CBSE Class 10 NCERT Maths Chapter 3 PDF

Summary

This document is chapter 3 of a Class 10 NCERT Maths textbook in Hindi. It covers pair of linear equations in two variables. The chapter likely discusses various methods for solving these equations, including graphical, substitution, and elimination.

Full Transcript

42 xf.kr nks pj okys jSf[kd lehdj.k ;qXe 3 3.1 Hkwfedk vkius bl izdkj dh fLFkfr dk lkeuk vo'; fd;k gksxk] tSlh uhps nh xbZ gS% vf[ky...

42 xf.kr nks pj okys jSf[kd lehdj.k ;qXe 3 3.1 Hkwfedk vkius bl izdkj dh fLFkfr dk lkeuk vo'; fd;k gksxk] tSlh uhps nh xbZ gS% vf[kyk vius xk¡o osQ ,d esys esa xbZA og ,d pj[kh (Giant wheel) dh lokjh djuk pkgrh Fkh vkSj gwiyk (Hoopla) [ ,d [ksy ftlesa vki ,d LVky esa j[kh fdlh oLrq ij ,d oy; (ring) dks isaQdrs gSa vkSj ;fn og oLrq dks iw.kZ:i ls ?ksj ys] rks vkidks og oLrq fey tkrh gS ] [ksyuk pkgrh FkhA ftruh ckj mlus gwiyk [ksy [ksyk mlls vk/h ckj mlus pj[kh dh lokjh dhA ;fn izR;sd ckj dh lokjh osQ fy, mls 3 rFkk gwiyk [ksyus osQ fy, 4 [kpZ djus iM+s] rks vki oSQls Kkr djsaxs fd mlus fdruh ckj pj[kh dh lokjh dh vkSj fdruh ckj gwiyk [ksyk] tcfd mlus blosQ fy, oqQy 20 [kpZ fd,\ gks ldrk gS fd vki bls Kkr djus osQ fy, vyx&vyx fLFkfr;k¡ ysdj pysaA ;fn mlus ,d ckj lokjh dh] D;k ;g laHko gS\ D;k ;g Hkh laHko gS fd mlus nks ckj 2019-20 nks pj okys jSf[kd lehdj.k ;qXe 43 lokjh dh\ bR;kfnA vFkok vki d{kk IX osQ Kku dk mi;ksx djrs gq,] bu fLFkfr;ksa dks nks pjkas okys jSf[kd lehdj.kksa }kjk fu:fir dj ldrs gSaA vkb, bl izfØ;k dks le>saA vf[kyk }kjk lokjh djus dh la[;k dks x rFkk mlosQ }kjk gwiyk [ksy [ksyus dh la[;k dks y ls fu:fir dhft,A vc nh gqbZ fLFkfr dks nks lehdj.kksa }kjk O;Dr fd;k tk ldrk gS % 1 y= x (1) 2 3x + 4y = 20 (2) D;k ge bl lehdj.k ;qXe dk gy Kkr dj ldrs gSa\ bUgsa Kkr djus oQh dbZ fof/;k¡ gSa] ftudk ge bl vè;k; esa vè;;u djsaxsA 3.2 nks pjksa esa jSf[kd lehdj.k ;qXe d{kk IX ls ;kn dhft, fd fuEu lehdj.k nks pjksa osQ jSf[kd lehdj.kksa osQ mnkgj.k gSa% 2x + 3y = 5 x – 2y – 3 = 0 vkSj x – 0y = 2 vFkkZr~ x = 2 vki ;g Hkh tkurs gSa fd og lehdj.k] ftldks ax + by + c = 0 osQ :i esa j[kk tk ldrk gS] tgk¡ a, b vkSj c okLrfod la[;k,¡ gSa vkSj a vkSj b nksuksa 'kwU; ugha gSa] nks pjksa x vkSj y esa ,d jSf[kd lehdj.k dgykrk gSA (izfrca/ tSls a vkSj b nksuksa 'kwU; ugha gSa] ge izk;% a2 + b2 ≠ 0 ls izn£'kr djrs gSaA) vkius ;g Hkh ikus osQ fy,] ge oqQN vkSj mnkgj.k gy djrs gSa % mnkgj.k 12 : foyksiu fof/ dk iz;ksx djosQ] fuEu jSf[kd lehdj.k ;qXe osQ lHkh laHko gy Kkr dhft,% 2x + 3y = 8 (1) 4x + 6y = 7 (2) gy : pj.k 1 : lehdj.k (1) dks 2 ls rFkk lehdj.k (2) dks 1 ls] x osQ xq.kkadksa dks leku djus osQ fy,] xq.kk dfj,A rc ge fuEu lehdj.k ikrs gSa% 4x + 6y = 16 (3) 4x + 6y = 7 (4) pj.k 2 : lehdj.k (4) dks lehdj.k (3) esa ls ?kVkus ij] (4x – 4x) + (6y – 6y) = 16 – 7 vFkkZr~ 0 = 9, tks ,d vlR; dFku gSA vr%] lehdj.kksa osQ ;qXe dk dksbZ gy ugha gSA mnkgj.k 13 : nks vadksa dh ,d la[;k ,oa mlosQ vadksa dks myVus ij cuh la[;k dk ;ksx 66 gSA ;fn la[;k osQ vadksa dk varj 2 gks] rks la[;k Kkr dhft,A ,slh la[;k,¡ fdruh gSa\ gy : ekuk izFke la[;k dh ngkbZ rFkk bdkbZ osQ vad Øe'k% x vkSj y gSaA blfy,] izFke la[;k dks izlkfjr :i esa 10 x + y fy[k ldrs gSa [mnkgj.k osQ fy,] 56 = 10(5) + 6]A tc vad myV tkrs gSa] rks x bdkbZ dk vad cu tkrk gS rFkk y ngkbZ dk vadA ;g la[;k izlkfjr :i esa 10y + x gS [mnkgj.k osQ fy,] tc 56 dks myV fn;k tkrk gS] rks ge ikrs gSa% 65 = 10(6) + 5]A fn, gq, izfrca/ksa osQ vuqlkj] (10x + y) + (10y + x) = 66 vFkkZr~ 11(x + y) = 66 2019-20 nks pj okys jSf[kd lehdj.k ;qXe 63 vFkkZr~ x+y= 6 (1) gesa ;g Hkh fn;k x;k gS fd vadksa dk varj 2 gSA blfy,] ;k rks x–y= 2 (2) ;k y–x= 2 (3) ;fn x – y = 2 gS] rks (1) vkSj (2) dks foyksiu fof/ ls gy djus ij] x = 4 vkSj y = 2 izkIr gksrk gSA bl fLFkfr esa] gesa la[;k 42 izkIr gksrh gSA ;fn y – x = 2 gS] rks (1) vkSj (3) dks foyksiu fof/ ls gy djus ij] gesa x = 2 vkSj y = 4 izkIr gksrk gSA bl fLFkfr esa] gesa la[;k 24 izkIr gksrh gSA bl izdkj ,slh nks la[;k,¡ 42 vkSj 24 gSaA lR;kiu % ;gk¡ 42 + 24 = 66 vkSj 4 – 2 = 2 gS rFkk 24 + 42 = 66 vkSj 4 – 2 = 2 gSA iz'ukoyh 3.4 1. fuEu lehdj.kksa osQ ;qXe dks foyksiu fof/ rFkk izfrLFkkiuk fof/ ls gy dhft,A dkSu&lh fof/ vf/d mi;qDr gS\ (i) x + y = 5 vkSj 2x – 3y = 4 (ii) 3x + 4y = 10 vkSj 2x – 2y = 2 x 2y y (iii) 3x – 5y – 4 = 0 vkSj 9x = 2y + 7 (iv) + = − 1 vkSj x − = 3 2 3 3 2. fuEu leL;kvksa esa jSf[kd lehdj.kksa osQ ;qXe cukb, vkSj muosQ gy (;fn mudk vfLrRo gks) foyksiu fof/ ls Kkr dhft, % (i) ;fn ge va'k esa 1 tksM+ nsa rFkk gj esa ls 1 ?kVk nsa] rks fHkUu 1 esa cny tkrh gSA ;fn 1 gj esa 1 tksM+ nsa] rks ;g gks tkrh gSA og fHkUu D;k gS\ 2 (ii) ik¡p o"kZ iwoZ uwjh dh vk;q lksuw dh vk;q dh rhu xquh FkhA nl o"kZ i'pkr~] uwjh dh vk;q lksuw dh vk;q dh nks xquh gks tk,xhA uwjh vkSj lksuw dh vk;q fdruh gSA (iii) nks vadksa dh la[;k osQ vadksa dk ;ksx 9 gSA bl la[;k dk ukS xquk] la[;k osQ vadksa dks iyVus ls cuh la[;k dk nks xquk gSA og la[;k Kkr dhft,A (iv) ehuk  2000 fudkyus osQ fy, ,d cSad xbZA mlus [ktk¡ph ls  50 rFkk  100 osQ uksV nsus osQ fy, dgkA ehuk us oqQy 25 uksV izkIr fd,A Kkr dhft, fd mlus  50 vkSj  100 osQ fdrus&fdrus uksV izkIr fd,A 2019-20 64 xf.kr (v) fdjk, ij iqLrosaQ nsus okys fdlh iqLrdky; dk izFke rhu fnuksa dk ,d fu;r fdjk;k gS rFkk mlosQ ckn izR;sd vfrfjDr fnu dk vyx fdjk;k gSA lfjrk us lkr fnuksa rd ,d iqLrd j[kus osQ fy,  27 vnk fd,] tcfd lwlh us ,d iqLrd ik¡p fnuksa rd j[kus osQ  21 vnk fd,A fu;r fdjk;k rFkk izR;sd vfrfjDr fnu dk fdjk;k Kkr dhft,A 3.4.3 otz&xq.ku fof/ vc rd] vkius i

Use Quizgecko on...
Browser
Browser