please tell me the answer
Answers
Step-by-step explanation:
Solution :-
i)
Given that
(2/x)+(3/y) = 13--------(1)
(5/x)-(4/y) = -2 -------(2)
Put 1/x = a and 1/y = b then
2a + 3b = 13 ----------(3)
On multiplying with 5 then
10a +15 b = 65 -------(4)
and
5a -4b= -2 ----------(5)
On multiplying with 2 then
10 a - 8b = -4 -------(6)
On Subtracting (6) from (4) then
10a + 15b = 65
10a - 8b = -4
(-)
_____________
0 + 23 b = 69
______________
=> 23b = 69
=> b = 69/23
=> b =3
On Substituting the value of b in (5)
5a -4b= -2
=> 5a -4(3) = -2
=> 5a -12 = -2
=> 5a = -2+12
=> 5a = 10
=> a = 10/5
=> a = 2
Now we have
a = 2
=>1/x = 2
=> x = 1/2
and
b = 3
=>1/y = 3
=> y = 1/3
Answer :-
The solution for the given problem is (1/2,1/3)
Check:-
If x = 1/2 and y = 1/3 then
LHS = (2/x)+(3/y)
= 2/(1/2)+3/(1/3)
= (2×2)+(3×3)
= 4+9
= 13
= RHS
LHS=RHS is true
and
LHS=(5/x)-(4/y)
=> 5/(1/2)- 4/(1/3)
=> (5×2)-(4×3)
=> 10-12
=> -2
=> RHS
LHS = RHS is true
ii)
Given that:
(1/2x)+(1/3y) = 2 --------(1)
(1/3x)+(1/2y) = 13/6-----(2)
Put 1/x = a and 1/y = b then
(1) becomes
(a/2 )+( b/3) = 2
=> (3a+2b)/6 = 2
=> 3a+2b = 2×6
=> 3a+2b = 12 --------(3)
(2) becomes
(a/3)+(b/2)=13/6
=> (2a+3b)/6 = 13/6
=> 2a+3b=13----------(4)
On adding (3)&(4) then
3a+2b=12
2a+3b=13
(+)
_________
5a+5b = 25
_________
=> 5a+5b = 25
=> 5(a+b)=25
=> a+b=25/5
=> a+b=5 -------------(5)
On subtracting (3) from (4)
2a+3b=13
3a+2b=12
(-)
_________
-a+b = 1
_________
=> -a+b = 1
=> b = 1+a --------(6)
On Substituting the value of b in (5) then
a+1+a = 5
=>2a+1 = 5
=> 2a = 5-1
=> 2a = 4
=> a = 4/2
=> a = 2
On Substituting the value of a in (6) then
b = 1+2
b = 3
Now,
a = 2
=> 1/x=2
=> x = 1/2
and
b= 3
=> 1/y = 3
=>y = 1/3
Therefore, x= 1/2 and y= 1/3
Answer:-
The solution for the given problem is (1/2,1/3)
Check:-
If x = 1/2 and y = 1/3 then
LHS = (1/2x)+(1/3y)
= 1/2(1/2)+1/3(1/3)
= 1/(2/2)+1/(3/3)
= (1/1)+(1/1)
= 1+1
=2
= RHS
LHS=RHS is true
and
LHS=(1/3x)+(1/2y)
=> 1/3(1/2)+ 1/2(1/3)
=> 1/(3/2)+1/(2/3)
=> (2/3)+(3/2)
=> (4+9)/6
=> 13/6
=> RHS
LHS = RHS is true
Used Method:-
- Reducible to linear equations method
- Substitution method