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22-son 4 –to’plam fevral 2022
Sahifa: 361
Ikkala tenglik ham to'g'ri tenglik. Shunday qilib, (3) sistema birgina yechimga ega:
x
=1,
y
=2.
(3) sistemani yechishning ko'rib chiqilgan bu usuli o'rniga qo'yish usuli deyiladi. U
quyidagilardan iborat:
1) sistemaning bir tenglamasidan (qaysinisidan bo'lsa ham farqi yo'q) bir noma'lumni
ikkinchisi orqali, masalan,
y
ni
x
orqali ifodalash kerak;
2) hosil qilingan ifodani sistemaning ikkinchi tenglamasiga qo'yish kerak - bir
noma'lumli tenglama hosil bo'ladi;
3) bu tenglamani yechib,
x
ning qiymatini topish kerak;
4)
x
ning topilgan qiymatini y uchun ifodaga qo'yib,
y
ning qiymatini topish kerak.
3-masala.
Tenglamalar sistemasini yeching:
.
5
3
5
,
16
2
3
y
x
y
x
1) Birinchi tenglamadan -2
y
=16-3
x
,
,
2
3
16
x
y
ya‘ni
x
y
2
3
8
ekanini topamiz.
2)
x
y
2
3
8
ni sistemaning ikkinchi tenglamasiga qo'yamiz:
.
5
2
3
8
3
5
x
x
3) Bu tenglamani yechamiz:
,
5
2
9
24
5
x
x
,
19
2
19
x
.
2
x
4)
2
x
ni
x
y
2
3
8
tenglikka qo'yib, quyidagini topamiz:
.
5
2
2
3
8
y
Endi umumiy holdagi
2
2
2
1
1
1
,
c
y
b
x
a
c
y
b
x
a
chiziqli tenglamalar sistemasini o‘rniga qo‘yish
usulida yechish usulini ko‘rsatamiz:
1)
1
1
1
c
y
b
x
a
,
,
1
1
1
y
b
c
x
a
;
1
1
1
1
y
a
b
a
c
x
www.pedagoglar.uz
22-son 4 –to’plam fevral 2022
Sahifa: 362
2)
,
2
2
2
c
y
b
x
a
,
2
2
1
1
1
1
2
c
y
b
a
b
a
c
a
,
2
2
1
1
2
1
1
2
c
y
b
a
b
a
a
c
a
,
1
1
2
2
1
1
1
2
2
1
a
c
a
c
a
y
a
b
a
b
a
;
1
2
2
1
1
2
2
1
b
a
b
a
c
a
c
a
y
3)
1
2
2
1
1
2
1
1
1
2
1
1
2
2
1
1
1
1
2
2
1
1
1
1
2
1
2
1
1
2
2
1
1
1
1
2
2
1
1
1
1
1
2
2
1
1
2
2
1
1
1
1
1
b
a
b
a
a
c
b
a
c
b
a
b
a
b
a
a
c
b
a
c
b
a
c
b
a
c
b
a
b
a
b
a
a
c
b
a
c
b
a
a
c
b
a
b
a
c
a
c
a
a
b
a
c
x
;
1
2
2
1
2
1
1
2
1
2
2
1
1
2
1
1
2
1
1
2
2
1
1
2
1
1
1
2
1
b
a
b
a
c
b
c
b
b
a
b
a
a
c
b
c
b
a
b
a
b
a
a
c
b
a
c
b
a
Bu yerda
0
1
2
2
1
b
a
b
a
bo‘lishi kerak. Bundan
2
1
2
1
b
b
a
a
shart bajarilishi kerak degan
xulosaga kelamiz.
Bundan tashqari x va y ning hosil bo‘lgan qiymatlaridan
2
1
2
1
2
1
с
с
b
b
a
a
,
2
1
2
1
2
1
с
с
b
b
a
a
bo‘lganda tenglamalar sistemasi yechimlari qanday bo‘ladi degan muhim savol kelib
chiqadi?
Quyida biz bu savollarga batafsil izoh berib o‘tamiz.
1) Agar
2
1
2
1
b
b
a
a
shart bajarilsa, tenglamalar sistemasi yagona yechimga ega;
2) Agar
2
1
2
1
2
1
с
с
b
b
a
a
shart bajarilsa, tenglamalar sistemasi, yechimga ega emas;
3) Agar
2
1
2
1
2
1
с
с
b
b
a
a
shart bajarilsa, tenglamalar sistemasi cheksiz ko‘p yechimga ega.
Javob:
;
1
2
2
1
2
1
1
2
b
a
b
a
c
b
c
b
x
.
1
2
2
1
1
2
2
1
b
a
b
a
c
a
c
a
y
5-masala: Tenglamalar sistemasini yeching:
.
4
2
3
,
7
5
y
x
y
x
Yechish:
1
a
=1,
1
b
=5,
1
c
=7,
2
a
=3,
2
b
= -2,
2
c
=4
;
2
17
34
15
2
20
14
5
3
)
2
(
1
4
5
7
2
1
2
2
1
2
1
1
2
b
a
b
a
c
b
c
b
x
www.pedagoglar.uz
22-son 4 –to’plam fevral 2022
Sahifa: 363
.
1
17
17
15
2
21
4
5
3
)
2
(
1
7
3
4
1
1
2
2
1
1
2
2
1
b
a
b
a
c
a
c
a
y
Javob:
,
2
x
.
1
y
6-masala: a va b ning qanday qiymatlarida
7
6
5
,
10
3
y
ax
by
x
tenglamalar sistemasi
cheksiz ko‘p yechimga ega?
Yechish:
1
a
=3,
1
b
=-b,
1
c
=10,5,
2
a
=a,
2
b
= -6,
2
c
=7
2
1
2
1
2
1
с
с
b
b
a
a
shartga ko‘ra
7
5
,
10
6
3
b
a
, bundan esa
,
2
a
9
b
ekani kelib chiqadi.
Javob:
,
2
a
9
b
.
7-masala: m ning qanday qiymatlarida
0
8
2
,
0
4
2
my
x
y
mx
tenglamalar sistemasi
yechimga ega emas?
Yechish: Avval tenglamalar sistemasini
2
2
2
1
1
1
,
c
y
b
x
a
c
y
b
x
a
ko‘rinishga keltirib olamiz.
8
2
,
4
2
my
x
y
mx
, endi esa 2) shartni tekshiramiz:
1
a
=
m
,
1
b
=2,
1
c
=-4,
2
a
=2,
2
b
=
m
,
2
c
=8.
2
1
2
1
2
1
с
с
b
b
a
a
dan
8
4
2
2
m
m
,
.
4
m
Javob:
.
4
m
IV. Mavzuni mustahkamlash:
Mustahkamlash uchun savollar:
1.
Chiziqli tenglamalar sistemasi deb nimaga aytiladi?
2.
Chiziqli tengamalar sistemasini yechishni qanday usullari mavjud?
3.
Chiziqli tenglamalar sistemasiga misollar keltiring?
Mustaqil yechish uchun misollar:
VI.
Chiziqli tenglamalar sistemasini o‘rniga qo‘yish usulida yeching.
www.pedagoglar.uz
22-son 4 –to’plam fevral 2022
Sahifa: 364
1)
3
3
5
,
11
12
y
x
y
x
2)
;
5
2
3
,
0
3
2
y
x
y
x
3)
;
3
8
2
3
,
3
3
2
y
x
y
x
4)
.
6
8
7
6
5
,
3
4
5
3
2
y
x
y
x
2. Chiziqli tenglamalar sistemasini qo‘shish usulida yeching.
1)
;
7
15
24
,
2
21
18
y
x
y
x
2)
;
3
1
5
,
4
2
,
4
2
1
3
y
y
x
y
x
y
x
3)
.
24
1
7
3
4
,
20
3
2
5
y
y
x
y
x
x
x
y
y
x
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