Question

Solve the following systems of equations by method of cross-multiplication
$
  \dfrac{5}{{x + y}} - \dfrac{2}{{x - y}} = - 1 \\
  \dfrac{{15}}{{x + y}} + \dfrac{7}{{x - y}} = 10 \\
  {\text{where x}} \ne {\text{0,y}} \ne {\text{0 and x}} \ne {\text{y}} \\
 $

Answer 1

Hint – In this question use the concept that if we have two general question of the form $ax + by + c = 0{\text{ and }}px + qy + r = 0$, then using the cross-multiplication rule the relative between x, y and the coefficients of the two equations can be written as $\dfrac{x}{{br - cq}} = \dfrac{y}{{cp - ar}} = \dfrac{1}{{aq - bp}}$. Before that put $\dfrac{1}{{x + y}} = a$ and $\dfrac{1}{{x - y}} = b$. This will help getting the values of x and y.

Complete Step-by-Step solution:
Let
$\dfrac{1}{{x + y}} = a$......................... (1)
And
$\dfrac{1}{{x - y}} = b$........................... (2)
So the given system of equations becomes,
$ \Rightarrow 5a - 2b = - 1$....................... (3)
And
$ \Rightarrow 15a + 7b = 10$................................ (4)
Now let us consider two equations
$ax + by + c = 0$
And
$px + qy + r = 0$
So according to cross-multiplication rule the solution of these equation is
$\dfrac{x}{{br - cq}} = \dfrac{y}{{cp - ar}} = \dfrac{1}{{aq - bp}}$
$ \Rightarrow x = \dfrac{{br - cq}}{{aq - bp}},y = \dfrac{{cp - ar}}{{aq - bp}}$
So this is the rule of cross-multiplication.
Now convert equation (3) and (4) in the standard form we have,
$ \Rightarrow 5a - 2b + 1 = 0$
And
$ \Rightarrow 15a + 7b - 10 = 0$
Now solve these system of equation using cross-multiplication rule which is described above,
  $ \Rightarrow \dfrac{a}{{ - 2\left( { - 10} \right) - \left( 1 \right)\left( 7 \right)}} = \dfrac{b}{{1\left( {15} \right) - 5\left( { - 10} \right)}} = \dfrac{1}{{5\left( 7 \right) - \left( { - 2} \right)\left( {15} \right)}}$
Now simplify the above equation we have,
$ \Rightarrow \dfrac{a}{{20 - 7}} = \dfrac{b}{{15 + 50}} = \dfrac{1}{{35 + 30}}$
$ \Rightarrow \dfrac{a}{{13}} = \dfrac{b}{{65}} = \dfrac{1}{{65}}$
$ \Rightarrow a = \dfrac{{13}}{{65}} = \dfrac{1}{5}$
And
$b = \dfrac{{65}}{{65}} = 1$
Now substitute the value of (a) and (b) in equation (1) and (2) we have,
$ \Rightarrow \dfrac{1}{{x + y}} = \dfrac{1}{5}{\text{ and }}\dfrac{1}{{x - y}} = 1$
$ \Rightarrow x + y = 5$, $ \Rightarrow x + y - 5 = 0$
And
$ \Rightarrow x - y = 1$, $x - y - 1 = 0$
Now again apply cross-multiplication rule we have,
$ \Rightarrow \dfrac{x}{{1\left( { - 1} \right) - \left( { - 5} \right)\left( { - 1} \right)}} = \dfrac{y}{{ - 5\left( 1 \right) - 1\left( { - 1} \right)}} = \dfrac{1}{{1\left( { - 1} \right) - \left( 1 \right)\left( 1 \right)}}$
Now simplify the above equation we have,
$ \Rightarrow \dfrac{x}{{ - 1 - 5}} = \dfrac{y}{{ - 5 + 1}} = \dfrac{1}{{ - 1 - 1}}$
$ \Rightarrow \dfrac{x}{{ - 6}} = \dfrac{y}{{ - 4}} = \dfrac{1}{{ - 2}}$
$ \Rightarrow x = \dfrac{{ - 6}}{{ - 2}} = 3$
And
$ \Rightarrow y = \dfrac{{ - 4}}{{ - 2}} = 2$

So the solution of the given system of equations is
$ \Rightarrow \left( {x,y} \right) = \left( {3,2} \right)$
So this is the required answer.

Note – In this question it is being specifically asked to use cross-multiplication, if this would not have been the case then we could have used two other methods to solve, first one is substitution. In this one variable is evaluated in terms of another variable and then substituted into another equation. The other method of elimination in this one variable of both the equations is eliminated by making their coefficients same in both the equations and then performing basic operations of addition or subtraction.