Solve the following simultaneous equations. 1/(3x + y) + 1/(3x - y) = 3/4;

Question

Solve the following simultaneous equations.

\(\frac{1}{3x + y}+\frac{1}{3x - y} =\frac{3}{4};\)

1/(3x + y) + 1/(3x - y) = 3/4;

\(\frac{1}{2(3x + y)}+\frac{1}{2(3x - y)} =-\frac{1}{8};\)

1/2(3x + y) + 1/2(3x - y) = -1/8:

Answer 1

The given simultaneous equations are

1/(3x + y) + 1/(3x - y) = 3/4;

\(\therefore\) Equations (i) and (ii) become

Multiplying equation (iv) by 2, we get

p - q = \(-\frac{1}{4} \) ......(v)

Adding equations (iii) and (v), we get

Substituting p = 1/4 in equation (iii), we get

Resubstituting the values of p and q, we get

Adding equations (vi) and (vii), we get

Substituting x = 1 in equation (vi), we get

3(1) + y = 4 

∴ 3 + y = 4 

∴ y = 4 – 3 = 1 

∴ (x, y) = (1, 1) is the solution of the given simultaneous equations.