Question

Solve the following systems of equations:
$
  \dfrac{{7x - 2}}{{xy}} = 5 \\
  \dfrac{{8x + 7y}}{{xy}} = 15 \\
 $

Answer 1

Hint – In this question take xy of the denominator to the right hand side, compute the value of xy from one of the equations and substitute it back into another, use the relationships formed between x and y to get their values.

Complete Step-by-Step solution:
Given system of equations is written as
$ \Rightarrow \left( {7x - 2} \right) = 5xy$................. (1)
$ \Rightarrow 8x + 7y = 15xy$................... (2)
Now from equation (1) we have,
$ \Rightarrow xy = \dfrac{{7x - 2}}{5}$................... (3)
Now from equation (2) we have,
 $ \Rightarrow 8x + 7y = 15 \times \left( {\dfrac{{7x - 2}}{5}} \right)$
Now simplify the above equation we have,
$ \Rightarrow 8x + 7y = 21x - 6$
$ \Rightarrow 13x - 7y = 6$....................... (4)
Now from equation (3) calculate the value of y in terms of x we have,
$ \Rightarrow y = \dfrac{{7x - 2}}{{5x}}$..................... (5)
Now substitute this value in equation (4) we have,
\[ \Rightarrow 13x - 7\left( {\dfrac{{7x - 2}}{{5x}}} \right) = 6\]
Now simplify the above equation by multiplying (5x) throughout we have,
$ \Rightarrow 65{x^2} - 49x + 14 = 30x$
$ \Rightarrow 65{x^2} - 79x + 14 = 0$
Now factorize this equation we have,
$ \Rightarrow 65{x^2} - 65x - 14x + 14 = 0$
$ \Rightarrow 65x\left( {x - 1} \right) - 14\left( {x - 1} \right) = 0$
$ \Rightarrow \left( {x - 1} \right)\left( {65x - 14} \right) = 0$
$ \Rightarrow x = 1,\dfrac{{14}}{{65}}$
Now from equation (5) we have,
When x = 1
$ \Rightarrow y = \dfrac{{7\left( 1 \right) - 2}}{{5\left( 1 \right)}} = \dfrac{5}{5} = 1$
When x = (14/65)
$ \Rightarrow y = \dfrac{{7\left( {\dfrac{{14}}{{65}}} \right) - 2}}{{5\left( {\dfrac{{14}}{{65}}} \right)}} = \dfrac{{\dfrac{{98 - 130}}{{65}}}}{{\dfrac{{70}}{{65}}}} = \dfrac{{ - 32}}{{70}} = \dfrac{{ - 16}}{{35}}$
So the solutions of given system of equation are
$ \Rightarrow \left( {u,v} \right) = \left( {1,1} \right){\text{ and }}\left( {\dfrac{{14}}{{65}},\dfrac{{ - 16}}{{35}}} \right)$
So this is the required solution.
So this is the required answer.

Note – We could have used another method to factorize the formed quadratic equation in the middle of the solution. Dharacharya formula that is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$helps finding out the roots of any quadratic equation of the form $a{x^2} + bx + c = 0$.