Question

Solve the pair of linear equations
${\displaystyle \frac{1}{3x+y}}+{\displaystyle \frac{1}{3x-y}}={\displaystyle \frac{3}{4}}$
${\displaystyle \frac{1}{2(3x+y)}}-{\displaystyle \frac{1}{2(3x-y)}}=-{\displaystyle \frac{1}{8}}$
$3x+y\ne 0$ and $3x-y\ne 0$

Answer 1

Hint: Assume $a={\displaystyle \frac{1}{3x+y}}$ and $b={\displaystyle \frac{1}{3x-y}}$. Hence form two linear equations in a and b. Solve for a and b. Now revert to original variables and hence form two linear equations in x and y. Solve for x and y. Remove extraneous roots, i.e. those roots at which either a or b is not defined.

Complete step-by-step answer:
Let $a={\displaystyle \frac{1}{3x+y}}$ and $b={\displaystyle \frac{1}{3x-y}}$.
Since ${\displaystyle \frac{1}{3x+y}}+{\displaystyle \frac{1}{3x-y}}={\displaystyle \frac{3}{4}}$, we have
$a+b={\displaystyle \frac{3}{4}}\text{ (i)}$
Also since ${\displaystyle \frac{1}{2(3x+y)}}-{\displaystyle \frac{1}{2(3x-y)}}=-{\displaystyle \frac{1}{8}}$ , we have
${\displaystyle \frac{a}{2}}+{\displaystyle \frac{b}{2}}={\displaystyle \frac{-1}{8}}$
Multiplying both sides by 2, we get
$\begin{array}{rl}& {\displaystyle \frac{2a}{2}}-{\displaystyle \frac{2b}{2}}={\displaystyle \frac{-2}{8}}\\ & \Rightarrow a-b={\displaystyle \frac{-1}{4}}\text{ (ii)}\end{array}$
Adding equation (i) and equation (ii), we get
$\begin{array}{rl}& a+b+a-b={\displaystyle \frac{3}{4}}-{\displaystyle \frac{1}{4}}\\ & \Rightarrow 2a={\displaystyle \frac{1}{2}}\end{array}$
Dividing both sides by 2, we get
$a={\displaystyle \frac{1}{2\times 2}}={\displaystyle \frac{1}{4}}$
Substituting the value of “a” in equation (i), we get
$\begin{array}{rl}& a+b={\displaystyle \frac{3}{4}}\\ & \Rightarrow {\displaystyle \frac{1}{4}}+b={\displaystyle \frac{3}{4}}\\ & \Rightarrow b={\displaystyle \frac{3}{4}}-{\displaystyle \frac{1}{4}}={\displaystyle \frac{2}{4}}={\displaystyle \frac{1}{2}}\end{array}$
Hence, we have $a={\displaystyle \frac{1}{4}}$ and $b={\displaystyle \frac{1}{2}}$
Reverting to original variables, we get
$\begin{array}{rl}& {\displaystyle \frac{1}{3x+y}}={\displaystyle \frac{1}{4}}\\ & \end{array}$
Taking reciprocals on both sides, we get
$3x+y=4\text{ (iii)}$
and
${\displaystyle \frac{1}{3x-y}}={\displaystyle \frac{1}{2}}$
Taking reciprocals on both sides, we get
$3x-y=2\text{ (iv)}$
Adding equation (iii) and equation (iv), we get
$\begin{array}{rl}& 3x+y+3x-y=2+4\\ & \Rightarrow 6x=6\\ & \Rightarrow x=1\end{array}$
Substituting the value of x in equation (iii), we get
$\begin{array}{rl}& 3x+y=4\\ & \Rightarrow 3+y=4\\ & \Rightarrow y=4-3=1\end{array}$
Hence we have x=1 and y=1.
At x=1 ,y =1, 3x+y = 4 and 3x-y =2
Since $3x+y\ne 0$ and $3x-y\ne 0$, we have
(x,y) = (1,1) is the solution of the given system of equations.
Note: Visualising graphically.

If we plot the curves ${\displaystyle \frac{1}{3x+y}}+{\displaystyle \frac{1}{3x-y}}={\displaystyle \frac{3}{4}}$ and ${\displaystyle \frac{1}{2(3x+y)}}-{\displaystyle \frac{1}{2(3x-y)}}=-{\displaystyle \frac{1}{8}}$ on the same graph paper, then these curves will intersect at the point (1,1).

In the above graph, the blue curve is of ${\displaystyle \frac{1}{3x+y}}+{\displaystyle \frac{1}{3x-y}}={\displaystyle \frac{3}{4}}$ equation and the pink curve of ${\displaystyle \frac{1}{2(3x+y)}}-{\displaystyle \frac{1}{2(3x-y)}}=-{\displaystyle \frac{1}{8}}$ equation. As can be seen from the graph, the two curves intersect at (1,1).