek sasta check hai.

Example 3.1NCERT Example 1 · p. 26

Kya pair consistent hai? Graph se solve karo

Graphs ki madad se tay karo ki $2 x + 3 y = 12$ aur $x - y = 1$ ka koi common solution hai ya nahi, aur agar hai to use nikaalo.

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

4/4Lines $C (3, 2)$ par milti hain, yahi unique solution hai.

Fig. 3.6Do points har line ko fix karte hain; crossing point pair ko solve karta hai.NCERT Fig. 3.1 · p. 27

Ek grid jo across minus 1 se 7 aur upar minus 2 se 5 tak hai. Step 1 pehli equation se A (0, 4) aur B (6, 0) plot karta hai. Step 2 A aur B se hokar solid blue line kheenchta hai, jis par 2x + 3y = 12 likha hai. Step 3 P (1, 0) aur Q (5, 4) plot karta hai aur dashed orange line x minus y = 1 kheenchta hai. Step 4 crossing point C (3, 2) ko ring karta hai aur batata hai ki lines wahan milti hain.

Example 3.2NCERT Example 2 · p. 27

Coincident pair pehchaanna

Pair $6 x - 9 y + 3 = 0$ aur $\frac{x}{2} - \frac{3 y}{4} + \frac{1}{4} = 0$ ke kitne solutions hain? Fractions hata kar tay karo, phir kuch points se confirm karo.

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

Example 3.3NCERT Example 3 · p. 27

Kaun si bicycle shop sasti hai?

Manali mein Shop P bicycles ₹60 plus ₹20 per hour par kiraye par deti hai, aur Shop Q ₹100 plus ₹10 per hour par. Har shop par $x$ hours ki cost ko equation ki tarah likho, aur graphs se pata karo ki dono costs kab barabar hoti hain.

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

2/2Lines $(4, 140)$ par milti hain: dono shops 4 hours ke ₹140 leti hain.

Fig. 3.7Shop P sasti shuru hoti hai par tezi se badhti hai.NCERT Fig. 3.2 · p. 28

Hours x across 0 se 8 tak hain aur cost y rupees mein upar 0 se 240 tak. Step 1 Shop P ki solid blue line dikhata hai, jo 60 se shuru hokar har hour 20 badhti hai, aur Shop Q ki dashed orange line, jo 100 se shuru hokar har hour 10 badhti hai. Step 2 unke crossing point (4, 140) ko ring karta hai, neeche 4 hours tak aur across 140 rupees tak dashed guide lines ke saath, aur batata hai ki 4 hours par cost same hai.

Graph milne wale point se zyada dikhata hai. $x = 4$ ke left mein Shop P ki line neeche hai, isliye chhoti rides ke liye P sasti hai. Uske right mein Q sasti hai. Graphs aise comparisons ek nazar mein dikhane mein bahut achhe hain.

Quick check

4 mein se 1

Lines ke har pair ko uske solutions ki sankhya se match karo.

Left mein koi item tap karo, phir right mein uska match.

Answer dekho

Answer

Intersecting : exactly ek
Parallel : koi nahi
Coincident : infinitely many

Solutions common points hi hain. Cross karti lines ka ek point common hota hai, parallel lines ka koi nahi, aur coincident lines ke saare points common hote hain.

3.3 Algebraic Methods of Solving a Pair of Linear Equations

Jab lines kisi grid point par milti hain to graphs achha kaam karte hain, lekin asli problems itni meharbaan kam hi hoti hain. Fig. 3.9 mein pair $5 x - 3 y = 9$ aur $x + 2 y = 6$ dekho. Lines saaf-saaf cross karti hain, par exactly kahan? Kahin $(2.8, 1.6)$ ke aas-paas, aur magnified view bhi isse zyada nahi bata sakta.

3/3Algebra exact point $(\frac{36}{13}, \frac{21}{13})$ deta hai (Example 3.4).

Fig. 3.9Drawing crossing point ko sirf andaaze se locate kar sakti hai.

Step 1 lines 5x minus 3y = 9 aur x + 2y = 6 ko grid lines ke beech cross karte dikhata hai, ek question mark aur (2.8, 1.6) ke aas-paas wale guess ke saath. Step 2 x = 2 se 3 aur y = 1 se 2 tak ka ek magnified square jodta hai jismein tenths mark hain; crossing point phir bhi across 2.7 aur 2.8 ke beech aur upar 1.6 aur 1.7 ke beech rehta hai. Step 3 ek box mein exact answer, 36 over 13 aur 21 over 13, dikhata hai.

Pencil ki line ki thickness hoti hai, graph paper ke squares limited hote hain, aur hamari aankhen exact nahi hotin. $\frac{36}{13}$ jaise answers pakke taur par nikaalne ke liye humein algebra chahiye. Agle do subsections aise do methods dete hain.

3.3.1 Substitution Method

Substitution method (प्रतिस्थापन विधि) ka idea simple hai. Ek equation ko aise rearrange karo ki ek variable ek side par akela rahe, jaise $x = 6 - 2 y$ . Phir doosri equation mein us variable ki jagah yeh rakh do. Jo bachta hai usmein sirf ek unknown hota hai, aur use solve karna tum pehle se jaante ho.

Example 3.4NCERT Example 4 · p. 30

Substitution se solve karna

$5 x - 3 y = 9$ (1) aur $x + 2 y = 6$ (2) ko substitution se solve karo.

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

Key idea

Substitution method

Express. Ek equation ko rearrange karke ek variable ko akela karo, jaise $x = 6 - 2 y$ . 1 ya $- 1$ coefficient ho to yeh aasaan hota hai.
Substitute. Doosri equation mein us variable ki jagah yeh expression rakho. Ab sirf ek unknown bachta hai, use solve karo.
Back-substitute. Step 1 wala expression ab seedha doosra unknown de deta hai. Aakhir mein dono original equations mein check karo.

Kabhi-kabhi step 2 mein variable cancel ho jaata hai aur sirf numbers bachte hain, jaise $7 = 7$ ya $0 = 4$ . $7 = 7$ jaisa hamesha sach result batata hai ki ek line ka har point doosri par hai: infinitely many solutions. $0 = 4$ jaisa impossible result batata hai ki lines kabhi nahi milti: koi solution nahi. Example 3.6 aur Example 3.7 dono dikhate hain.

Naam step 2 se aata hai, jahan ek expression variable ki jagah leta hai: use uski jagah substitute kiya jaata hai.

Example 3.5NCERT Example 5 · p. 31

Dadaji aur pote ki ages

Dadaji apne pote Kabir se 63 saal bade hain. 8 saal baad Dadaji ki age Kabir ki age ki chaar guna hogi. Unki present ages nikaalo.

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

3/3Gap ab bhi 63 saal hai: $(g + 8) - (k + 8) = g - k$ .

Fig. 3.10Dono ages utne hi saal badhti hain, isliye gap kabhi nahi badalta.

Ek timeline arrow jis par do marks hain, abhi aur 8 saal baad. Dadaji aur Kabir ki rows mein unki ages hain. Step 1 present ages g aur k dikhata hai, ek bracket 63 saal ka gap mark karta hai aur equation g minus k = 63 hai. Step 2 8 saal baad ki ages, g + 8 aur k + 8, aur condition g + 8 = 4 times (k + 8) jodta hai. Step 3 future ages par bhi bracket lagata hai aur batata hai ki gap ab bhi 63 saal hai, kyunki dono ages utni hi badhti hain.

$k$ across aur $g$ upar wale graph par lines (1) aur (2) $(13, 76)$ par milti. Substitution bina kuch draw kiye usi point tak pahunch jaata hai.

Example 3.6NCERT Example 6 · p. 32

Variable gayab ho jaata hai: ek true statement

Ek bakery mein Neha 4 cupcakes aur 6 cookies ke liye ₹90 deti hai. Arjun 6 cupcakes aur 9 cookies ke liye ₹135 deta hai. Kya tum ek cupcake aur ek cookie ka price nikaal sakte ho?

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

Example 3.7NCERT Example 7 · p. 32

Variable gayab ho jaata hai: ek false statement

Ravi kehta hai ki school ki stationery shop par 2 notebooks aur 6 pens ₹90 ke hain. Sana kehti hai ki usi shop par 1 notebook aur 3 pens ₹50 ke hain. Kya dono sahi ho sakte hain?

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

Fig. 3.11 pichhle do examples ko saath-saath rakhta hai. $135 = 135$ jaisa true statement batata hai ki dono lines ek doosre ke upar padi hain. $100 = 90$ jaisa false statement batata hai ki woh kabhi nahi milti.

1/3Gayab variable ya to hamesha sach ya kabhi sach na hone wala statement chhodta hai.

Fig. 3.11Variable cancel hone ke baad jo bachta hai, woh case bata deta hai.

Do panels saath-saath. Left panel ke upar 135 = 135, hamesha sach, likha hai aur usmein ek solid line hai jiske theek upar ek dashed line padi hai: same line, infinitely many solutions. Right panel ke upar 100 = 90, kabhi sach nahi, likha hai aur usmein ek solid line aur ek dashed line parallel chalti hain: koi solution nahi. Opening step dono panels dikhata hai; true step left panel ko highlight karta hai aur false step right panel ko.

Quick check

3 mein se 1

Substitution method ke steps ko sahi order mein lagao.

$y$ se $x$ nikaalo
Dono equations check karo
$x$ ko $y$ ke terms mein likho
Doosri equation mein substitute karo
$y$ nikaalo

Answer dekho

Answer

$x$ ko $y$ ke terms mein likho
Doosri equation mein substitute karo
$y$ nikaalo
$y$ se $x$ nikaalo
Dono equations check karo

Ek variable express karo, use doosri equation mein substitute karo, solve karo, back-substitute karo, aur aakhir mein dono original equations mein check karo.

Practice · 9 questions

Practice 3.2: answers ke saath

1
Short answerEasyNCERT Exercise 3.2, Q1
Har pair ko substitution method se solve karo.
1. $x + y = 11$ aur $x - y = 5$
  Solution dikhao
  Answer $x = 8$ , $y = 3$
  Pehli se, $x = 11 - y$ .
  $11 - y - y = 5$ , isliye $y = 3$ .
  Tab $x = 8$ .
2. $p - q = 6$ aur $\frac{p}{4} + \frac{q}{3} = 5$
  Solution dikhao
  Answer $p = 12$ , $q = 6$
  Doosri ko 12 se multiply karo: $3 p + 4 q = 60$ . Pehli se, $p = q + 6$ .
  $3 (q + 6) + 4 q = 60$ se $7 q = 42$ milta hai, isliye $q = 6$ .
  Tab $p = 12$ .
3. $0.4 x + 0.3 y = 2.5$ aur $0.2 x + 0.5 y = 2.3$
  Solution dikhao
  Answer $x = 4$ , $y = 3$
  Dono ko 10 se multiply karo: $4 x + 3 y = 25$ aur $2 x + 5 y = 23$ . Doosri se, $2 x = 23 - 5 y$ .
  $2 (23 - 5 y) + 3 y = 25$ se $46 - 7 y = 25$ milta hai, isliye $y = 3$ .
  Tab $2 x = 23 - 15$ , isliye $x = 4$ .
4. $2 x - y = 4$ aur $6 x - 3 y = 12$
  Solution dikhao
  AnswerInfinitely many solutions hain.
  Pehli se, $y = 2 x - 4$ .
  $6 x - 3 (2 x - 4) = 12$ , isliye $12 = 12$ .
  Yeh hamesha sach hai: $y = 2 x - 4$ ka har point ek solution hai.
Short answerHardNCERT Exercise 3.2, Q1(v)
$\sqrt{5} x + \sqrt{2} y = 0$ aur $\sqrt{2} x - \sqrt{5} y = 0$ ko substitution se solve karo.
Solution dikhao
Answer $x = 0$ , $y = 0$
$y = - \frac{\sqrt{5}}{\sqrt{2}} x$ ko doosri equation mein substitute karo:
$\sqrt{2} x + \frac{5}{\sqrt{2}} x = 0$ , isliye $\frac{7}{\sqrt{2}} x = 0$ .
To $x = 0$ , aur phir $y = 0$ . Dono lines origin se guzarti hain aur alag-alag lines hain, isliye origin hi akela solution hai.
Short answerMediumNCERT Exercise 3.2, Q2
Pehle $3 x + 2 y = 4$ aur $5 x - 2 y = 12$ solve karo. Solution line $y = m x + 5$ par bhi hai: $m$ kya hai?
Solution dikhao
Answer $x = 2$ , $y = - 1$ , $m = - 3$
Pehli se, $2 y = 4 - 3 x$ . Ise doosri mein substitute karo:
$5 x - (4 - 3 x) = 12$ se $8 x = 16$ milta hai, isliye $x = 2$ .
Tab $2 y = 4 - 6$ , isliye $y = - 1$ . Aakhir mein $- 1 = 2 m + 5$ , isliye $m = - 3$ .
Short answerEasyNCERT Exercise 3.2, Q3(i)
Do numbers ka sum 96 hai, aur unmein se ek doosre ka paanch guna hai. Numbers nikaalo.
Solution dikhao
Answer80 aur 16
Maan lo numbers $x$ aur $y$ hain, jahan $x + y = 96$ aur $x = 5 y$ . Tab $6 y = 96$ , isliye $y = 16$ aur $x = 80$ .
Short answerEasyNCERT Exercise 3.2, Q3(ii)
Do angles complementary hain. Bada angle chhote ke do gune se 12° zyada hai. Dono angles nikaalo.
Solution dikhao
Answer64° aur 26°
Maan lo angles $x °$ aur $y °$ hain: $x + y = 90$ aur $x = 2 y + 12$ .
Tab $3 y + 12 = 90$ , isliye $y = 26$ aur $x = 64$ .
Short answerMediumNCERT Exercise 3.2, Q3(iii)
Ek sports teacher 5 footballs aur 8 skipping ropes ₹2,740 mein kharidti hai. Baad mein woh 3 footballs aur 10 skipping ropes ₹2,060 mein kharidti hai. Ek football aur ek skipping rope ka price nikaalo.
Solution dikhao
AnswerFootball ₹420 ka, skipping rope ₹80 ki.
Maan lo ek football ₹ $f$ ka hai aur ek rope ₹ $r$ ki.
$5 f + 8 r = 2740$ aur $3 f + 10 r = 2060$ .
Doosri se, $3 f = 2060 - 10 r$ . Pehli ko 3 se multiply karo aur substitute karo:
$5 (2060 - 10 r) + 24 r = 8220$ se $26 r = 2080$ milta hai, isliye $r = 80$ .
Tab $3 f = 2060 - 800 = 1260$ , isliye $f = 420$ .
Short answerMediumNCERT Exercise 3.2, Q3(iv)
Ek water-tanker service ek fixed delivery fee aur har hazaar litres ka ek rate leti hai. 6 hazaar litres ki delivery ₹1,040 ki hai aur 10 hazaar litres ki ₹1,600 ki. Fee aur rate nikaalo. 15 hazaar litres ki cost kya hogi?
Solution dikhao
AnswerFee ₹200, rate ₹140 per hazaar litres; 15 hazaar litres ki cost ₹2,300.
Maan lo fee ₹ $f$ hai aur rate ₹ $r$ .
$f + 6 r = 1040$ aur $f + 10 r = 1600$ .
Pehli se, $f = 1040 - 6 r$ . Tab $1040 + 4 r = 1600$ , isliye $r = 140$ aur $f = 200$ .
15 hazaar litres ke liye: $200 + 2100 = 2300$ rupees.
Short answerMediumNCERT Exercise 3.2, Q3(v)
Agar kisi fraction ke numerator aur denominator dono mein se 3 ghataya jaaye, to woh $\frac{1}{2}$ ban jaata hai. Agar dono mein 1 joda jaaye, to woh $\frac{2}{3}$ ban jaata hai. Fraction nikaalo.
Solution dikhao
Answer $\frac{7}{11}$
Pehli condition: $2 (x - 3) = y - 3$ , isliye $y = 2 x - 3$ . Doosri: $3 (x + 1) = 2 (y + 1)$ , isliye $3 x - 2 y = - 1$ .
$3 x - 2 (2 x - 3) = - 1$ se $- x + 6 = - 1$ milta hai, isliye $x = 7$ .
To $y = 11$ . Check: $\frac{4}{8} = \frac{1}{2}$ aur $\frac{8}{12} = \frac{2}{3}$ .
Short answerMediumNCERT Exercise 3.2, Q3(vi)
Meher ki aunt Meher se 26 saal badi hain. Teen saal pehle aunt ki age Meher ki age ki teen guna thi. Unki present ages nikaalo.
Solution dikhao
AnswerAunt 42 saal, Meher 16 saal.
Maan lo aunt $a$ saal ki hain aur Meher $m$ saal ki.
$a - m = 26$ aur $a - 3 = 3 (m - 3)$ .
Pehli se $a = m + 26$ milta hai. Ise doosri mein substitute karo: $m + 23 = 3 m - 9$ , isliye $m = 16$ aur $a = 42$ .
Check: $42 - 16 = 26$ , aur teen saal pehle woh 39 aur 13 ki thin.

3.3.2 Elimination Method

Substitution tab sabse tez hai jab kisi variable ka coefficient 1 ho. Jab har coefficient bada ho, to yeh fractions le aata hai. Elimination method (विलोपन विधि) inse bachata hai. Hum equations ko sahi numbers se multiply karte hain taaki ek variable ka coefficient dono mein same ho jaaye. Phir equations ko subtract (ya add) karne se woh variable poori tarah mit jaata hai, aur eliminate ka matlab yahi hai.

Example 3.8NCERT Example 8 · p. 34

Sahi aur galat answers ke marks

Ek quiz mein har sahi answer par kuch marks milte hain aur har galat answer par kuch marks katte hain. Asha ke 14 sahi aur 4 galat the aur usne 48 score kiye. Bilal ke 11 sahi aur 5 galat the aur usne 34 score kiye. Sahi answer par kitne marks milte hain, aur galat par kitne katte hain?

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

4/4 $r = 4$ ko (1) mein rakhne se $w = 2$ milta hai.

Fig. 3.12Ek coefficient match karo, subtract karo, aur woh variable gayab ho jaata hai.

Equations ki rows, term ke hisaab se aligned. Step 1 equation (1), 14r minus 4w = 48, par times 5 aur equation (2), 11r minus 5w = 34, par times 4 likha dikhata hai. Step 2 equations (3), 70r minus 20w = 240, aur (4), 44r minus 20w = 136, jodta hai, dono minus 20w terms highlighted. Step 3 dono w-terms ko kaat deta hai, ek line kheenchta hai aur (3) minus (4) dikhata hai: 26r = 104, isliye r = 4, aur batata hai ki w chala gaya. Step 4 r = 4 ko (1) mein rakhne ka result jodta hai: w = 2.

Hum $r$ ko bhi eliminate kar sakte the, lekin uske liye $154 r$ tak pahunchne ko multipliers 11 aur 14 chahiye. Woh variable chunne se jiske coefficients ka common multiple chhota ho, numbers chhote rehte hain. Substitution bhi kaam karta, lekin woh ek fraction, $r = \frac{48 + 4 w}{14}$ , se shuru hota hai, isliye yahan elimination natural choice hai.

Example 3.9NCERT Example 9 · p. 35

Elimination mein false statement

Elimination se $3 x - 5 y = 4$ (1) aur $9 x - 15 y = 7$ (2) ke saare solutions nikaalo.

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

Example 3.10NCERT Example 10 · p. 35

Two-digit number aur uska reverse

Ek two-digit number aur uske digits ulat kar bane number ka difference 36 hai. Number ke digits ka sum 10 hai. Aise saare numbers nikaalo.

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

3/3Difference hamesha digits ke difference ka 9 guna hota hai.

Fig. 3.13Digits ulatne se unki place values aapas mein badal jaati hain.

Do slots jin par tens aur ones likha hai. Step 1 mein tens slot mein ek blue tile x aur ones slot mein ek orange tile y hai, value 10x + y ke saath. Step 2 tiles ko ek doosre ke paar khiskata hai taaki y tens slot mein aur x ones slot mein aa jaaye, value 10y + x aur note digits swapped ke saath. Step 3 difference (10x + y) minus (10y + x) = 9x minus 9y = 9 times (x minus y) dikhata hai, example 73 minus 37 = 9 times 4 = 36 ke saath.

Practice · 6 questions

Practice 3.3: answers ke saath

1
Short answerMediumNCERT Exercise 3.3, Q1
Har pair ko elimination se solve karo. Phir apna answer substitution se check karo.
1. $x + y = 9$ aur $3 x - 2 y = 2$
  Solution dikhao
  Answer $x = 4$ , $y = 5$
  Pehli ko 2 se multiply karo aur doosri mein add karo:
  $5 x = 20$ , isliye $x = 4$ .
  Tab $y = 5$ . Check: $12 - 10 = 2$ .
2. $2 x + 5 y = 16$ aur $3 x - 2 y = 5$
  Solution dikhao
  Answer $x = 3$ , $y = 2$
  $\pm 10 y$ paane ke liye 2 aur 5 se multiply karo:
  $4 x + 10 y = 32$ aur $15 x - 10 y = 25$ .
  Add karo: $19 x = 57$ , isliye $x = 3$ . Tab $5 y = 10$ , isliye $y = 2$ .
3. $4 x - 3 y - 7 = 0$ aur $6 x = 5 y + 9$
  Solution dikhao
  Answer $x = 4$ , $y = 3$
  Ise $4 x - 3 y = 7$ aur $6 x - 5 y = 9$ ki tarah likho. 3 aur 2 se multiply karo:
  $12 x - 9 y = 21$ aur $12 x - 10 y = 18$ .
  Subtract karo: $y = 3$ . Tab $4 x = 16$ , isliye $x = 4$ .
4. $\frac{x}{3} + \frac{y}{2} = 4$ aur $2 x - \frac{y}{4} = 11$
  Solution dikhao
  Answer $x = 6$ , $y = 4$
  Fractions hatao (6 aur 4 se multiply karo):
  $2 x + 3 y = 24$ aur $8 x - y = 44$ .
  Doosri ko 3 se multiply karke add karo: $26 x = 156$ , isliye $x = 6$ . Tab $y = 48 - 44 = 4$ .
Short answerMediumNCERT Exercise 3.3, Q2(i)
Agar kisi fraction ka numerator double kiya jaaye aur denominator 1 badhaya jaaye, to fraction 1 ban jaata hai. Agar sirf numerator mein se 2 ghataya jaaye, to woh $\frac{1}{3}$ ban jaata hai. Fraction nikaalo.
Solution dikhao
Answer $\frac{5}{9}$
Maan lo fraction $\frac{x}{y}$ hai. Tab $2 x = y + 1$ aur $3 (x - 2) = y$ .
$2 x - y = 1$ aur $3 x - y = 6$ .
Subtract karo: $x = 5$ , isliye $y = 9$ . Check: $\frac{10}{10} = 1$ aur $\frac{3}{9} = \frac{1}{3}$ .
Short answerMediumNCERT Exercise 3.3, Q2(ii)
Chhe saal pehle Gurpreet ki age uske bhatije Arman ki age ki paanch guna thi. Ab se baarah saal baad uski age Arman ki tab ki age ki do guna hogi. Abhi dono kitne saal ke hain?
Solution dikhao
AnswerGurpreet 36 ki hai aur Arman 12 ka.
Chhe saal pehle: $g - 6 = 5 (a - 6)$ . Baarah saal baad: $g + 12 = 2 (a + 12)$ .
$g - 5 a = - 24$ aur $g - 2 a = 12$ .
Doosri mein se pehli subtract karo: $3 a = 36$ , isliye $a = 12$ aur $g = 36$ . Check: 30 aur 6; 48 aur 24.
Short answerHardNCERT Exercise 3.3, Q2(iii)
Ek two-digit number ke digits ka sum 6 hai. Number ka saat guna, uske digits ulat kar bane number ke chaar gune ke barabar hai. Number nikaalo.
Solution dikhao
Answer24
Maan lo digits $x$ (tens) aur $y$ (units) hain, isliye $x + y = 6$ . Doosri condition simplify hoti hai:
$7 (10 x + y) = 4 (10 y + x)$ se $66 x = 33 y$ milta hai, isliye $2 x - y = 0$ .
Ise $x + y = 6$ mein add karo: $3 x = 6$ , isliye $x = 2$ aur $y = 4$ . Number 24 hai. Check: $7 \times 24 = 168 = 4 \times 42$ .
Short answerMediumNCERT Exercise 3.3, Q2(iv)
Ek dukaandaar ke drawer mein sirf ₹20 aur ₹50 ke notes hain: kul 40 notes jinki value ₹1,550 hai. Har tarah ke kitne notes hain?
Solution dikhao
Answer₹20 ke 15 notes aur ₹50 ke 25 notes.
Maan lo ₹20 ke $x$ notes hain aur ₹50 ke $y$ .
$x + y = 40$ aur $20 x + 50 y = 1550$ .
Pehli ko 20 se multiply karke subtract karo: $30 y = 750$ , isliye $y = 25$ aur $x = 15$ . Check: $300 + 1250 = 1550$ .
Short answerMediumNCERT Exercise 3.3, Q2(v)
Ek parking lot pehle 2 hours ki ek fixed fee aur har extra hour ka alag amount leta hai. 6 hours park ki gayi car ₹110 deti hai, aur 4 hours park ki gayi car ₹80. Fixed fee aur per extra hour charge nikaalo.
Solution dikhao
AnswerFixed fee ₹50; har extra hour ke ₹15.
Maan lo fee ₹ $f$ hai aur extra charge ₹ $e$ per hour.
$f + 4 e = 110$ aur $f + 2 e = 80$ .
Subtract karo: $2 e = 30$ , isliye $e = 15$ aur $f = 50$ .

3.4 Summary

Koi bhi pair solve karne se pehle tay karo ki kaise karoge. Fig. 3.14 plan karne ka ek tareeka hai, aur uske baad ke points chapter ke main ideas collect karte hain.

3/3Method koi bhi ho, answer ko dono original equations mein wapas rakh kar dekho.

Fig. 3.14Linear equations ke kisi bhi pair ke liye ek plan.

Teen stage ka ek flowchart. Step 1 pehla box dikhata hai, ratio test first, jo batata hai ki ek solution, koi nahi ya infinitely many expect karo. Step 2 teen branches jodta hai: agar x ya y ka coefficient plus ya minus 1 hai, to substitution use karo; agar coefficients match kiye ja sakte hain, to elimination use karo; agar picture ya estimate chahiye, to graphs use karo. Step 3 aakhri box jodta hai: dono equations mein check karo.

Recap

Summary aur key points

$x$ aur $y$ mein pair of linear equations ka general form (3.1) hai. Solution dono equations ko ek saath satisfy karta hai (Section 3.1).
Har equation ek straight line hai, isliye pair ke solutions woh points hain jo dono lines mein common hain (Section 3.2).
Intersecting lines: exactly ek solution, ek consistent pair (Fig. 3.4).
Parallel lines: koi solution nahi, ek inconsistent pair.
Coincident lines: infinitely many solutions, ek dependent pair, jo consistent bhi hai.
Ratio test: $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ matlab ek solution; teenon ratios barabar matlab infinitely many; sirf pehle do barabar matlab koi nahi (ratio test).
Graphs exact answers tabhi dete hain jab lines kisi grid point par milti hain; algebraic methods hamesha exact hote hain (Section 3.3).
Substitution: ek equation se ek variable express karo, use doosri mein rakho, solve karo, phir back-substitute karo (Section 3.3.1).
Elimination: ek variable ke coefficients match karne ke liye multiply karo, use hatane ke liye add ya subtract karo, solve karo, phir back-substitute karo (Section 3.3.2).
Agar bina variable wala statement aaye, to true matlab infinitely many solutions aur false matlab koi nahi (Fig. 3.11).
Word problems mein unknowns ko naam do, har condition ke liye ek equation likho, solve karo, aur answer ko original shabdon se check karo.

Flash cards · 7

Key terms

1 / 7

Board exam practice

Yeh questions CBSE Class 10 board paper ke formats use karte hain: one-mark multiple-choice aur assertion–reason items, short answers, ek case study aur ek graph question. Har ek ka solution kholne se pehle use khud try karo.

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 3.1, Q1	Practice 3.1 · Q1
NCERT Exercise 3.1, Q2	Practice 3.1 · Q2
NCERT Exercise 3.1, Q3	Practice 3.1 · Q3
NCERT Exercise 3.1, Q4	Practice 3.1 · Q4
NCERT Exercise 3.1, Q5	Practice 3.1 · Q5
NCERT Exercise 3.1, Q6	Practice 3.1 · Q6
NCERT Exercise 3.1, Q7	Practice 3.1 · Q7
NCERT Exercise 3.2, Q1	Practice 3.2 · Q1
NCERT Exercise 3.2, Q1(v)	Practice 3.2 · Q2
NCERT Exercise 3.2, Q2	Practice 3.2 · Q3
NCERT Exercise 3.2, Q3(i)	Practice 3.2 · Q4
NCERT Exercise 3.2, Q3(ii)	Practice 3.2 · Q5
NCERT Exercise 3.2, Q3(iii)	Practice 3.2 · Q6
NCERT Exercise 3.2, Q3(iv)	Practice 3.2 · Q7
NCERT Exercise 3.2, Q3(v)	Practice 3.2 · Q8
NCERT Exercise 3.2, Q3(vi)	Practice 3.2 · Q9
NCERT Exercise 3.3, Q1	Practice 3.3 · Q1
NCERT Exercise 3.3, Q2(i)	Practice 3.3 · Q2
NCERT Exercise 3.3, Q2(ii)	Practice 3.3 · Q3
NCERT Exercise 3.3, Q2(iii)	Practice 3.3 · Q4
NCERT Exercise 3.3, Q2(iv)	Practice 3.3 · Q5
NCERT Exercise 3.3, Q2(v)	Practice 3.3 · Q6

Quick check: MCQs

3.1 Introduction3 questions
Pair ko graph se solve karna4 questions
3.3.1 Substitution Method3 questions
3.3.2 Elimination Method4 questions

Glossary · 11

Key terms (English + Hindi)

coincident linesसंपाती रेखाएँsampati rekhayen: Do lines jo bilkul ek doosre ke upar padi hon, isliye har point common ho.Chapter mein pehli baar yahan aaya →
consistent pairसंगतsangat: Linear equations ka aisa pair jiska kam se kam ek solution ho.Chapter mein pehli baar yahan aaya →

dependent pairआश्रितaashrit: Aisa pair jismein ek equation doosri ka multiple hai, isliye dono same line draw karti hain aur infinitely many solutions hote hain; yeh consistent bhi hai.Chapter mein pehli baar yahan aaya →

elimination methodविलोपन विधिvilopan vidhi: Ek variable ke coefficients barabar karke, phir use hatane ke liye equations ko add ya subtract karke pair solve karna.Chapter mein pehli baar yahan aaya →

inconsistent pairअसंगतasangat: Linear equations ka aisa pair jiska koi solution nahi; iski lines parallel hoti hain.Chapter mein pehli baar yahan aaya →
intersecting linesप्रतिच्छेदी रेखाएँpratichchhedi rekhayen: Do lines jo exactly ek point par cross karti hain.Chapter mein pehli baar yahan aaya →

linear equation in two variablesदो चरों वाला रैखिक समीकरणdo charon wala rekhik samikaran: Aisi equation jise $a x + b y + c = 0$ ki tarah likha ja sake, jahan $a$ , $b$ , $c$ real hain aur $a$ , $b$ dono zero nahi. Iska graph ek straight line hota hai.Chapter mein pehli baar yahan aaya →

pair of linear equationsरैखिक समीकरण युग्मrekhik samikaran yugm: Same do variables mein do linear equations, jinhe saath mein dekha jaata hai.Chapter mein pehli baar yahan aaya →
parallel linesसमांतर रेखाएँsamantar rekhayen: Plane mein aisi lines jo kitni bhi door tak draw karo, kabhi nahi milti.Chapter mein pehli baar yahan aaya →

solution of a pairहलhal: $x$ aur $y$ ki woh values jo pair ki dono equations ko ek hi time par satisfy karti hain.Chapter mein pehli baar yahan aaya →
substitution methodप्रतिस्थापन विधिpratisthapan vidhi: Ek variable ko doosre ke terms mein likh kar aur us expression ko doosri equation mein rakh kar pair solve karna.Chapter mein pehli baar yahan aaya →

Pair of Linear Equations in Two Variables: Class 10 Maths Chapter 3 Hinglish mein

Is chapter mein aap kya seekhenge

3.1 Introduction

3.2 Graphical Method of Solution of a Pair of Linear Equations

Coefficients se case predict karna

Pair ko graph se solve karna

Practice 3.1: answers ke saath

3.3 Algebraic Methods of Solving a Pair of Linear Equations

3.3.1 Substitution Method

Practice 3.2: answers ke saath

3.3.2 Elimination Method

Practice 3.3: answers ke saath

3.4 Summary

Summary aur key points

Board exam practice

Practice questions answers ke saath (NCERT “Board exam practice”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye