[40/(x+y)] + [2/(x-y)] = 5 [25/(x+y)] - [3/(x-y)] = 1

Mukund Manohar   8 years AGO

Let 1/x+y = P and 1/x-y = Q
40P+2Q=5 -----I
25P-3Q=1 -----II
MULTIPLYING I BY 3 AND II BY 2 AND ADDING IT
P=1/10
SUBSITUTING THE VALUE OF P IN I,WE GET Q = 1/2
NOW,
1/x+y = 1/10 and 1/x-y = 1/2
x+y=10 and x-y=2
ON SOLVING,
x=6 and y=4

Like? Yes (8) | No (2)  

Neetin Kumar   8 years AGO

[40/(x+y)]+[2/(x-y)]=5.........(i)
[25/(x+y)]-[3/(x-y)]=1..........(ii)

equation (i)*3+(ii)*2

[120/(x+y)]+[6/(x-y)]=15.......(iv)
[50/(x+y)]-[6/(x-y)]=2............(v)

Add (iv)+(v)

[170/(x+y)]=17
x+y=10.....(vi)

put (x+y) in (i)

[40/(x+y)]+[2/(x-y)]=5

40/10+2/(x-y)=5
x-y=2........(vii)

from vi and vii

x=6 and y=4

Like? Yes (10) | No (5)  

Shagun Chaudhary   8 years AGO

put (1/(x+y))=u and (1/(x-y))=v
40u+2v=5
25u-3v=1
u=0.1,v=0.5
=>x+y=10
=>x-y=2
On solving,
=>x=6,y=4

Like? Yes (6) | No (2)  

MRITUNJAY   8 years AGO

40/x+y+2/x-y=5
25/x+y-3/x-y=1
after solving the equation we get
x+y=10
from equation 1
x-y=2
then x=6,y=4
option 7

Like? Yes (3) | No   

Sanjay Karmakar   8 years AGO

let x+y=p and x-y=q
then the given equations become
(40/p)+(2/q)=5
(25/p)-(3/q)=1
now, equating these two equations we have p=10 and q=2.
Now, equating x+y=10 & x-y=2, we have x=6 and y=4.

Like? Yes (3) | No   

A Dinesh Kumar Reddy   8 years AGO

A=1/x+y B=1/x-y
40 A+2 B=5---------1
25 A-3 B=1----------2
multiply eq 1 by 3 and eq 2 by 2 and add them
120 A+6 B+50 A-6 B=15+2
170 A=17
A=1/10
So then
40/10+2 B=5
4+2 B=5
2 B=1
B=1/2
So
1/x+y=1/10
x+y=10------------3
1/x-y=1/2
x-y=2--------------4
Add eq 3 and eq 4
x+y+x-y=10+2
2x=12
x=6
Then
6+y=10
y=4

Like? Yes (3) | No   

yedukondalu .t   8 years AGO

substitute
or
simlify

Like? Yes (3) | No   

Siva   8 years AGO

40A+2B=5,25A-3B=1 by solving these equattions we get A=1/10&B=1/2; this means x+y=10,x-y=2 by solving these two equations we will get x=6,y=4.

Like? Yes (2) | No   

Testing Account   8 years AGO

by simplify x/y=102/68
=>x/y=3/2
=>x=6;y=4

Like? Yes (2) | No   

Ankita   8 years AGO

Using the method of substitution put 1/(x+y)=A and 1/(x-y)=B
Then on solving simultaneous equation answer we get as 1/(x+y)=1/10 and 1/(x-y)=1/2
on solving these equation we get x=6 and y=4

Like? Yes (2) | No   

Ranvir Singh   8 years AGO

We need x2 - y2 = positive number
So x>y and difference between x&y a small number

Like? Yes (2) | No   

sandhya kumari   8 years AGO

(21*x-19*y)/(11*x-14*y)=5 gives x=1.5*y
put in one of given equations to get x=6 and y=4

Like? Yes (2) | No   

mangobind kumar kishan   8 years AGO

6+4=10
40/10=4
6-4=2
2/2=1
4+1=5

Like? Yes (2) | No   

Akshat Mokashi   8 years AGO

consider x+y=a and x-y=b and their product (x+y)(x-y)=ab and solve the equation. you will get y:x=4:6. Therefore y=4 and x=6.

Like? Yes (2) | No   

Maneesh Gupta   8 years AGO

assume (1/(x+y))=u and
(1/(x-y))=v
40u+2v=5
25u-3v=1
u=0.1
v=0.5
=x+y=10
=x-y=2
further On solving,
=>x=6,y=4

Like? Yes (3) | No (2)  

Siddharth Jain   8 years AGO

Let 1/(x+y) = a
and 1/(x+y) = b
so 40a + 2b = 5 ---(1)
and 25a - 3b =1----(2)
solving (1) and (2)
a = 1/(x+y) = 1/10 ------(3)
b = 1/ (x-y) = 1/2 ------(4)

so, solving eq. (3) and (4)
we get
x =6, y = 4 ANSWER

Like? Yes (1) | No   

K.NagaSugunaManasa   8 years AGO

substitute x=6 and y=4 in equation 1 and 2 then we get for 1st equation 5 and for second 1 so option is 7

Like? Yes (1) | No   

Surabhi Yadav   8 years AGO

By using the given options in the place of x & y
x=6,y=4

Like? Yes (1) | No   

Amit ambastha   8 years AGO

assume 1/x+y=p and 1/x-y= q then the equation will be 40p+2q=5 and 25p-3q =1 solve the equation then we get p=1/10 and q=1/2 then put the value of p and q in 1/x+y and 1/x-y respectively and we get by solving x=6 and y=4

Like? Yes (1) | No   

vikash kumar das   8 years AGO

put 1/x+y=u and 1/x-y=v..then the eqn becomes
40u+2v=5 and 25u-3v=1 and we get u=1/10 v=1/2
then x+y=10 and x-y=2 and after solving these two eqns we get x=6 and y=4

Like? Yes (1) | No   

dev   8 years AGO

40/(x+y) + 2/(x-y) = 5 ---------(1)
25/(x+y) - 3/(x-y) = 1 ----------(2)

(1) * 3 --> > 120/(x+y) + 6/(x-y) = 15 ---------(3)
(2) * 2 --> > 50/(x+y) - 6/(x-y) = 2 ---------(4)

adding eq(3) + eq(4) ,,,we get ----
120 / (x+y) + 50 / (x+y) = 15+2
=> 170 / (x+y) = 17
=> x+y =10 ---------(5)

putting value of (x+y) in eq (1) ,,,we get
4+2 / (x-y) = 5
=>2 / (x-y) = 1
=> x - y = 2 -------------(6)

adding eq (5) and (6) ,,we will get
x + y + x - y = 10+2
=> 2x = 12
=> x = 6
by putting value of x in eq(6) ,,,,we will get
6 - y = 2
=> y = 6 - 2 = 4

so option is (7)............

Like? Yes (1) | No   

Akshat Sood   8 years AGO

Both the equations equate to (x+y)(x-y)
so we can find value of x which comes out to be 3/2y
put it in eq and we get x=6 and y=4

Like? Yes (1) | No   

sachin sawant   8 years AGO

Let x+y=a &x-y=b
40/a+2/b=5 ---(1)
25/a-3/b=1 ---(2)
Solving we geta=a=10& b= =2
Putting value of a & b in x+y & x-y we get
x=6& y=4
Option7

Like? Yes (1) | No   

Elamaruthu   8 years AGO

[40/(6+4)]+[2/(6-4)]=5
[25/(6+4)]-[3/(6-4)]=1

Like? Yes (1) | No   

S.MANJULA   8 years AGO

40(x-y)+2(x+y)=5((x^2)-(y^2)) --------(1)
25(x-y)-3(x+y)=((x^2)-(y^2)) ---------------(2)
(1)*3==> 120(x-y)+6(x+y)=15((x^2)-(y^2))
(2)*2==> 50(x-y)-6(x+y)=2((x^2)-(y^2))
(1)+(2)==> x+y=10------(3)
similarly (1)-(2)==> x-y=2 -----(4)
solving for (3) and (4) we get,
x=6 and y=4

Like? Yes (1) | No   

mcamanjeet   8 years AGO

let 1/x+y=p,1/x-y=q and solve equation
you will get p=1/10,q=1/2
now x+y=10,x-y=2
hence final solution is x=6,y=4

Like? Yes (1) | No   

CHANDAN KUMAR   8 years AGO

by solving the equations
we get x=6
and y=4.

Like? Yes (1) | No   

shubhrika   8 years AGO

Let 1/(x+y)=w and 1/(x-y)=z , 40w+2z=5;25w-3z=1
w=1/10 z=1/2
x+y=10
x-y=2
x=6 y=4

Like? Yes (1) | No   

gunjan   8 years AGO

If we put x =6 and y =4 in both equation,it satisfy the condition.

Like? Yes (1) | No   

Saleel Kulkarni   8 years AGO

Method 1:
Solve the two equations simultaneously and yield answer.
Method 2:
Trial and error method:
Each time it is not feasible to do complete calculations in short span of time.
For the first equation [40/(x/y)]+[2/(x-y)]=5
we can predict that 4+1=5
so to get 4 the x+y must be equal to 10
So the options are now 7 and 10
but for x-y we require 2 so x must be 6 and y must be 4
Apply the x and y values for equation 2 for cross checking
Hence Option 7

Like? Yes (1) | No (1)  

Jamil Ahmed   8 years AGO

[40/(x+y)] + [2/(x-y)] = 5
or, [120/(x+y)] + [6/(x-y)] = 15 -------- (eqn. 1)

[25/(x+y)] - [3/(x-y)] = 1
or, [50/(x+y)] - [6/(x-y)] = 2 ------- (eqn.2)

Adding eqns. 1 & 2, we get,
[120/(x+y)] + [50/(x+y)] = 15+2
or, x+y = 10 ---------------- eqn.(a)

Again, [40/(x+y)] + [2/(x-y)] = 5
or, [200/(x+y)] + [10/(x-y)] = 25 ------------- eqn.3

[25/(x+y)] - [3/(x-y)] = 1
or, [200/(x+y)] - [24/(x-y)] = 8 -------------- eqn.4

Subtracting eqn.4 from eqn.3, we get,
[10/(x-y)] + [24/(x-y)] = 25-8
or, x-y = 2 -------------- eqn. (b)

From eqns. (a) & (b), x = 6 & y = 4

Like? Yes | No (1)  

Nikunj Agarwal   8 years AGO

if x=6 & y=4 then, (40/(x+y)=4) & 2/(x-y)= 1, which gives rhs values, & satisfies both equation

Like? Yes | No (1)  

VEMULA HIMAVARSHA   8 years AGO

(40/x+y)+(2/(x-y))=5=>40x-40y+2x+2y=>42x-38y=5x^2-5y^2---->eq1
(25/x+y)-(3/x-y)=1=>25x-25y-3x-3y=>22x-28y=x^2-y^2-------->eq2
multiply eq2 to with 5
110x-140y=5x^2-5y^2---->eq3
subtract eq1 from eq3then
-68x+102y=0=>68x=102y
divided by 34 on both sides
2x=3y=>by substituting x=6 and y=4 we will get equal values i.e
2*6=3*4=>12=12
so option 7

Like? Yes | No (1)  

Mathi Vanan P   8 years AGO

assume X-Y=A and X+Y=B
(40/B)+(2/A)=5
assume the above eqn 1
(25/B)-(3/A)=1
assume the above eqn 2
multiply 3 with eqn 1
eqn 1 becomes,
(120/B)+(6/A)=15
multiply 2 with eqn 2
eqn 2 becomes
(50/B)-(6/A)=2
add the above two eqns
(170/B)=17
B=10
put B=10 in eqn 1
4+(2/A)=5
2/A=1
A=2
X-Y=2
X+y=10
the answer is X=6,Y=4

Like? Yes (1) | No (2)  

Jebas Gnana Prakash E   8 years AGO

put x=6 and y=4

in eqn 1weget
40/10 + 2/2
=4+1
=5

in eqn 2

(25/10) - (3/2)
=5/2 -3/2
5-3=2
2=2

Like? Yes | No (1)  

sanjeet kumar   8 years AGO

Solving these two equation, we get
2x^2 - 10x =2y^2 - 5y.
By trail we find x=6 and y=4

Like? Yes | No (1)  

Amit Kumar   8 years AGO

to get the ans 5 in 1st we need to look that 40 is getting divided by such a number that sum of x and y must be multiple of 5 or 10 or 8.this left us with only two option forhit n trial

Like? Yes | No (1)  

pooja   8 years AGO

multiply the both equation by (x^2-y^2) and eliminate x+y in both equations we get x+y=10 similarly for x-y hence we get x=6 and x=4

Like? Yes | No (2)