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
<u>Answer</u><u>:</u><u>-</u> Hence, the value of x and y with be 1/2 and 1/3 respectively.
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
<u>also</u><u> read</u><u> similar</u><u> questions</u><u>:</u> Simultaneous equations: 1. 2x+2y=10 X+2y=6 2. 3x+y=18 2x+y=13 3. X+y=1 X-y=5 4. 3x+4y=29 X-4y=-17 5. 4c+3y=11 2x+y=7
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