CHUYN - TON LP 7

CC BI TON V T L THC
TNH CHT CA DY T S BNG NHAU.

A. Kin thc c bn.

I. T l thc.
1. nh ngha: T l thc l ng thc ca hai t s
Dng tng qut:

hoc a:b=c:d

Cc s hng a v d gi l ngoi t; b v c gi l trung t

2. Tnh cht.
a) Tnh cht 1 (Tnh cht c bn)
=> ad = bc (vi b,d0)
b) Tnh cht 2 (Tnh cht hon v)
T t l thc

(a,b,c,d0) ta c th suy ra ba t l

thc khc bng cch:

- i ch ngoi t cho nhau
- i ch trung t cho nhau
- i ch ngoi t cho nhau v i ch trung t cho nhau

C th: T

Bi 15.

(a,b,c,d0)

Qung ng AB di 76m, ngi th nht i t A n B

v ngi th hai i t B n A. Vn tc ca ngi th nht ch
4
vn tc ca ngi th hai (n lc gp nhau). Thi gian
5
10
ca ngi th nht ch bng thi gian ca ngi th hai. Tnh
11

bng

qung ng mi ngi i c ?
II. Tnh cht ca dy t s bng nhau.
1) Tnh

cht

1:

thc

suy

ra

(bd)
ta suy ra

2) Tnh cht 2:

(Gi thit cc t s u c ngha)

* Nng cao.
1. Nu
2. T

=k th
=> +)

43

Bi 10.
Bit

+)

x
y
z
t

y z t z t x t x y x y z

Tnh P

(Tnh cht ny gi l tnh cht tng hoc hiu t l)

* Ch : Cc s x, y, z t l vi cc s a, b, c =>

x y y z z t t x

z t t x x y y z

Ta cn vit x:y:z = a:b:c

Bi 11.

B. Cc dng ton v phng php gii.

Tm s t nhin c ba ch s, bit rng s l bi ca 72

v cc ch s ca n xp t nh n ln th t l vi 1 ;2 ;3

Dng 1: Tm thnh phn cha bit trong t l thc, dy t

s bng nhau
Dng 2: Chng minh t l thc

Bi 12 :
Tm hai phn s ti gin bit hiu ca chng l

Dng 3: Tnh gi tr biu thc

3
v cc
196

t tng ng t l vi 3 v 5 , cc mu tng ng t l vi 4 v 7

Dng 4: ng dng tnh cht ca t l thc, dy t s bng

nhau vo gii bi ton chia t l.

Bi 13.

Dng 5: Tnh cht ca t l thc p dng trong bt ng

Cho ABC cc gc ngoi ca tam gic ti A,B,C t l vi

4 ;5 ;6 . Cc gc trong tng ng t l vi cc s no ?
Bi 14.

thc

Dng 1: TM THNH PHN CHA BIT TRONG T L

Trong mt t lao ng, ba khi 7,8,9 chuyn c

912m t. Trung bnh mi hc sinh khi 7,8,9 theo th t lm
c 1, 2m3 ;1, 4m3 ;1, 6m3 . S hc sinh khi 7 v khi 8 t l vi 1
v 3, s hc sinh khi 8 v 9 t l vi 4 v 5. Tnh s hc sinh
mi khi ?
3

42

THC, DY T S BNG NHAU

Bi 1: Tm x bit:
a)
3

Bi 6.

b)

Cho dy t s bng nhau :

Gii

a
a1 a2 a3

2014 Cmr ta c
a2 a3 a4
a2015

ng thc
a) T

=> 7(x-3) = 5(x+5). Gii ra x = 23

b) Cch 1. T

a a a a2014
a1
1 2 3

a2015 a2 a3 a4 a2015

=> (x-1)(x+3) = (x+2)(x-2)

Bi 7.

(x-1).x + (x-1).3 = (x+2).x (x+2).2

- x + 3x 3 =

+ 2x 2x 4

2014

Cho

a c
cc s x, y, z, t tha mn ax yb 0 v zc td 0
b d

a v 2x = -1 => x =
Cch 2:

Cmr :

x 1
x2
+1=
+1
x2
x3

Bi 8.

2x 1 2x 1
=
x2
x3
2x+1=0 x= -

Cho t l thc
1
(Do x+2 x+3)
2

Bi 2: Tm x, y, z bit:

v x 3y + 4z = 62
Gii

Cch 1 (t gi tr chung)
t

xa yb xc yd

za tb
zc td

2a 13b 2c 13d
a c
Cmr :

3a 7b
3c 7 d
b d

Bi 9.
Cho

a
a
a1 a2 a3

n 1 n
a2 a3 a4
an
a1

Tnh : 1) A

=>
2) B

a12 a22 an2

a1 a2 an

a19 a29 an9

a1 a2 an

M x 3y + 4z = 62 => 4k 3.3k + 4.9k = 62

4

41

( a1 a2 an 0 )

x 3 y 5
; v 2 x 3 y 5 z 1
y 2 z 7

a.

4k 9k + 36k = 62

b,

31k = 62 => k = 2

1 4 y 1 6 y 1 8y

13
19
5x
2x 1 y 2 2x 3 y 1
d,

5
7
6x
y z 1 x z 2 y x 3
1

x
y
z
x yz

Do

c.

Vy x = 8; y= 6; z = 18
Cch 2 (S dng tnh cht ca dy t s bng nhau)
p dng tnh cht ca dy t s bng nhau ta c:

Bi 4.
Cho t l thc

a c
. Chng minh rng ta c t l thc sau
b d

=>

( vi gi thit cc t s u c ngha )
2 a 7b 2c 7 d

3a 4b 3c 4d
2015a 2016b 2015c 2016 d

2016c 2017 d 2016 a 2017b

a.

Cch 3 (Phng php th)

T

ab
a b
c.
2
2
cd

b,

c d

d,

ab 2a 3b

cd 2c 3d

=> x=

e,

=> y=

7 a 5ac 7b 5bd

7 a 2 5ac 7b 2 5bd

M x 3y + 4z = 62 =>

Bi 5.

ua v 31z =

558 => z = 18
a
b

Cho a c 2b v 2bd c b d ; b, d 0 CMR :

c
d

Do x =

; y=

Vy x = 8; y = 6 v z =18
40

b.
Bi 3: Tm x, y, z bit:

x y z

10 6 21

v 5 x y 2 z 28

c. 4 x 3 y ; 7 y 5 z v 2 x 3 y z 6
a)

v 2x + 3y z = 186
d. x : y : z 12 : 9 : 5 v xyz 20

b) 2x = 3y = 5z v

=95
Gii

a) Cch 1: T
V

=>

=>

=>

=>

e.

10
6
14
v xyz 6720

x 5 y 9 z 21

f.

x 16 y 25 z 9
v 2 x3 1 15

9
16
25

Bi 2.
Tm cc s x,y,z bit rng

=>

(*)
a. x : y : z 3 : 4 : 5 v 5 z 2 3x 2 2 y 2 594

Ta c:

b. 3 x 1 2 y 2 ; 4 y 2 3 z 3 v 2 x 3 y z 50

=>

c.

12 x 15 y 20 z 12 y 15 y 20 z
v x y z 48

7
9
11

Vy x=45; y=60 v z=84

d.

2x 3 y 4z
v x y z 49

3
4
5

Cch 2: Sau khi lm n (*) ta t

=k

(Sau gii nh cch 1 ca bi 2)

Bi 3.
Tm cc s x,y,z bit :

Cch 3: Sau khi lm n (*) dng phng php th gii nh

cch 3 ca bi 2.
6

39

d
d
d c

6
d+a+b+c d a b a b c d

Cng bt ng thc kp (3); (4); (5); (6) theo tng v th

b) V 2x = 3y = 5z =>

Bi 2. Cho

a c
a ab cd c
v b; d 0 CMR: 2

b d
b b d2 d

=>

x y z 95

x y z 95

c:
a
b
c
d
1

2
abc bcd cd a d ab

+) Nu x+y-z= 95
Ta c

=>

+) Nu x + y z = - 95
Ta c

=>

Gii:
Ta c

a c
a.b c.d
ab cd
v b; d 0 nn

2 2
b d
b.b d.d
b
d

Theo tnh cht (2) ta c:

ab ab cd cd
a ab cd c
2
2 2

2
2
b
b d
d
b b d2 d

C. BI TP P DNG

Vy:

Bi 4: Tm x, y, z bit:
a)

v x + z = -196

b)

v 5z 3x 4y = 50

Bi 1.
c)

Tm cc s x,y,z bit rng

a.

4
3
2

v x + y z = - 10
3x 2 y 2 z 4 x 4 y 3z

Gii

x2 x4

x 1 x 7

a) V
38

=>
=>
=>

a. Nu

th

b. Nu

th

Bi 1. Cho a; b; c; d > 0.

CMR: 1
Ta c

=>

a
b
c
d

2
abc bcd cd a d ab

Gii:
Vy x = 231; y = 28 v z = 35
+ T
b) Ta c

a
1 theo tnh cht (3) ta c:
abc

ad
a

1 (do d>0)
abcd abc

Mt khc:

+ T (1) v (2) ta c:

a
a
ad

3
abcd abc abcd

Tng t ta c:

Vy x = 5; y = 5 v z = 17
c) V

a
a

2
abc abcd

4
3
2

=
3x 2 y 2 z 4 x 4 y 3 z

b
b
ba

4
abcd bcd abcd
c
c
cb

5
abcd cd a cd ab

37

+ C:

ad bc ad bc
a c

b 0; d 0 bd db
b d

Tnh cht 2: Nu b > 0; d > 0 th t

a c
a ac c
(Bi

b d
b bd d

=>

=>

5/33 SGK 7)
Gii:
a c

+ b d
ad bc(1) thm vo 2 v ca (1) vi ab ta
b 0; d 0

c:

x y z x y z 10

10
2 3 4 23 4
1

=>

Vy x = - 20; y = -30 v z = -40

ad ab bc ab
a b d b c a

a ac

2
b bd

Bi 5: Tm x. y, z bit:

+ Thm vo hai v ca (1) dc ta c:

a) x: y: z = 2: 3: 5 v xyz = 810

1 ad dc bc dc
d a c c b d

b)

= - 650

ac c
3
bd d

Gii
a) V x: y: z = 2: 3: 5 =>

+ T (2) v (3) ta c:
T

Cch 1 (t gi tr chung)

a c
a ac c

(pcm)
b d
b bd d

=>

Tnh cht 3: a; b; c l cc s dng nn

36

M xyz = 810 => 2k.3k.5k = 810 => 30

=810 =>

=27 => k

=>

2x 2 y 2z

12 9 10

=>

x y z
x y z 186

6
12 9 10 12 9 10 31

=3
=>

Vy x = 6; y = 9 v z = 15

Cch

2:

=>

x 72
=> y 54 ( Tha mn iu kin )
z 60

Vy trong ngy ca hng bn s mt vi cun th

nht, th hai, th ba ln lt l : 24; 36; 24 (mt).

=> x = 6 thay vo bi tm ra y = 9 ; z = 15
Dng 5: TNH CHT CA T L THC

Vy x = 6; y = 9 v z = 15
Cch 3: (Phng php th) Lm tng t cch 3 ca bi 2
b) T

=>

=>

Cch 1: (t gi tr chung)
t
M

=
+2

Tnh cht 1: (Bi 3/33 GK 7) Cho 2 s hu t

b> 0; d >0.

= k =>
3

P DNG TRONG BT NG THC

CM:

= - 650 => 4

=>-26

+ 2.9

a c
ad bc
b d

Gii:
a c

ad cb
+ C b d

ad bc

bd db

b 0; d 0

Nu k = 5=>
10

35

a
c
v vi
b
d

th nht, th hai, th ba ln lt t l vi 2; 3; 2. Tnh xem trong

ngy ca hng bn c bao nhiu mt vi mi cun.

Nu k = -5 =>

Vy

Li gii
Gi chiu di cun vi th nht, th hai, th ba ln lt l
x, y, z (m)

Cch 2 (S dng tnh cht ca dy t s bng nhau)

V

=>

K: 0< x, y, z < 186

+) Tng chiu di ba cun vi l 186m => x + y + z =

=>

186
+ Sau khi bn c mt ngy ca hng cn li

cun th

x 10; y 15; z 20

nht, cun th hai, cun th ba

=> Trong ngy ca hng bn c s mt vi cun
th nht, th hai, th ba ln lt l

Theo bi suy ra x,y,z cng du

Vy
x 10; y 15; z 20
Cch 3 (Phng php th)

x 2 y 2z
(mt)
, ,
3 3 5

+) S tin bn c ca ba cun th nht, th hai, th ba ln

Bi 6: Tm x, y, z bit:

lt t l vi 2; 3; 2 v gi tin mi mt vi ca ba cun nh

(1)

nhau.
Gii:

=> S mt vi bn c ca ba cun th nht, th hai, th ba

* Nu

ln lt t l vi 2; 3; 2
=>

x 2 y 2z
:
:
2:3: 2
3 3 5
34

11

Ta c

Gi vn tc d nh l x, vn tc mi tng l y ( x,y > 0)

Ta c y

120
y 6
x =>
100
x 5

(2)

Gi C l trung im ca AB. t n B sm hn d nh

T (1) v (2) ta c x + y + z =

10 pht l nh tng vn tc t im C.
Nu t i t C n B vi vn tc x mt thi gian l
Nu t i t C n B vi vn tc y mt thi gian l

=>

thay vo bi ta c:
Th x. = y.
=>

y
x

=>

y 6
t1
m
x 5
t2

t 60
t t
t t
t1 6
=> 1 2 1 2 10 => 1

6 5
65
t2 5
t2 50

=>Thi gian t i na ng AB vi vn tc tng ht

Hay

50 pht
Thi gian t i na ng AB vi vn tc d nh ht 60

+)

=> 2x =

=> 3x =

=> x =

pht.
Vy thi gian t i t A n B l 60 + 50 = 110 (pht)

+)

=> 2y =

=> 3y =

=> y =
Bi 7: Mt ca hng c ba cun vi, tng chiu di ba

+) C x + y + z =

, m x =

v y =

cun vi l 186m, gi tin mi mt vi ca ba cun l nh

nhau. Sau khi bn c mt ngy ca hng cn li

cun th

nht, cun th hai, cun th ba. S tin bn c ca ba cun

12

33

Gi a, b, c l di ba cnh ca mt tam gic v

ln

lt l cc chiu cao tng ng.

Din tch ca tam gic l:
b.

= c.

=>z=
a.ha b.hb c.hc

2
2
2

=> a.

b
3

* Nu x + y + z = 0 ta c:

(1)

(1) =>

c
4

=> k ( k o )

=> x = y = z = 0

=> a = 2k, b = 3k v c = 4k
(1) =>2k.
=> 2
=>

=3

= 3k.
=4

ha hb hc

=>
6 4 3

Vy

+) c a, b, c t l vi 2; 3; 4
a
2

Vy

= 4k.
=>

2ha 3hb 4hc

12 12 12

t l vi 6; 4 ; 3

Bi 7: Tm x, y bit:
a)

Vy di ba cnh ca mt tam gic t l vi 2; 3; 4 th ba

chiu cao tng tng vi ba cnh t l vi 6; 4; 3.

b)
Gii

Bi 6: Mt t phi i t A n B trong mt thi gian d

a) V

=> 24(1+2y) = 18(1+4y)

nh. Sau khi i c qung ng th t tng vn tc thm

=>24 +48y = 18 +72y
20%. Do t n B sm hn c 10 pht. Tnh thi gian
t i t A n B.
Li gii
32

13

a v 24y = 6 => y =

Li gii

thay vo bi ta c

Gi ba phn s cn tm l

a , b, c , d , e , g Z

vi

b, d , g 0

Theo u bi ta c
=>
Ta c

a : c : e = 3:4 :5, b : d : g =5:1:2

= 18. => 18x = 90 => x = 5

1 3y
12

+) a:c:e= 3 :4 :5 =>

a c e
3
3
b d g
70

a c e
k vi k Z
3 4 5

a=3k ,c =4k , e =5k

b d g
t vi t Z , t o
5 1 2

+) b : d : g = 5 : 1 : 2 =>
=>1+3y = -12y => 15y = -1 => y =

thay vo

b=5t, d=t, g=2t

+)

Ta c

=> 5x .

=>

Vy x = 2 v y =

=> x = 2

3k 4k 5k 213
a c e
3
=>
3
5t
t
2t
70
b d g
70

k 71 213
k 3
=>
.
t 10
70
t
7

a 9 c 12 e 15

,
,
g 14
b 35 d
7

Vy ba phn s cn tm l

9 12 15
,
,
35
7
14

Bi 5: di ba cnh ca mt tam gic t l vi 2; 3; 4. Ba

chiu cao tng ng vi ba cnh t l vi ba s no?
Dng 2: CHNG MINH T L THC
14

Li gii
31

Li gii

chng minh t l thc

Gi s hc sinh t mt, t hai, t ba ca lp 7A ln lt l

x, y, z.(hc sinh)

ta thng dng mt s

phng php sau:

) Phng php 1: Chng t rng A.D = B.C

K: x, y, z N * , x, y, z 52

) Phng php 2: Chng t hai t s

+) Lp 7A c 52 hc sinh => x + y + z = 52
+) Nu t mt bt i 1 hc sinh, t hai bt i 2 hc sinh, t

c cng gi tr

) Phng php 3: S dng tnh cht ca t l thc

ba thm vo 3 hc sinh th s hc sinh t mt, hai, ba t l

* Mt s kin thc cn ch

nghch vi 3; 4; 2

Nn ta c 3.(x 1) = 4.(y 2) = 2.(z + 3)

3 x 1 4 y 2 2 z 3

12
12
12

2 z 3

3
6

(n

0)

=>

(n

Bi 1: Cho t l thc

x 1 16
x 17

y 2 12 y 14 (Tha mn iu kin)
z 3 24
z 21

Vy s hc sinh t mt, t hai, t ba ca lp 7A ln lt l

17 hc sinh, 14 hc sinh, 21 hc sinh.

Chng minh rng

GII
Cch 1 (pp1):

. Bit t ca

chng t l vi 3; 4; 5 cn mu ca chng t l vi 5; 1; 2.
30

N*)

Sau y l mt s bi tp minh ha ( gi thit cc t s cho

u c ngha)

x 1 y-2 z 3 x y z 52

4
4
3
6
13
13

Bi 4: Tm ba phn s c tng bng

15

K: x, y, z N * , x, y, z 144
Ta c:

+) Ba lp 7A,7B,7C c tt c 144 hc

sinh

=> x y z 144
(a+b).(c-d) = (a b).(c+d)

+) Nu rt lp 7A i

sinh, rt lp 7C i
Cch 2 (pp2):
t

1
hc sinh th s hc sinh cn li ca
3

3 lp bng nhau.

= k =>

Nn ta c

1
1
hc sinh, rt lp 7B i hc
4
7

3
6
2
x y z
4
7
3

3
6
2
x y z x y z 144
x
y

6
24
42
18 z
8 7 9 8 7 9 24

x 48

y 42 (Tha mn iu kin)
z 54

Vy s hc sinh lc u ca cc lp 7A, 7B, 7C ln lt l

Cch 3 (pp3):

48 hc sinh, 42 hc sinh, 54 hc sinh.

T
Bi 3: Lp 7A c 52 hc sinh c chia lm ba t. Nu t
mt bt i 1 hc sinh, t hai bt i 2 hc sinh, t ba thm

Ta c:

vo 3 hc sinh th s hc sinh t mt , hai, ba t l nghch

vi 3; 4; 2. Tm s hc sinh mi t.
16

29

a
1

b
3

c
2

+) a, b, c t l vi 1; 3; 2 =>

abc
6

Cch 4: T

=>

=>a + b + c 6
9

Li c

<=>a + b + c 9

M 1 a b c 27
Nn a + b + c = 18
=>

a b c
3
1 3 2

=>

=>

(Tha mn iu
Bi 2: Cho t l thc

Chng minh rng

kin)
(1)
Nu a, b, c t l vi 3; 1; 2 =>

(Tha mn
GII

iiu kin)

Cch 1:

Vy s t nhin c 3 ch s cn tm l 396; 936.

Bi 2: Ba lp 7A, 7B, 7C c tt c 144 hc sinh. Nu rt
lp 7A i

1
1
s hc sinh, rt lp 7B i s hc sinh, rt
4
7

lp 7C i

1
hc sinh th s hc sinh cn li ca c 3 lp
3

bng nhau. Tnh s hc sinh mi lp ban u.

Li gii
Gi s hc sinh ban u ca lp 7A,7B.7C ln lt l x,y,

Cch 2:

z (hc sinh)
28

17

= k =>

thay vo 2 v ca (1) chng minh 2

Vi a, b, c 0 ta c :

v c cng gi tr

ab bc ca
1 1 1 1 1 1

ab
bc
ca
b a c b a c

1 1 1
a b c P 1
a b c

Cch 3:

=>

ab
bc
ca

ab bc ca

Dng 4: NG DNG TNH CHT CA T L THC,

DY T S BNG NHAU VO GII BI TON CHIA

T L

B i 3: chng minh rng nu

th
Bi 1: Tm s t nhin c ba ch s bit rng s chia

a)
b)

ht cho 18 v cc ch s ca n chia ht cho t l vi 1;2;3.

Li gii

Gi s t nhin c 3 ch s cn tm l
GII

( K

: a, b, c N * ,1 a 9, 0 b, c 9 )
=> 1 a b c 27

a) T

+)

18

<=>

( do 18=2.9 v

=>
CLN(2;9)=1 )
b) T

+) Cc ch s ca s cn tm t l vi 1; 2; 3
M

2 => c 2

=>a, b, c t l vi 1;3; 2 hoc a; b; c t l vi 3; 1; 2

18

27

Cho a , b ,c i mt khc nhau v tha mn

=>

ab bc ca

c
a
b
a

Tnh gi tr ca biu thc P 1 1

1
b
c
a

Bi gii:

ab bc ca
ab
bc
ca
T

1
1
1
c
a
b
c
a
b

=>

abc abc abc

(*)

c
a
b

+) Xt a b c 0 a b c; a c b; b c a
P

a b b c a c c a b abc

1
b
c
a
b c a
abc

Bi 4: Cho b2 = ac; c2 = bd. Chng minh rng:

1)

+) Xt a b c 0 T (*) ta c :
2)

a bc P 8

Bi 5 :

GII

Cho cc s a;b;c khc 0 tha mn

2

Tnh gi tr ca biu thc P

ab
bc
ca

ab bc ca

ab bc ca
a 3 b3 c3

1) V

Bi gii:
26

19

a
b
c
d
Tnh gi tr ca biu thc

bcd acd abd bca

Vy

ab bc cd d a

cd ad ab bc

Bi gii:
2) C:

Bi 5: Cho a, b, c tha mn

a
b
c
d

bcd acd abd bca

a
b
c
d
1
1
1
1
bcd
acd
abd
bca

abcd abcd abcd abcd

(*)
bcd
acd
abd
bca

+) Xt a b c d 0 a b (c d ); b c (a d )
Chng minh: 4(a-b)(b-c) =
M 4

GII
+) Xt a b c d 0 T (*) ta c :
T

bcd acd abd bca

abcd M 4

Bi 6: Bit
CMR: abc +

Bi 4:

=0
20

25

3x
1
3x y 3
3
y
T

x
x y 4
1 4
y

GII
x
3a 1
3
t = a
=
a 1
4
y

Bi 2:
Cho

yzx
x y z
. Tnh gi tr ca biu thc P =
2 3 4
x yz

=> ab +

Nhn c hai v ca (1) vi c ta c: abc +

(2)
Ta c :

Cch 1:

(1)

=> bc +

(3)
t

x y z
= k x = 2k ; y = 3k ; z = 4k ( k 0)
2 3 4

P=

3k 4k 2k 5k 5

2k 3k 4k 3k 3

Vy P =

5
3

abc +

abc +

x y z
yzx yzx x yz x yz
=

2 3 4 3 4 2
5
23 4
3

yzx x yz
yzx 5

5
3
x yz 3

Vy P =

ta c:

Cng c hai v ca (2) v (4) ta c:

Cch 2 :
C

Nhn c hai v ca (3) vi

(4)

=
=0

Bi 7: Cho

(1)

CMR:
GII

5
3

Nhn thm c t v mu ca (1) vi a hoc b; c

Bi 3 :

T (1) ta c:

Cho dy t s bng nhau

24

21

=
=0

Dng 3 : TNH GI TR BIU THC

Bi 1 :
Cho t l thc
Bi 8: CMR: Nu a(y+z) = b(z+x) = c(x+y)
(1)

3x y 3
x
. Tnh gi tr ca t s
x y 4
y

Bi gii:

Trong a,b,c l cc s khc nhau v khc 0 th:

Cch 1 :
T

GII
V a,b,c 0 nn chia cc s ca (1) cho abc ta c:

3x y 3
4(3x y) = 3(x+y) 12x 4y = 3x + 3y
x y 4

12x 3y = 3(x+y) 9x = 7y

Vy

7
x
=
9
y

Cch 2:
=
22

23