2.14 KPPP gSvP PgtU rĪz(Solving of Simultaneous Linear Equations):

 

dUtvz DgAsz rz PɼV ĸAiģ rĪz E PAiİzê.

v vAzAi Ml Ai 55 U. 16 U Avg vAzAi Aiĸ AiĹ JgqgUĪzzg, FU AiĸĔ?

 

UU x+1 = 5, 2a+6 =10, F jwAi PgtU rĪz Pwzê. EU MAz ZgPgz. EAv PgtU gSvP PgtUɣvê.

 

FU MAz Pgt x+y = 5 vUzPƼ.Ezg x v y JA Jgq ZgPgU. FU F Pgtz

x v y UU gɨg ɯU PqĪ. DU (x=1,y=4), (x=2,y=3), (x=3,y=2), (x=0,y=5), (x= -2, y=7) F jw ɯ Egz ɯU UAU Pgt vȦۥrv۪. x v y UU CASv ɯU Uv۪. Uzg, Ez KP? Pgtz x PAvjzU, y = 5-x DUvz. x Aiiz ɯU yU vP ɯU zgAiv۪. F jwAi PgtU r U x v yU߼UAq EAz Pgt P.

 

Jgq ZgPgUgĪ gSvP PgtP CAv ɯU UĪzjAz,EAx PgtU r Cz ZgPgU AgĪ EAz Pgt P.

 

FU ûv UAz Dlz P̪ PqvۣAz s. ê jAivg ýzg, C AiĹ ASAi .r.U U Pqvۣ. F Az ê Pjwg?

 

2.14 ĸ1 (Azz P): v vAzAi Ml Ai 55 U. 16 U Avg vAzAi Aiĸ AiĹ JgqgUĪzzg, FU Aiĸ?

 

CAzfAz P̪ ê iqz. ûv aP UĪ DUz EgĪzjAz Cv Aiĸ 9 UAz DgA ĸAiģ rĪ.

 

FU(Ml=55)

16 U Avg

ûv Aiĸ

C vAzAi Aiĸ

ûv Aiĸ

C vAzAi Aiĸ

9

46

25

62

10

45

26

61

11

44

27

60

12

43

28

59

13

42

29

58

14

41

30

57

15

40

31

56

 

 

ð v:S۬Az wz gĪzãAzg, ûv Aiĸ FU 13 Uzg 15 U Avg, C vAzAi Aiĸ(58) Dv AiĹ(29) JgqgUz.FU ê ûvAz 13 .r.U Püz.

Dzg dn ĸU F PĪ Cĸj zs嫮. Uzg EzPAz AiĪħzުz PīzAi?

 

jg:

ûv FV Aiĸ = y UVg.

C vAzAi Aiĸ = x UVg.

Cjg AiĹ v 55 UzjAz,

x+y =55

16 U Avg ûv Aiĸ = y+16

ûv vAzAi Aiĸ = x+16.

ĸAiİ PlAv, x+16 =2*(y+16)

x+16 = 2y+ 32 (ĮPjz.)

x-2y = 32-16 =16 (PAvjz.)

 

PƣAiİ VU Jgq PgtU zgv:

(1) x+y =55

(2) x-2y = 16

 

gSvP PgtU r Pgtz MAz ZgPg P.DzjAz MAz ZgPg PgtAz UrP. U?

zv PgtU:

 

x+y =55 == (1)

x-2y=16 == (2)

----------

(2) (1) jAz Pɬj 0+3y =39 == (3)

-----------

3y = 39

y=13

x+y =55 == (1)

x = 55-y (PAvjz.)

ð Pgtz y=13JAz DzòzU,

x=55-13

=42

ûv Aiĸ = 13

C vAzAi Aiĸ = 42

 

v:

FU ûv Aiĸ = 13 , C vAzAi Aiĸ = 42 DzU Cg Ml Ai 55 U.

16 U Avg, ûv Aiĸ = 29 , C vAzAi Aiĸ =58( DU C Aiĸ vAzAi AiĹ JgqgUĪz)

2. x v yU ɯU (1) g DzòzU: x+y = 42+13 = 55 , (2) g DzòzU: x-2y = 42-26 = 16

 

2.14 ĸ2: MAz PA nUAi ɯAi MAz ɤߣ ɯVAv g.18 e. CzsPg U 240 g U Pl 5 PA nU v 10 ɣU vg ýzg, MAz PA nU v MAz ɤߣ PAi PAqĻr.

 

jg:

MAz PA nUAi PAi = y DVg

MAz ɤߣ PAi = x DVg

(1) y = x+18 ==(1)

(2) 5y+10x = 240 ==(2)

 

(1)   jAiiz gƥz gzU, y-x =18 ===(3)

(2)   5 jAz sVzU, y+2x= 48 ===(4)

(3)   g (4) PzU -----------------

-3x =-30 ------(2)

X = -30/-3 =10 ------(3)

MAz ɤߣ PAi = 10 g.

MAz PA nUAi PAi = 28 g

Cs: x v y Ai ɯU Pgt (1) v (2)g Dzò, vɣr.

 

2.14 ĸ 3: r: 2x+2y =4 v x+y =2

 

jg:

2x+2y =4 ===(1)

x+y = 2 ===(2)

(2) 2 jAz Ut¹.

2x+2y=4 ===(3)

 

(1) jAz(3) Pɬj 0 =0 Ez AiiU v.

x v y UU Aiiz ɯU Pqz. CUU ֪z ɯU( KPAzg Jgq Pgt zAizg Czszz)

 

2.14 ĸ 4: r: 2x+2y =4 v x+y = 3

 

jg:

 

zv PgtU: 2x+2y =4 ====(1)

x+y = 3 =====(2)

 

(2) 2 jAz Ut¹zU,

2x+2y=6 =====(3)

 

(3) jAz (1) PzU, 0 =2 Ez d.

DzjAz x v y U Aii ɯU Pl PgtU vȦۥrĪŢ.

 

S: Jgq ZgPgUļ Jgq gSvP PgtU MnU vUzPAqU, CU KPPP gSvP PgtUĔ(simultaneous linear equations) Jvê.

CU , a1 x+ b1 y = c1 v a2 x+b2 y = c2

E a1, b1, a2, b2, c1 ,c2 U gAU, x v y U ZgPgU ( EU ɯAiģ PAqĻrAiĨPzz.)

 

VgĪ KPPP gSvP PgtU U rzê?

 

CvU Uƽ, UrĪ P (Method of Elimination by equating co efficients):

 

CĸjPz AvU:

 

1.       ZgPgU AS CvU Uƽ, PgtU (MAz Cx Jgq) P۪z ASUAz Ut¹.

2.      zU AS CvU Az ewAi aU qɢzg PgtU PrP. ewAi aU Azg PAiĨP.

3.      F jw qz Pgt r Czg MAz ZgPgz ɯAiģ PAqĻrAiĨP.

4.      Qz ZgPgz ɯAiģ AiizzgAz Pgtz Dzò, EAz ɯAiģ PAqĻrAiĨP.

 

 

UĤ:

J AzsU KPPP Pgt r zs嫮.

1. a1 x+ b1 y = c1

2. a2 x+b2 y = c2

 

1. (a1 / a2) = (b1 / b2) (c1 / c2) Dzg svA E.

2. (a1 / a2) = (b1 / b2) = (c1 / c2) Dzg CASv jgU.

3. (a1 / a2) (b1 / b2) DzU iv svA qAiĮ zs.

 

2.14 ĸ 5:

 

r: x+y =2xy ------(1)

x-y = 6xy -----(2)

 

jg:

Jgq PgtU Pr, 2x = 8xy

CAzg 1 = 4y

y = 1/4

yAi ɯAiģ Pgt (1) g DzòzU,

x+ 1/4 = 2x/4 = x/2

PAvjzU, x-x/2 = - 1/4

x = -1/2

 

v:

x,y ɯU DzòzU:

x+y = -1/2+1/4 = -1/4

2xy = 2*(-1/2)*(1/4) = -1/4

x+y =2xy

x-y = -1/2-1/4 = -3/4

6xy= 6*(-1/2)*(1/4) = -3/4

x-y = 6xy

 

2.14 ĸ 6: MAz jPAiİ GwtvU CwtvU EgĪ Cĥv 4:1( Gwtgzg Cwtgzg 4 l).

MAz ü jPU PĽvg 30 zyU jPU dgUzݰ, 20 zyU Pr GwtgUwzg. DU D Cĥv 5:1 DVgwv. ( Gwtgzg Cwtgzg 5 l) Uzg jPU dgz zyU AS PAqĻr.

 

jg:

jPAiİ Gwtgz zyU: x DVg.

Cwtgz zyU = y DVg.

x=4y ------(1)

jPU PĽvg= x+y

30 A Pr dgzU, 20 A (x-20) Pr GwtgUwzg. DU

1) dgUgĪ Ml zyU = x+y-30

2) GwtgUgĪ zyU = (x+y-30) (x-20)

= y-10

 

3) GwtvU CwtvU EgĪ Cĥv 5:1 DUwv.

(x-20) = 5(y-10) -----(2)

 

FU U Jgq PgtU Qz: ɯAiģ 1 Pgtz DzòzU

4y-20 = 5(y-10) ----- (x ɯAiģ 2 Pgtz DzòzU)

= 5y-50

4y-20 -4y+20 = 5y-50-4y+20 (Jgq Az 4y Pz 20 PrzU)

0= y-30

30=y

x=4*30( y ɯAiģ 1 Pgtz DzòzU)

=120

 

v:

Gwtg: Cwtg = 120:30 (Ez 4:1 g Cĥvz Ez)

z jPU PĽvg= 120+30=150

30 A dgUzݰ jPU PĽvg = 150-30 =120 v 20 A Pr GwtgUwzg. DU

Gwtg = 120-20 =100

Cwtg = 120-100 = 20

Gwtg: Cwtg = 100:20 (Ez 5:1 g Cĥvz Ez)

 

2.14 ĸ 7: Jgq CAQ ASAi r DAPU v 9. F ASAi 9 g, r DAQU Cz z irzU zgv Ƹ ASAi JgqgP Ezg CAQU AiiŪ?

jg:

x CAQAi vg ܣz y Ai r ܣz Eg. DU ASAi(xy) ɯ 10x+y. Ezg ܣU zzU zgAiĪ ASAi (yx) ɯ 10y+x.

zvAzAv:

x+y = 9 (xy)

9(10x+y) = 2(10y+x)

 

Cs:

F PgtU r(x =1 and y=8). AS 18 DVgvz.

1+8 = 9

9*18 =2*81

 

2.14 ĸ 8: vAi evU HjU UPVz JAz sj, ê zyAiiVgĪzjAz U nPm ɯAiİ 50% Prv Egvz, Dzg PjĪ Į z jAiiw EgĪŢ. PjĪ Į j vAi nPm 2125 g Ez g nPm U 3200 g Dzg, M Aiĸ̣

nPm ɯ v PjĪ Į K?

 

jg:

x nPm ɯ v y PjĪ Į Eg. ê zy DVgĪzjAz nPm ɯ (1/2)x

x+y = 2125 ----(1)

ê zy DVgĪzjAz nPm ɯ (1/2)x . PjĪ Įz w Ez EgĪzjAz Cz EjU y Ai DVgvz.

g nPm ɯ = {(1/2)x+y} + (x +y) =3200

(3/2)x+2y =3200 = 3200(Jgq Pq Ut¹zU:)

3x+4y =6400 ----(2)

PgtU rzU x= 2100, 7= 25JAz zgAivz.

v:

2125 = 2100+25

3200 = 2100+25+1050+25

 

 

 

 

2.14 Pv SAU

 

 

P.A.

Pv SAU

1

( a1 x+ b1 y = c1, a2 x+b2 y = c2)F jwAi KPPP PgtU g PgtUV jw, rĪz.

2

J KPPP gSvP PgtU r zs嫮.

 

aѣ iwU

 

AiiAi zs(Alternate method): DzñAz UrĪz(Method of elimination by substitution)

 

KPPP gSvP PgtU EAz zsAz rz:

 

1.      AiizzgAz Pgtz MAz ZgPgz ɯAiģ (y) EAz ZgPgz (x) gƥz gɬj.

2.      ð Avz qz (y) Ai ɯAiģ EAz Pgtz Dzò,(x) ɯ PAqĻrj.

3.      (x) ɯAiģ AiizzgAz Pgtz Dzò,(y) Ai ɯ PAqĻrj.

 

FU 2.14.2 ĸAiģ F zsz rĪ.

r:

5y+10x =240 ----(1)

5y -5X = 90 ----(2)

5y= 5x + 90 (Pgt 2 g 5x PAvjz)

5 jAz sV, y = x+18

y Ai ɯAiģ Pgt 1 g Dzò,

5y +10x =240

5(x+18)+10x=240

5x+90+10x=240

15x= 240-90

=150

CAzg 150 = 15x

x = 10

x ɯAiģ 1g Dzò,

5y+10*10 = 240

CAzg 5y = 240-100=140

y = 28

F ɯU Gz.2.14.2 g Pq zgw.