Question

Solve the following system of equation:
\[\dfrac{3}{{x + y}} + \dfrac{2}{{x - y}} = 2\], \[\dfrac{9}{{x + y}} - \dfrac{4}{{x - y}} = 1\]

Answer 1

Hint: In these types of questions substitute \[\dfrac{1}{{x + y}} = u\] and \[\dfrac{1}{{x - y}} = v\] then use it to find the values of u and v then use the values of u and v to find the values of x and y.

Complete step-by-step answer:
Let \[\dfrac{1}{{x + y}} = u\] and \[\dfrac{1}{{x - y}} = v\]
So the equations are
\[3u + 2v = 2\] and \[9u - 4v = 1\]
Let \[3u + 2v = 2\] equation 1 and \[9u - 4v = 1\] equation 2
Multiplying equation 1 by 2 and equation 2 by 1
$ \Rightarrow $\[2 \times 3u + 2 \times 2v = 2 \times 2\] and \[1 \times 9u - 1 \times 4v = 1 \times 1\]
$ \Rightarrow $\[6u + 4v = 4\] and \[9u - 4v = 1\]
Adding both the equations with each other
$ \Rightarrow $\[6u + 4v + 9u - 4v = 4 + 1\]
$ \Rightarrow $\[u = \dfrac{1}{3}\]
Putting value of u in equation 1
$ \Rightarrow $\[3 \times \dfrac{1}{3} + 2v = 2\]
$ \Rightarrow $\[v = \dfrac{1}{2}\]
Now using the value of u for finding the value of x
\[\dfrac{1}{{x + y}} = \dfrac{1}{3}\]
$ \Rightarrow $\[x + y = 3\] (Equation 3)
Now using the value of v for finding the value of y
\[\dfrac{1}{{x - y}} = \dfrac{1}{2}\]
$ \Rightarrow $\[x - y = 2\] (Equation 4)
Adding equation 3 and 4
$ \Rightarrow $\[x + y + x - y = 3 + 2\]
$ \Rightarrow $\[x = \dfrac{5}{2}\]
Putting the value of x in equation 4
$ \Rightarrow $\[\dfrac{5}{2} - y = 2\]
$ \Rightarrow $\[y = \dfrac{1}{2}\]
Hence the value of x and y are $\dfrac{5}{2},\dfrac{1}{2}$

Note: The equation is in complex form we can simplify the equation by substituting \[\dfrac{1}{{x + y}} = u\] and \[\dfrac{1}{{x - y}} = v\] in the equation and then we can compare then we can apply any operation to find the value of u and v since we got the values of variables which we needed to find the value of x and y.