Question

Solve the following systems of equation-
$
  \dfrac{1}{{2x}} + \dfrac{1}{{3y}} = 2 \\
    \\
  \dfrac{1}{{3x}} + \dfrac{1}{{2y}} = \dfrac{{13}}{6} \\
$

Answer 1

Hint- The equations above in the question are not linear equations so we will first make them linear using the method shown in the solution below. Then we will use the method of substitution in order to solve the question further.

Complete step-by-step answer:
Let the equations given the question be equation 1 and equation 2-
$ \Rightarrow \dfrac{1}{{2x}} + \dfrac{1}{{3y}} = 2$ (equation 1)
$ \Rightarrow \dfrac{1}{{3x}} + \dfrac{1}{{2y}} = \dfrac{{13}}{6}$ (equation 2)
Now, as we can see that the equations given to us are not linear equations because y is in the denominator in both the equations. So, in order to make the linear, we will-
Let, $\dfrac{1}{x} = v,\dfrac{1}{y} = w$
Now, substituting the value of v and w into equation 1 and equation 2-
$ \Rightarrow \dfrac{v}{2} + \dfrac{w}{3} = 2$
$ \Rightarrow \dfrac{v}{3} + \dfrac{w}{2} = \dfrac{{13}}{6}$
Let the above equation be equation 3 and equation 4-
 $ \Rightarrow \dfrac{v}{2} + \dfrac{w}{3} = 2$ (equation 3)
 $ \Rightarrow \dfrac{v}{3} + \dfrac{w}{2} = \dfrac{{13}}{6}$ (equation 4)
Now, we notice that the coefficient of v in equation 3 is equal to the coefficient of w in equation 4 and that the coefficient of w in equation 3 is equal to the coefficient of v in equation 4. In such cases, we add and subtract the equations respectively.
Adding equation 3 and equation 4, we get-
$
   \Rightarrow \dfrac{v}{2} + \dfrac{w}{3} + \dfrac{v}{3} + \dfrac{w}{2} = 2 + \dfrac{{13}}{6} \\
    \\
   \Rightarrow \left( {\dfrac{1}{2} + \dfrac{1}{3}} \right)v + \left( {\dfrac{1}{2} + \dfrac{1}{3}} \right)w = \dfrac{{12 + 13}}{6} \\
    \\
   \Rightarrow \left( {\dfrac{1}{2} + \dfrac{1}{3}} \right)\left( {v + w} \right) = \dfrac{{12 + 13}}{6} \\
    \\
   \Rightarrow \dfrac{5}{6}\left( {v + w} \right) = \dfrac{{25}}{6} \\
    \\
   \Rightarrow v + w = 5 \\
$
Now, subtracting the equation 3 and equation 4, we get-
$
   \Rightarrow \dfrac{v}{2} + \dfrac{w}{3} - \dfrac{v}{3} - \dfrac{w}{2} = 2 - \dfrac{{13}}{6} \\
    \\
   \Rightarrow \left( {\dfrac{1}{2} - \dfrac{1}{3}} \right)v + \left( {\dfrac{1}{3} - \dfrac{1}{2}} \right)w = 2 - \dfrac{{13}}{6} \\
    \\
   \Rightarrow \dfrac{1}{6}v - \dfrac{1}{6}w = - \dfrac{1}{6} \\
    \\
   \Rightarrow v - w = - 1 \\
$
Naming the above equations as equation 5 and equation 6-
$ \Rightarrow v + w = 5$ (equation 5)
$ \Rightarrow v - w = - 1$ (equation 6)
Now, as we can see, equation 5 and equation 6 are linear equations. So, we will add equation 5 and equation 6 in order to get the value of v-
$
   \Rightarrow v + w + v - w = 5 - 1 \\
    \\
   \Rightarrow 2v = 4 \\
    \\
   \Rightarrow v = 2 \\
$
Substituting this value of v into equation 5-
$
   \Rightarrow v + w = 5 \\
    \\
   \Rightarrow 2 + w = 5 \\
    \\
   \Rightarrow w = 3 \\
$
Now, as we already knew $\dfrac{1}{x} = v,\dfrac{1}{y} = w$, putting the value of v and w will get us the value of x and y-
$
   \Rightarrow \dfrac{1}{x} = 2 \\
    \\
   \Rightarrow x = \dfrac{1}{2} \\
$
And,
$
   \Rightarrow \dfrac{1}{y} = 3 \\
    \\
   \Rightarrow y = \dfrac{1}{3} \\
$
Thus, the values of x and y are $x = \dfrac{1}{2},y = \dfrac{1}{3}$.

Note: Whenever we notice that coefficient of v in equation 3 is equal to the coefficient of w in equation 4 and that the coefficient of w in equation 3 is equal to the coefficient of v in equation 4 in any given question, add and subtract the equations respectively. Use the method of substitution as well to get a final answer.