Solve the following pairs of equations by reducing
them to pair of
linead equations:
1/3x+y+1/(3x-y)=3/4 1/2(3x+y)-1/2(3x-y)=-1/8
Answers
Step-by-step explanation:
Given:-
1/(3x+y)+1/(3x-y)=3/4 and 1/2(3x+y)-1/2(3x-y)=-1/8
To find:-
Solve the following pairs of equations by reducing them to pair of linead equations:
1/(3x+y)+1/(3x-y)=3/4 and 1/2(3x+y)-1/2(3x-y)=-1/8
Solution:-
Given that
1/(3x+y)+1/(3x-y)=3/4 ---------------(1)
1/2(3x+y)-1/2(3x-y)=-1/8--------------(2)
Let 1/(3x+y) = a and 1/(3x-y)=b then
The above equations become
a+b = 3/4
=>4(a+b)=3
=>4a+4b = 3 ----------------(3)
and 1/2(a)-1/2(b)=-1/8
=>1/2(a-b)=-1/8
=>(a-b)=-(1/8)×2
=>a-b = -2/8
=>a-b = -1/4
=>4(a-b)=-1
=>4a-4b = -1 ------------------(4)
On adding (3)&(4) then
4a + 4b = 3
4a - 4b = -1
(+)
___________
8a +0 = 2
___________
=>8 a = 2
=>a = 2/8
=>a = 1/4
The value of a = 1/4
=>1/(3x+y) = 1/4
=>3x + y = 4 -----------------------(5)
and Substituting the value of a in (3)
=>4(1/4)+4b = 3
=>(4/4)+4b = 3
=>1+4b = 3
=>4b = 3-1
=>4b = 2
=>b = 2/4
=>b = 1/2
=>1/(3x-y)=1/2
=>3x-y = 2 ----------------------(6)
On adding (5)&(6)
3x + y = 4
3x - y = 2
(+)
__________
6x + 0 = 6
___________
=>6x = 6
=>x = 6/6
=>x = 1
On Substituting the value of x in (5) then
3(1)+ y = 4
=>3 + y = 4
=>y = 4-3
=>y = 1
The value of x = 1 and y = 1
Answer:-
The solution for the given problem is
( 1 , 1 )
Check:-
If x = 1 and y = 1 then LHS of (1)
1/(3x+y)+1/(3x-y)
=>1/(3×1+1)+1/(3×1-1)
=>1/(3+1)+1(3-1)
=>(1/4)+(1/2)
=>(1+2)/4
=>3/4
RHS
LHS = RHS is true for x=1 and y=1
If x=1 and y=1 then LHS of (2)
1/2(3x+y)-1/2(3x-y)
=>1/2 (3×1+1) - 1/2 (3×1-1)
=> 1/2 (3+1) - 1/2 (3-1)
=>1/(2×4) -1/(2×2)
=>(1/8)-(1/4)
=>(1-2)/8
= -1/8
=>RHS
LHS = RHS is true for x = 1 and y =1
Verified the given relations