Question

Solve the following:
 \[\dfrac{{2xy}}{{x + y}} = \dfrac{3}{2}\]
 \[\dfrac{{xy}}{{2x - y}} = - \dfrac{3}{{10}}\]

Answer 1

Hint: We need to solve for \[x\] and \[y\] from the given two equations by solving them simultaneously. We will first do the reciprocal of both the equations and then split their denominators. After this, we will simplify and obtain the equations in terms of \[\dfrac{1}{x}\] and \[\dfrac{1}{y}\] . Then we will let \[\dfrac{1}{x}\] and \[\dfrac{1}{y}\] be equal to \[u\] and \[v\] respectively. Now, we will have the two equations in terms of \[u\] and \[v\] . We will then solve for \[u\] and \[v\] from the two equations and then equate them to \[\dfrac{1}{x}\] and \[\dfrac{1}{y}\] respectively to obtain the values of \[x\] and \[y\] .

Complete step by step answer:
We are given,
 \[\dfrac{{2xy}}{{x + y}} = \dfrac{3}{2} - - - - - (1)\]
 \[\dfrac{{xy}}{{2x - y}} = - \dfrac{3}{{10}} - - - - - (2)\]
Doing Reciprocal of both the equations i.e. changing the numerator and denominator on both left and right side of the equations.
After reciprocal equation (1) becomes \[\dfrac{{x + y}}{{2xy}} = \dfrac{2}{3}\]
Now, Multiplying with \[2\] on both the sides , we get \[2 \times \left( {\dfrac{{x + y}}{{2xy}}} \right) = 2 \times \dfrac{2}{3}\]
Which is equal to \[\dfrac{{x + y}}{{xy}} = \dfrac{4}{3}\]
Splitting the denominator, we get
 \[\dfrac{x}{{xy}} + \dfrac{y}{{xy}} = \dfrac{4}{3}\]
Now, cancelling the terms from numerator and denominator
 \[\dfrac{1}{y} + \dfrac{1}{x} = \dfrac{4}{3} - - - - - (3)\]
After reciprocal equation (2) becomes \[\dfrac{{2x - y}}{{xy}} = - \dfrac{{10}}{3}\]
Now splitting the denominator
 \[\dfrac{{2x}}{{xy}} - \dfrac{y}{{xy}} = - \dfrac{{10}}{3}\]
After cancelling the terms from numerator and denominator, we get
 \[\dfrac{2}{y} - \dfrac{1}{x} = - \dfrac{{10}}{3} - - - - - (4)\]
Now, we got two equations in terms of \[\dfrac{1}{x}\] and \[\dfrac{1}{y}\] .
Letting \[\dfrac{1}{x} = u\] and \[\dfrac{1}{y} = v\] , equation (3) and (4) becomes
 \[v + u = \dfrac{4}{3} - - - - - (5)\]
 and
 \[2v - u = - \dfrac{{10}}{3} - - - - - (6)\]
respectively.
Adding these two equations we get
 \[u + v + 2v - u = \dfrac{4}{3} + \left( { - \dfrac{{10}}{3}} \right)\]
 \[u + v + 2v - u = \dfrac{4}{3} - \dfrac{{10}}{3}\]
Now clubbing like terms on left side together and taking LCM on the right side, we get
 \[(u - u) + (v + 2v) = \dfrac{{4 - 10}}{3}\]
 \[0 + 3v = \dfrac{{ - 6}}{3}\]
Now cancelling terms on the right hand side.
 \[3v = - 2\]
Dividing both the sides by \[3\] , we get
 \[\dfrac{{3v}}{3} = \dfrac{{ - 2}}{3}\]
 \[v = \dfrac{{ - 2}}{3} - - - - - (7)\]
Now putting \[v = \dfrac{{ - 2}}{3}\] in equation (5), we will solve for \[u\] .
 \[ - \dfrac{2}{3} + u = \dfrac{4}{3}\]
Adding \[\dfrac{2}{3}\] both sides, we have
 \[ - \dfrac{2}{3} + u + \dfrac{2}{3} = \dfrac{4}{3} + \dfrac{2}{3}\]
Clubbing like terms on the left hand side and taking LCM on the right side.
 \[u + \left( {\dfrac{2}{3} - \dfrac{2}{3}} \right) = \dfrac{{4 + 2}}{3}\]
 \[u + 0 = \dfrac{6}{3}\]
Solving the right hand side and using \[a + 0 = a\] on the left hand side.
 \[u = 2 - - - - - (8)\]
We had \[\dfrac{1}{x} = u\] and \[\dfrac{1}{y} = v\] . We will now substitute the values of \[u\] and \[v\] to obtain the values of \[x\] and \[y\] .
From \[\dfrac{1}{x} = u\] and (8), we get
 \[\dfrac{1}{x} = 2\]
As \[\dfrac{a}{1} = a\] , we can write the above equation as
 \[\dfrac{1}{x} = \dfrac{2}{1}\]
By Cross Multiplying
 \[2x = 1 \times 1\]
 \[2x = 1\]
Dividing by \[2\] both the sides,
 \[\dfrac{{2x}}{2} = \dfrac{1}{2}\]
 \[x = \dfrac{1}{2}\]
Using \[\dfrac{1}{y} = v\] and (7), we get
 \[\dfrac{1}{y} = - \dfrac{2}{3}\]
By Cross Multiplying
 \[3 \times 1 = - 2y\]
 \[3 = - 2y\]
Dividing both the side by \[ - 2\] , we get
 \[\dfrac{3}{{ - 2}} = \dfrac{{ - 2y}}{{ - 2}}\]
Using \[\dfrac{a}{{ - b}} = - \dfrac{a}{b}\] on left hand side and cancelling terms on right hand side,
 \[ - \dfrac{3}{2} = y\]
Hence, we get
  \[x = \dfrac{1}{2}\] and \[y = - \dfrac{3}{2}\]

Note:
We normally forget to carry the negative sign forward in the next steps which will result in the wrong answer. Also, we need to take care that we have to solve for \[x\] and \[y\] and not for \[u\] and \[v\] . So, we have to substitute back the values of \[u\] and \[v\] obtained in order to find the values of \[x\] and \[y\] .