Question

Solve the following systems of equations: $\dfrac{1}{{5x}} + \dfrac{1}{{6y}} = 12$, \[\dfrac{1}{{3x}} - \dfrac{3}{{7y}} = 8\], \[x \ne 0,y \ne 0\]

Answer 1

Hint: In the given question multiply the equations by 1/3 and 1/5 and simplify them to find the values of x and y.

Complete step-by-step answer:
Let $\dfrac{1}{{5x}} + \dfrac{1}{{6y}} = 12$ be the equation 1 and \[\dfrac{1}{{3x}} - \dfrac{3}{{7y}} = 8\] be the equation 2
Multiplying equation 1 by 1/3 and equation 2 by 1/5
So we get\[\dfrac{1}{3} \times (\dfrac{1}{{5x}} + \dfrac{1}{{6y}}) = \dfrac{1}{3} \times (12)\], \[\dfrac{1}{5} \times (\dfrac{1}{{3x}} - \dfrac{3}{{7y}}) = \dfrac{1}{5} \times (8)\]
\[ \Rightarrow \]\[\dfrac{1}{{15x}} + \dfrac{1}{{18y}} = 4\], \[\dfrac{1}{{15x}} - \dfrac{3}{{35y}} = \dfrac{8}{5}\]now subtracting both the equation
We get \[\dfrac{1}{{15x}} + \dfrac{1}{{18y}} - \dfrac{1}{{15x}} + \dfrac{3}{{35y}} = 4 - \dfrac{8}{5}\]
\[ \Rightarrow \]\[\dfrac{{89}}{{630y}} = \dfrac{{12}}{5}\]
\[ \Rightarrow \]\[y = \dfrac{{89}}{{1512}}\]
Now using the value of y to find the value of x
Putting the value of y in equation 2
\[ \Rightarrow \]\[\dfrac{1}{{3x}} - \dfrac{{3 \times 1512}}{{7 \times 89}} = 8\]
\[ \Rightarrow \]\[x = \dfrac{{89}}{{4080}}\]
Therefore the value of x and y are $\dfrac{{89}}{{4080}},\dfrac{{89}}{{1512}}$

Note: In these types of questions first let $\dfrac{1}{{5x}} + \dfrac{1}{{6y}} = 12$ be the equation 1 and \[\dfrac{1}{{3x}} - \dfrac{3}{{7y}} = 8\] be the equation 2 then multiply by some constant value to make coefficient equal then subtract both the equation and find the value of y then with the help of the value of y find the value of x.