Question

Solve the following systems of equations by method of cross-multiplication
$
  \dfrac{x}{a} + \dfrac{y}{b} = a + b \\
  \dfrac{x}{{{a^2}}} + \dfrac{y}{{{b^2}}} = 2 \\
$

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 $\dfrac{x}{{br - cq}} = \dfrac{y}{{cp - ar}} = \dfrac{1}{{aq - bp}}$. This will help getting the values of x and y.

Complete Step-by-Step solution:
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 given system of equation are
$\dfrac{x}{a} + \dfrac{y}{b} = a + b$, $ \Rightarrow \dfrac{{bx + ay}}{{ab}} = a + b$, $ \Rightarrow bx + ay = {a^2}b + a{b^2}$
And
$\dfrac{x}{{{a^2}}} + \dfrac{y}{{{b^2}}} = 2$, $ \Rightarrow \dfrac{{{b^2}x + {a^2}y}}{{{a^2}{b^2}}} = 2$, $ \Rightarrow {b^2}x + {a^2}y = 2{a^2}{b^2}$
Now convert these system of equation in standard form we have,
$ \Rightarrow bx + ay - \left( {{a^2}b + a{b^2}} \right) = 0$
And
$ \Rightarrow {b^2}x + {a^2}y - 2{a^2}{b^2} = 0$
Now solve these system of equation using cross-multiplication rule which is described above,
  $ \Rightarrow \dfrac{x}{{a\left( { - 2{a^2}{b^2}} \right) - \left( { - \left( {{a^2}b + a{b^2}} \right)} \right)\left( {{a^2}} \right)}} = \dfrac{y}{{ - \left( {{a^2}b + a{b^2}} \right)\left( {{b^2}} \right) - b\left( { - 2{a^2}{b^2}} \right)}} = \dfrac{1}{{b\left( {{a^2}} \right) - \left( a \right)\left( {{b^2}} \right)}}$
Now simplify the above equation we have,
$ \Rightarrow \dfrac{x}{{ - 2{a^3}{b^2} + {a^4}b + {a^3}{b^2}}} = \dfrac{y}{{ - {a^2}{b^3} - a{b^4} + 2{a^2}{b^3}}} = \dfrac{1}{{b{a^2} - a{b^2}}}$
$ \Rightarrow x = \dfrac{{ - 2{a^3}{b^2} + {a^4}b + {a^3}{b^2}}}{{b{a^2} - a{b^2}}}$
\[ \Rightarrow x = \dfrac{{{a^4}b - {a^3}{b^2}}}{{b{a^2} - a{b^2}}} = \dfrac{{{a^2}\left( {b{a^2} - a{b^2}} \right)}}{{b{a^2} - a{b^2}}} = {a^2}\]
And
$ \Rightarrow y = \dfrac{{ - {a^2}{b^3} - a{b^4} + 2{a^2}{b^3}}}{{\left( {b{a^2} - a{b^2}} \right)}}$
$ \Rightarrow y = \dfrac{{ - a{b^4} + {a^2}{b^3}}}{{\left( {b{a^2} - a{b^2}} \right)}} = \dfrac{{{b^2}\left( {b{a^2} - a{b^2}} \right)}}{{b{a^2} - a{b^2}}} = {b^2}$
So the required solution of the given system of equation is
$ \Rightarrow \left( {x,y} \right) = \left( {{a^2},{b^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.