Question

1/3x+y+1/3x-y=3/4; 1/2(3x+y)-1/2(3x-y)=-1/8 Solve the simultaneous equation

Answer 1

1/(3x + y) + 1/(3x - y) = 3/4
1/2(3x + y) - 1/2(3x - y) = -1/8
put (3x + y) = P and (3x - y) = Q
then, equations convert into
1/P + 1/Q = 3/4 -----(1)
1/2P - 1/2Q = -1/8-----(2)

multiply 2 with equation (2) and then add with equation (1),
1/P + 1/Q + 2/2P - 2/2Q = 3/4 - 2/8
= > 1/P + 1/Q + 1/P - 1/Q = 3/4 - 1/4
=> 2/P = 2/4
=> 2/P = 1/2 , P = 4 put it in equation (1),
1/4 + 1/Q = 3/4
=> 1/Q = 3/4 - 1/4 = 1/2
so, Q = 2

now, P = 3x + y = 4 ----(3)
Q = 3x - y = 2 --------(4)
adding equations (3) and (4),
3x + y + 3x - y = 4 + 2
6x = 6 => x = 1
y = 1

hence, x = 1 and y = 1

Answer 2

Solution:-

given by :-
$\frac{1}{3x + y} + \frac{1}{3x - y} = \frac{3}{4} \\ 4( \frac{1}{3x + y} ) + 4( \frac{1}{3x - y} ) = 3 \\ assume \: \: ( \frac{1}{3x + y} ) \: as \: m \: and \: ( \frac{1}{3x - y} ) \: as \: n \: we \: get \: \\ 4m + 4n = 3..........(1) \\ \frac{1}{2(3x + y)} - \frac{1}{2(3x - y)} = \frac{ - 1}{8} \\ 4( \frac{1}{3x + y} ) - 4( \frac{1}{3x - y} ) = - 1 \\ 4m - 4n = - 1...........(2) \\ adding \: eq(1) \: and \: (2) \: we \: have \\ 4m + 4n = 3..........(1) \\ 4m - 4n = - 1........(2) \\ \\ 8m = 2 \\ m = \frac{1}{4} \\ put \: m = \frac{1}{4} in \: eq(1). \\ 4 \times \frac{1}{4} + 4m = 3 \\ 4m = 2 \\ m = \frac{2}{4} = \frac{1}{2} \\ here \: \\ m = ( \frac{1}{3x + y} ) = \frac{1}{4} \\ 3x + y = 4........(3) \\ n = ( \frac{1}{3x - y} ) = \frac{1}{2} \\ 3x - y = 2......(4) \\ additing \: eq(3) \: and \: eq(4) \\ 3x + y = 4........(3) \\ 3x - y = 2.........(4) \\ \\ 6x = 6 \\ x = 1 \\ put \: x \: = 1 \: in \: eq(3). \\ 3 \times 1 + y = 4 \\ y = 4 - 3 \\ y = 1$
here (x , y) = ( 1 , 1)

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