Question

Solve the following equations: \[\dfrac{6}{{x + y}} = \dfrac{7}{{x - y}} + 3\] , \[\dfrac{1}{{2(x + y)}} = \dfrac{1}{{3(x - y)}}\]where \[x + y \ne 0\]

Answer 1

Hint: In these types of questions simplify the equation 1 i.e. \[\dfrac{6}{{x + y}} = \dfrac{7}{{x - y}} + 3\] and then use the equation 2 i.e. \[\dfrac{1}{{2(x + y)}} = \dfrac{1}{{3(x - y)}}\]to find out the value of \[x = 5y\] and use it to find the value of x and y.

Complete step-by-step answer:
Let take\[\dfrac{6}{{x + y}} = \dfrac{7}{{x - y}} + 3\] as equation 1 and \[\dfrac{1}{{2(x + y)}} = \dfrac{1}{{3(x - y)}}\]as equation 2
Simplifying equation 1
\[\dfrac{6}{{x + y}} - \dfrac{7}{{x - y}} = 3\]
\[ \Rightarrow \dfrac{{6(x - y) - 7(x + y)}}{{(x + y)(x - y)}} = 3\]
\[ \Rightarrow \]\[6(x - y) - 7(x + y) = 3(x + y)(x - y)\]
\[ \Rightarrow \]\[6x - 6y - 7x - 7y = 3({x^2} - {y^2})\]
\[ \Rightarrow \]\[3{x^2} - 3{y^2} + x + 13y = 0\] (Equation 3)
Now simplifying equation 2
\[\dfrac{1}{{2(x + y)}} = \dfrac{1}{{3(x - y)}}\]
\[ \Rightarrow \]\[3x - 3y = 2x + 2y\]
\[ \Rightarrow \]\[x = 5y\] (Equation 4)
Putting the value of x in equation 3
\[ \Rightarrow \]\[3{(5y)^2} - 3{y^2} + 5y + 13y = 0\]
\[ \Rightarrow \]\[75{y^2} - 3{y^2} + 18y = 0\]
\[ \Rightarrow \]\[72{y^2} + 18y = 0\]
\[ \Rightarrow \]\[18y(4y + 1) = 0\]
So the value of y are $0,\dfrac{{ - 1}}{4}$
Putting value of y in equation 4
Values of x are 0, \[\dfrac{{ - 5}}{4}\]
Since x and y are in the denominator in the given equation so we can’t consider the values of x and y as 0 because the function will be not defined.
Hence the values of x and y are$\dfrac{{ - 1}}{4},\dfrac{{ - 5}}{4}$.

Note: In these types of questions first simplify the equation 1 \[\dfrac{6}{{x + y}} = \dfrac{7}{{x - y}} + 3\] and then simplify the equation 2 \[\dfrac{1}{{2(x + y)}} = \dfrac{1}{{3(x - y)}}\] and use the value of \[x = 5y\] from equation 2 and then put the value in equation 1 and find the value of y and then use the value of y to find out the value of x.