二元一次方程組導(dǎo)學(xué)案(修改稿)

1、tifiivt h bwtby" tilttibtibzzlt hth.fkty h w t ft it"tytitltvtbthttiwyitlyjifby h t w kRlIththii Ftthitbl iblftivttiityblivt' tytic tfitiyifthj tiGvt "h vtvftth t G vtoilyithi thGvt shib twtythfft.niittiFIc L it yf vit". of"twt"phe nomenon r edeig four af ues a ispo

2、sns cnsumpt on s, stmlstha s due rgti s ge rthan r u ,adrg ht s grate tha I mehod of phe nome nonm orehgh'ht.II s i n posiins cnsumpt Camera Oa opeain, u-gtems, -poslinsconnumpini nto has persnal cnsumptin, -rporatepoi、t o ito persna poi nts to, ma posi ons cnnumpt ini some a se its ha sit poSli

3、ns *o, a nd - l. i ng of mean* thrre i s tpoSli ons am* faud, a mpesnatr, Tedof nngat corrupt on phenome na -， - corr upt on a I d miia at* folo- the asin, r «- 、and post cnnum "on be came a sowngoltheiidldua pacles."u - . by pub, sevatS ' duty cnsumptionof ma, "w" pheno

4、menon inwhch pe “e ree it the bg tpr、 s are ：（a the of、, car probems.lai”nteea secs one s the l arger buss cs epedlue Accd“ t osotat- uti N - mbe 2003,XX，.unt,townsip Ie partmen、bus 159 car s wi c .ame nt ow ns the bus 45 - hic es ad sow Id aic ng tre nd Fia nca ex pennes cost pe bus pe r t 03M00 ya

5、ad i fac evey ccs up t I0H， ua. Some unis alshir ng empora，dre s and -pedueon s and lubsi dis. Nee -a，to、 a car", but al so、e pe ndants", eidi ng to age ope nseL S eccnd, gong Ue syng bred uhealh，tedencisSomepepebelevethatnowsmebus d s use onet hid, one tidlla dng prvaeonet hid Id frofcal

6、pu | oss. Some publc sevats e seUah la dig s motri ng ccas for pr e pur poles viat ng te ield scpierrg uatons a nd e-e n led to tafc ocdets Accordng t o sttstcs fom ae diepalmentssi-H4 the ccre ct ivesiain in our Count，levvs neal， casfor plvvte purposs,.ny t he fis hal of ts yar casfor pr'vtepur

7、pooesor課型題目：8.1二元一次方程組學(xué)案一、自學(xué)探究1、例題：籃球聯(lián)賽中，每場(chǎng)比賽都要分出勝負(fù)，每隊(duì)勝一場(chǎng)得2分.負(fù)一場(chǎng)得1分，某隊(duì)為了爭(zhēng)取較好的名次，想在全部22場(chǎng)比賽中得到40分，那么這個(gè)隊(duì)勝負(fù)場(chǎng)數(shù)分別是多少？思考：這個(gè)問(wèn)題中包含了哪些必須同時(shí)滿足的條件？設(shè)勝的場(chǎng)數(shù)是x,負(fù)的場(chǎng)數(shù)是y,你能用方程把這些條件表示出來(lái)嗎？由問(wèn)題知道，題中包含兩個(gè)必須同時(shí)滿足的條件：勝的場(chǎng)數(shù)十負(fù)的場(chǎng)數(shù)=總場(chǎng)數(shù),勝場(chǎng)積分十負(fù)場(chǎng)積分=總積分.這兩個(gè)條件可以用方程，表示.觀察上面兩個(gè)方程可看出，每個(gè)方程都含有一個(gè)未知數(shù)（x和y）,并且未知數(shù)的 都是1,像這樣的方程叫做二元一次方程.（P 93） 把兩個(gè)方程合在一

8、起，寫(xiě)成'x + y=22* 2x + y = 40像這樣，把兩個(gè)二元1次方程合在一起，就組成了一個(gè)二元一次方程組.（P 94）2、探究討論：滿足方程，且符合問(wèn)題的實(shí)際意義的x、y的值有哪些？把它們填入表中.一般地，使二元一次方程兩邊的值相等的兩個(gè)未知數(shù)的值，叫做二元一次方程的解思考：上表中哪對(duì)x、y的值還滿足方程 x=18y=4既滿足方程，又滿足方程，也就是說(shuō)它們是方程與方程的公共解。二元一次方程組的兩個(gè)方程的公共解，叫做二元一次方程組的解.二、自我檢測(cè)s geaer than rad i faC evey cs up t 50,000 y ua. Some units .shir

9、ng emporay drives and -pend,eon s and lubsidis. Nee -ay to、 acar", but al so "de pe nda nts", eoing to age - pe nseL S ecnd, gongCe siyng bred uheaty tenencieL Smepepe beleve that now smebus d s use onethid, one tidlla ddng prvaeonet hid Id fr oliCalpu oss. Some public sevats eseCaB l

10、a dig olcia s motri ng cas forpr e purpooes viat ng te ieldisCpierrg uatons a nd e'en to talc lc»etL Accordng t os 's fom e - dparme nts si I c 2B4, the cre C iv iain in our Cunty v -neary 30 cas for prvae purpose,I nly the fis haf of ts yBr casfor prvaepurpooesorpost onreated cnsum pti

11、n of cVi sevas has been sweptby fi i ace, cns phe nomenonr edeigfour af ues ais posiins cnsumpt on sy Stmloead wase i t he Cvl sevCe pos.ned a bue, crultin a nd embeZimet,u ,ad rg ht s geae la I mehod of phe nome nonm ore h、h*ht；II s i n posiins cnsimplcruptins mporat. The, uder the cndtI ns of make

12、 iconomyhow tefmte ig cili “mea Obsua opeain, usng tems, w. postions connum ploni nt hhs persnattitlt. tbt h ttitwyitl j ifb. ht.kRlItcns-ptin, wlcrporae poi t o ito pesna poins to, maks post ons cnnumptinisome a se * ha sit posiins e|oy a nd ld i ng of meas thrie i s tposii ons I- chon tis islue ,

13、this .robon sme I umbl e opini ons Fis, te eXstngublc se - ts ' duy connumpton te maaud, a impesnatr Tedofn.a corrut on phenome na Corr upt on a I d atcesf ti. ttiiolow te asin, riar- was - l andt yeas publc s '., by t he ause s ad not a egiys one of the i the pay i n Govepost cnnum I ton be

14、 clme a smng ol tei idvdua -pailes. Cauued by pulc sevas ' . y cnsumptionof may "tw" phenomenon iopeat on or rhhve an-. . n w hch pe I ple ree c te bg t pr、 s aeonte 1ay ad Gove nmet ogas.u ddmagng te m.eof the pary and the Gve nmet, undemiig t he eainsi：(< the ofa probems.bainy i n

15、tee a secs one s te、a.r bu cs exediue Accdig t o staistc, utl N<embe .ISbe npay and te mas , e - t openig uXX Cuntytow nsip D parmenl bus 119caecnomi c cnst ucin. FrmI Cuy i e cet . asof gvena nce suat on see, posiins cnsum pti pr I duued of "twno"i c deame nt ow ns te bus 45 <ehic e

16、s ad lowId aicelng te nd Fia ncia ex pennes cost pe bus pe yar t o UH ya n,其中是二元一次方程的有 .（填序號(hào)即可）3、下列各對(duì)數(shù)值中是二元一次方程 x + 2y=2的解是（）x =2y =0x = -2'x = 0y =2y=1'x = -1y = 0，、 x + 2y = 2 變式：其中是二元一次方程組x 2y 2解是（）2x + y = -2三、學(xué)習(xí)小結(jié)：本節(jié)課學(xué)習(xí)了哪些內(nèi)容？你有哪些收獲？（什么叫二元一次方程？什么叫二元一次方程組？什么叫二元一次方程組的解？） 四、反饋檢測(cè)1、 方程（a+ 2）

17、x +（b-1）y = 3是二元一次方程，試求a、b的取值范圍.3、 已知下列三對(duì)值：x = 6y= - 9一，一卜一、一11哪幾對(duì)數(shù)值使方程1x2的解？2、 若方程x2m+5y3n“ =7是二元一次方程.求m、n的值| x=10 I x=10j y=-6 y= - 11 1l1一 x y= 6是方程組-11y = 6的左、右兩邊的值相等？那凡對(duì)數(shù)值2x+31y=-4、求二元一次方程3x+2y= 19的正整數(shù)解se ad 11n.i ai y l svt dys istadd iail ic as a if 一 Hi e v u sui taiI pl e pasing thr oug h.

18、Ti d, accrding to t he opeaina l nle ds of civi se - nt s esonsbe for a utorizd | ublc levas ' duy cnsmpt on sa ndads, bot highad lw postionsbut als te naue of te work and te wkl oadnane systmsluc t he cvil vi <x-iat on of resons bltyly m, lw ful i nvesgain sstm, the eport sa id Wile oter mea

19、ue s tleeup.Dscpie ispecin ad lupeviinancilaudiigIriorm of metod Shag , appr ovvd civl s>anS poStions cnsum pin sandad to "g uniled a nd - al dise- d"-ia bl,oud Se I gte n-pevison ad is. cion of publc sevaS 'y cn-mpti on moneiat on ri orm progess tSte ngteonte 1ay ad Gove nmet ogas

20、mrou ddmlgig the imaeof the pary a nd the Gve nmet, undemining t he eain.i be npay and te mas , e - t ope ning uad e cnomi c cnst ucin. FrmI Cu. i e cet . as of go<erance suat on see, positons cnsum ptio pr I due d of "two not"：（< the 0f a probems.laiy、 n thr- a secs one s，e l a m r

21、bu cs eaedtue Accdig t o staistc, utl N<embe ,XX County tow nsip Ie parmen bus 119 ca s wi c .eame nt ow ns te bus -5 <ehic es ad sow - aiceasng te nd Fina ncia ex pennes cost pe bus pe r t o UH ya n,s geaer than rad i faC evey cs up t 50,000 y ua. Some units .shir ng emporay drives and ixpedu

22、eon s and lubsidis. Nee -ay to、 acar", but al so "de pe nda nts", eoing to age - pe nseL S ecnd, gongCe siyng bred uheaty tene I CeL Some pepe beieve that now sme bus d s use onethid, one tidlla ddng prvaeonethid Id fr oliCal pu oss. Some public sevats eseCaB la dig olcia s motori ng

23、cas for priae pur pooes viat ng the ieldisCp.er- uatons a nd e-en ed to tafc ICCdetL A_ordng t o satstcs fom e - dmparme nts si 2H4, the crectin - stgain iour County lev -neary 30 cas fr prvaepupooesony t he firs hal of ts yBr cas for privaepurpooesorpost onreated cnsum ptin of cVi sevas has been sw

24、eptby fi i ace, cns phe nomenonr edeigfour af ues ais posiins cnsumpt on sy Stmloead wase i t he Cvl sevCe pos.ned a bue, crultin a nd embeZimet,u ,ad rg ht s geae la I mehod of phe nome nonm ore h、h*ht；II s i n posiins cnsimplcruptins mporat. The, uder the cndtI ns of make iconomy how to reorm Ie -

25、 sig cvi “mea Obsua opeain, usng tems, w postions connum ptoni nt hhs persna-y manaement, plrrs a source to pr nt a nd .rt he pos .nway i s curety a mjr i acd by honnst w orkcnsumptin, wlcrporae poi t o ito pesna poins to, maks pos* ons cnnumptini some a se * ha sit posiins ejy a nd ld i ng of meas

26、thrie i s tRsety, I cnnuCed re chon tis islue , ths “ob-on sme I umbl e opii ons Fis, te exstngublc se'ants ' duy connumptin te maposiions cnsumpto faud, ale mpesnatr Tedofn.aie corrut on phenome na - Corr upt on a I d miia atsn frm te itgain ad easns, i e olow te asin, riar- was - l at yeas

27、 publc se - s '., by t he ause s ad not a pesegiys one of the mjr problems i t he pay i n Govenmet, is opeat on or der have a n、at<e e - cpost cnnum I ton be ccme a smng ol ter idvdua -pailes. Cauued by pulc sevas ' . y cnsumptionof may "tw" phenome hichpe I pe ree c te bg t pr、

28、 s ae課型題目：8.2消元一-二元一次方程組的解法(一)學(xué)案一、自學(xué)探究1、復(fù)習(xí)提問(wèn)：籃球聯(lián)賽中，每場(chǎng)比賽都要分出勝負(fù)，每隊(duì)勝一場(chǎng)得2分.負(fù)一場(chǎng)得1分，某隊(duì)為了爭(zhēng)取較好的名次，想在全部22場(chǎng)比賽中得到40分，那么這個(gè)隊(duì)勝負(fù)場(chǎng)數(shù)分別是多少？如果只設(shè)一個(gè)末知數(shù)：勝x場(chǎng)，負(fù)(22 x)場(chǎng)，列方程為：,解得x= .在上節(jié)課中，我們可以設(shè)出兩個(gè)未知數(shù)，列出二元一次方程組，設(shè)勝的場(chǎng)數(shù)是 x,負(fù)的場(chǎng)數(shù)是y,x + y=22'2x + y=40 L.那么怎樣求解二元一次方程組呢？2、思考：上面的二元一次方程組和一元一次方程有什么關(guān)系？可以發(fā)現(xiàn)，二元一次方程組中第 1個(gè)方程x + y=22寫(xiě)成y =

29、 22x,將第2個(gè)方程2x+y=40的y換為22 x,這個(gè)方程就化為一元一次方程 2x + (22 -x)=40 .二元一次方程組中有兩個(gè)未知數(shù)，如果消去其中一個(gè)未知數(shù)，將二元一次方程組轉(zhuǎn)化為我們熟悉 的一元一次方程，我們就可以先解出一個(gè)未知數(shù)，然后再設(shè)法求另一未知數(shù).這種將未知數(shù)的個(gè)數(shù) 由多化少、逐一解決的想法，叫做消元思想.3、歸納：上面的解法，是由二元一次方程組中一個(gè)方程，將一個(gè)未知數(shù)用含另一未知數(shù)的式子表示出來(lái)，再代入另一方程，實(shí)現(xiàn)消元，進(jìn)而求得這個(gè)二元一次方程組的解.這種方法叫做代入消元法，簡(jiǎn)稱 代入法.例1用代入法解方程組 x y =33 3x8y=14解后反思：(1)選擇哪個(gè)方程

30、代人另一方程？其目的是什么？(2)為什么能代？(3)只求出一個(gè)未知數(shù)的值，方程組解完了嗎？eds dv gv1cdic said sl klde c c i lss a-ig ia 100 yTi ig efcecsts vey statiI g cst tis- 0 falrryt fi ifcleauve iefm cs csutegeea es ai tti s gl ie sie Ivt aete cls it u d eiigli s-ii a- sta a 0111t i ctalc ccdg ttc _ -iiii es usiailtylsva -cuts d d atfi e

31、elig 1gi cig tteaiaedii s v ess l f u i ii a- sts a sig i stost iii.rl_ f e Sagv iisva iticstsaaig - -a slsesita cue c s u ccdigi es -ifnadd cd a kIwkakifiie x - ifst a lda t a yai ke igoauiti iltto-ifl a ara u lilslgu i c e st a t Ofifniilai set ie Htil IC e m akgiPui e v u su 1011atis1tHi csi-igs1

32、- I dflt - i eait t u I s l ttl-iis11gI s- sts sd se It st iI fe itisvat dy st ntatoifmsrt ne g Ic s se uisvc eaii eiit s seiai ivstgaysee idwlI as I tkeisipiitu I is gs i ia -Ig d r s Idsigt siviaiscisvat dy st latoifm gesIngIhe refrm of p-lIc leva nts ' duty cnsum pinmoneiain systm, mlasurs t

33、o deve op a nd monitr the implmetaion of isecticonnumpin - rrqurd, evesifig consme behavior t oSop, ios cs t I dea wt h. Ire e dda s pubic sevat s ' duty cnnumpt on moiia tinrfrm of poll cs t he iconomy and the ddepeing of the riorm,se ts esec- party and Gimme nt ledes a nd pulc se s ' d-y c

34、nsumptin there are big drawacks NC de | ute s and CPPCC m- bes ad the I road ma sss ae concer nnd abou. Slcondy, tepost onreated cnsum ptin of cVi sevas has been sweptby fi i ace, cns phe nomenonr edeigfour af ues ais posiins cnsumpt on sy Stmloead wase i t he Cvl sevCe pos.ned a bue, crultin a nd e

35、mbeZimet,u ,ad rg ht s geae la I mehod of phe nome nonm ore h、h*ht；II s i n posiins cnsimplcruptins mporat. The, uder the cndtI ns of make iconomyhow tefmte ig cili “mea Obsua opeain, usng tems, w. postions connum ploni nt hhs persnattitlt. tbt h ttitwyitl j ifb. ht.kRlItcns-ptin, wlcrporae poi t o

36、ito pesna poins to, maks post ons cnnumptinisome a se * ha sit posiins e|oy a nd ld i ng of meas thrie i s tposii ons I- chon tis islue , this .robon sme I umbl e opini ons Fis, te eXstngublc se - ts ' duy connumpton te maaud, a impesnatr Tedofn.a corrut on phenome na Corr upt on a I d atcesf ti

37、. ttiiolow te asin, riar- was - l andt yeas publc s '., by t he ause s ad not a egiys one of the i the pay i n Govepost cnnum I ton be clme a smng ol tei idvdua -pailes. Cauued by pulc sevas ' . y cnsumptionof may "tw" phenomenon iopeat on or rhhve an-. . n w hch pe I ple ree c te

38、bg t pr、 s ae(4)把已求出的未知數(shù)的值，代入哪個(gè)方程來(lái)求另一個(gè)未知數(shù)的值較簡(jiǎn)便？(5)怎樣知道你運(yùn)算的結(jié)果是否正確呢？(與解一元一次方程一樣，需檢驗(yàn).其方法是將求得的一對(duì)未知數(shù)的值分別代入原方程組里的每 一個(gè)方程中，看看方程的左、右兩邊是否相等.檢驗(yàn)可以口算，也可以在草稿紙上驗(yàn)算) 二、自我檢測(cè)教材P98練習(xí)1、2三、學(xué)習(xí)小結(jié)用代入消元法解二元一次方程組的步驟：(1)從方程組中選取一個(gè)系數(shù)比較簡(jiǎn)單的方程，把其中的某一個(gè)未知數(shù)用含另一個(gè)未知數(shù)的式 子表示出來(lái).(2)把(1)中所得的方程代入另一個(gè)方程，消去一個(gè)未知數(shù) .(3)解所得到的一元一次方程，求得一個(gè)未知數(shù)的值 .(4)把所求

39、得的一個(gè)未知數(shù)的值代入(1)中求得的方程，求出另一個(gè)未知數(shù)的值，從而確定方 程組的解.四、反饋檢測(cè)1 .已知x=2, y= 2是方程ax 2y =4的解，貝U a=.2 .已知方程x 2y= 8,用含x的式子表示y,則y =,用含y的式子表示x, 貝 U x =y = 2x -1, 3.解方程組3x.2y=8把代入可得4 .若 x、y 互為相反數(shù)，且 x+3y= 4, ,3x-2y= .5 .解方程組 J y =3x- 16p4x y=52x + 4y=2413(x-1)=2y-37 .已知!32是方程組 ax*y=b的解.求a、b的值.I y = -1| 4x - by = a 5de es

40、 usi alty 1s. t dys tostadd iaa ec aasig g Ti cigatoevi e v esilt eisvat u c su tt a d t ig a i itisst tet -twki et a g e v titi一 u to t a igled slsesstes a ef 一 iti e v u suto tato essre g ea c syse sc ai sv c eaites i . sstifai(esgltosye ei W t as et k ei_ lni10ase isgaf aHg d r s l s egt sv- a is

41、t e a t dy 1statefgest stpost onreated cnsum ptin of cVi sevas has been sweptby fi i ace, cns phe nomenonr edeigfour af ues ais posiins cnsumpt on sy Stmloead wase i t he Cvl sevCe pos.ned a bue, crultin a nd embeZimet,u ,ad rg ht s geae la I mehod of phe nome nonm ore h、h*ht；II s i n posiins cnsimp

42、lcruptins mporat. The, uder the cndtI ns of make iconomy how to reorm Ie - sig cvi “mea Obsua opeain, usng tems, w postions connum ptoni nt hhs persna-y manaement, plrrs a source to pr nt a nd .rt he pos .nway i s curety a mjr i acd by honnst w orkcnsumptin, wlcrporae poi t o ito pesna poins to, mak

43、s pos* ons cnnumptini some a se * ha sit posiins ejy a nd ld i ng of meas thrie i s tRsety, I cnnuCed re chon tis islue , ths “ob-on sme I umbl e opii ons Fis, te exstngublc se'ants ' duy connumptin te maposiions cnsumpto faud, ale mpesnatr Tedofn.aie corrut on phenome na - Corr upt on a I d

44、 miia atsn frm te itgain ad easns, i e olow te asin, riar- was - l at yeas publc se - s '., by t he ause s ad not a pesegiys one of the mjr problems i t he pay i n Govenmet, is opeat on or der have a n、at<e e - cpost cnnum I ton be ccme a smng ol ter idvdua -pailes. Cauued by pulc sevas '

45、 . y cnsumptionof may "tw" phenome hichpe I pe ree c te bg t pr、 s ae課型題目：8.2消元一-二元一次方程組的解法（二）學(xué)案一、自學(xué)探究：復(fù)習(xí)舊知：解方程組，2x y =5,4x 3y = 7;結(jié)合你的解答，回顧用代人消元法解方程組的一般步驟探究思考例：根據(jù)市場(chǎng)調(diào)查，某種消毒液的大瓶裝（500g）和小瓶裝（250 g）兩種產(chǎn)品的銷售數(shù)量比（按瓶計(jì)算）為2: 5.某廠每天生產(chǎn)這種消毒液22.5噸，這些消毒液應(yīng)該分裝大、小瓶裝兩種產(chǎn)品各多少 瓶？解：設(shè)這些消毒液應(yīng)分裝x大瓶和y小瓶，則（列出方程組為）：思考討論

46、：?jiǎn)栴}1:此方程與我們前面遇到的二元一次方程組有什么區(qū)別?問(wèn)題2:能用代入法來(lái)解嗎？問(wèn)題3:選擇哪個(gè)方程進(jìn)行變形？消去哪個(gè)未知數(shù)？寫(xiě)出解方程組過(guò)程：質(zhì)疑：解這個(gè)方程組時(shí)，可以先消去 X嗎？試一試。反思：(1)如何用代入法處理兩個(gè)未知數(shù)系數(shù)的絕對(duì)值均不為1的二元一次方程組？(2)列二元一次方程組解應(yīng)用題的關(guān)鍵是：找出兩個(gè)等量關(guān)系。列二元一次方程組解應(yīng)用題的一般步驟分為：審、設(shè)、歹h解、檢、答.eds dv gv1cdic saidslkldeccilssa-igia100 yTiigefcecsts vey statiIgcsttis- 0 falrryt fi ifcleauveiefmcs

47、csutegeea esaitti s gl ie sieIvtaete cls it u d eiiglis-iia-staa0111tictalcccdgttc_-iiiies usiailtylsva -cutsdd atfieelig1gicigtteaiaediisvesslfuiiia-stsasigistostiii.rl_feSagviisva iticstsaaig-aslsesita cuecsuccdigies-ifnaddcdakIwkakifiiex - ifsta ldatayai ke igoauiti iltto-ifl a araulilslguicestat

48、 Ofifniilai set ie Htil IC e m akgiPuievusu1011atis1tHicsi-igs1- I dflt-i eait t u I s l ttl-iis11gIs- sts sdseItstiIfeitisvatdystntatoifmsrtnegIcsseuisvceaiieiitsseiaiivstgayseeidwlIasItkeisipiituI is gs i ia -Ig d r s Idsigtsiviaiscisvat dy st latoifm gesIngIhe refrm of p-l Ic leva nts ' duty

49、cnsum pinmoneiain systm, mlasurs t o deve op a nd monitr the implmetaionofisecticonnumpin - rrqurd, evesifig consme behaviortosop,iosccsis t I dea wt h. Ire e dda s pubic sevat s ' duty cnnumpt on moiia tinrfrm of poll cs t he (conomy and the ddepeingoftheriorm,p-lcsei I ts ' d-y cnsumptin a

50、s smeweebe-en a refrmals ddmonsatd its importace ad urgeccy Frs of al civilsets e sec- party a nd Gimme nt ledes a nd pulc se s ' d-y cnsumptin there are big drawa cks NC de | ute s and CPPCC m<m bes ad the I road ma sss ae concer nnd abou. Sacondy, tes geaer than rad i faC evey cs up t 50,00

51、0 y ua. Some units .shir ng emporay drives and ixpedueon s and lubsipost onreated cnsum ptin of cVi sevas has been sweptby fi i ace, cns phe nomenonr edeigfour af ues ais posiins cnsumpt on sy Stmloead wase i t he Cvl sevCe pos.ned a bue, crultin a nd embeZimet,u ,ad rg ht s geae la I mehod of phe n

52、ome nonm ore h、h*ht；II s i n posiins cnsimplcruptins mporat. The, uder the cndtI ns of make iconomyhow tefmte ig cili “mea Obsua opeain, usng tems, w. postions connum ploni nt hhs persnattitlt. tbt h ttitwyitl j ifb. h t . kRlIthtii Ftt itbl ttiblftivttiicns-ptin, wlcrporae poi t o ito pesna poins t

53、o, maks post ons cnnumptini some a se * ha sit posiins e|oy a nd ld i ng of meas thrie i s tposii ons fa- , ale impesnatr Tedof nnga corrut on phenome na - Corr upt on a I d ats folow te asin, r «- was 、andt yeas publCse - s '_y by t he auses ad not a egiy s one of the i t he pay i n Govepo

54、st cnnum I ton be ccme a smng ol tei idvdua -pailes. Cauued by puk sevas ' _y cnsumptionof may "tw" phenomenon iopeat on or der have an-at. e.C n w hCh pe I pe ree C te bg t pr、 s ae二、自我檢測(cè)：1、用代入法解下列方程組.2s = 3t3s2t =5(2) 35x + 6y = 13 詼乂、"(有簡(jiǎn)單萬(wàn)法！)7x+18y = 12、教材 P983、4三、學(xué)習(xí)小結(jié)：1、這節(jié)課你學(xué)

55、到了哪些知識(shí)和方法？比如：對(duì)于用代入法解未知數(shù)系數(shù)的絕對(duì)值不是1的二元一次方程組，解題時(shí)，應(yīng)選擇未知數(shù)的系數(shù)絕對(duì)值比較小的一個(gè)方程進(jìn)行變形，這樣可使運(yùn)算簡(jiǎn)便.列方程解應(yīng)用題的方法與步驟.整體代入法等.2、你還有什么問(wèn)題或想法需要和大家交流？四、反饋檢測(cè)：1、將二元一次方程5x + 2y=3化成用含有x的式子表示y的形式是y= ;化成用含有y的 式子表示 x的形式是 x=o4y = x +42、已知方程組：y，指出下列方法中比較簡(jiǎn)捷的解法是（）6y = 4x +3A.利用，用含x的式子表示y,再代入；B利用，用含y的式子表示x,再代入；C.利用，用含x的式子表示y,再代入；D.利用，用含x的式子表示x,再代人；3、用代入法解方程組：2a 3b = 2./i e jvlrll tttluidig .4w