If 5x^2-3y^2=(11/2)*xy, then value of x:y is.... given that x and y are positive numbers.

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24 May 2013 expresssion

If 5x^2-3y^2=(11/2)*xy, then value of x:y is....
given that x and y are positive numbers.

OPtion

1) 1:2
2) 4:5
3) 3:4
4) 2:5
5) 11:2
6) 5:6
7) 3:2
8) 6:7
9) 7:8
10) None of these

Solution

5x^2-3y^2=(11/2)xy
10x^2-6y^2=11xy
(10x^2/y^2)-6=11x/y (Dividing each term by y^2)
let x/y=k
10k^2-11k-6=0
solve
k=3/2 or -2/5
Taking the positive value for k we get,
x/y=3/2
option 7

Correct Option: 7 Best Solution (19)

shivendra pratap singh   11 years AGO

5x^2-3y^2=(11/2)*xy
divide whole equation by xy
we get :
5x/y-3y/x=11/2
put x/y=t
then equation becomes 5t-3/t=11/2
solve this as quadratic equation
t=3/2
and we know t=3/2=x/y
so option 7)3:2 is correct one

Like? Yes (8) | No (1)  

IFFAT IRAM   11 years AGO

5x^2 - 3y^2 = (11/2)*xy
10x^2 - 11xy - 6y^2=0
10x^2 - 15xy + 4xy - 6y^2 = 0
5x(2x-3y) + 2y(2x-3y) = 0
(2x-3y)*(5x+2y) = 0
since, x and y are both positive
x:y= 3:2
option 7 is the right answer

Like? Yes (4) | No   

Dr.Dipin Singh   11 years AGO

5x^2-3y^2=(11/2)*xy,
5x^2-3y^2-(11/2)*xy=0,

10x^2-6y^2-11*xy=0

solving it for x, we get
x= (3/2 )*y
x/y=3/2

Like? Yes (4) | No   

Anita   11 years AGO

if x/y=k,
then x=ky
put x=ky in 5x^2-3y^2=(11/2)*xy
and splve for k after cancelling y
we get
k=3/2

Like? Yes (3) | No   

sanyam rastogi   11 years AGO

(5x^2 - 3y^2)/ xy = 11/2
5(x/y)- 3(y/x)=11/2
let x/y = a
5a-3/a= 11/2
10a^2 - 11a - 6= 0
(5a+2)(2a-3)=0
a=3/2 (+ve)
hence x/y == 3/2

Like? Yes (2) | No (1)  

vinod   11 years AGO

divide given equ by y^2
gives 10(x/y)^2 - 6 - 11(x/y)=0
(x/y)=ratio=a
gives 10a^2-6-11a=0
by solving a=3/2

Like? Yes (2) | No (1)  

ajitha   11 years AGO

The eq'n 5x^2-3y^2=(11/2)*xy
It would become 5(x/y)-3(y/x)=11/2
Let us take x/y=p
then the eq'n 5p-3/p=11/2
5p^2-3=11/2 p
10p^2-11p-6=0
(2p-3)(5p+2)=0
p=3/2 since x and y are _ve numbers
Hence x/y=3/2
x:y=3:2

Like? Yes (2) | No (1)  

sanjay singh   11 years AGO

solving the expression we get
(5x+2y)(2x-3y)=0
>x:y=-2:5 or >x:y=3:2
as x and are positive numbers,
x:y=3:2 is correct option

Like? Yes (2) | No (1)  

django   11 years AGO

factorise it

Like? Yes (1) | No   

Anil Kumar Bharti   11 years AGO

5x^2-3y^2=(11/2)*xy
5x^2-3y^2-(11/2)*xy=0
10x^2-6y^2-11*xy=0
10x^2-11*xy-6y^2=0
10x^2-15xy+4xy-6x^2=0
5x(2x-3y)+2y(2x-3y)=0
(2x-3y)(5x+2y)=0
now,
2x-3y=0
2x=3y
x/y=3/2
x:y=3:2 Ans (7).

Like? Yes (2) | No (1)  

Md Intshamuddin   11 years AGO

5x^2-3y^2=(11/2)*xy,or,10x^2-11xy-6y^2=0,or,x=(3/2)*y,or,x/y=3/2

Like? Yes (1) | No (1)  

Hasina Abdulhakkim   11 years AGO

5x^2-3y^2=(11/2)*xy,or,10x^2-11xy-6y^2=0,or,x=(3/2)*y,or,x:y=3:2

Like? Yes (1) | No (1)  

Shalu Jain   11 years AGO

on solving the above equation we get
(2x-3y)(5x+2y)=0
so,
either
2x-3y=0 or 5x+2y=0
but, given that x and y are positive numbers
so,
we are taking
2x-3y=0
=>x/y=3/2

so, x:y=3:2
QED...:)

Like? Yes (1) | No (1)  

ravi singh   11 years AGO

(y/x + 11/12)^2 = 5/3 + 121/144

Like? Yes (1) | No (1)  

krishnapriya   11 years AGO

put x=3 & y=2 in the equation
we get both sides equal
(ie)33=33

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NEHA    11 years AGO

simplyfying the equation we get..
10x^2-11xy+6y^2=o
After factorising we get,
(2x-3y)(x+2y)=0
x/y=3/2(because x and y are positive)

Like? Yes (1) | No (1)  

Devendra Marghade   11 years AGO

5x^2-3y^2=(11/2)*xy
Divide the eqn. by y^2, 5(x/y)^2-3=(11/2)(x/y)
Put x/y=a in the above equation
10a^2-11a-6=0
(5a+2)(2a-3)=0 or a=-2/5 or 3/2
Considering positive value, a=3/2=x/y

Like? Yes (1) | No (1)  

jitendra singh   11 years AGO

3/2

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Satyaranjan   11 years AGO

5x^2-3y^2=(11/2)*xy
multiplying 1/(y^2) with both sides we get
5(x^2/y^2)-(11/2)(x/y)-3=0
let x/y=k,then 5k^2-(11/2)k-3=0
By solving we get x/y=3/2

Like? Yes (1) | No (1)