Solve each of the following systems of equations by the method of cross-multiplication :
$a^{2} x+b^{2} y=c^{2}$
$b^{2} x+a^{2} y=d^{2}$
GIVEN:
$a^{2} x+b^{2} y=c^{2}$
$b^{2} x+a^{2} y=d^{2}$
To find: The solution of the systems of equation by the method of cross-multiplication:
Here we have the pair of simultaneous equation
$a^{2} x+b^{2} y-c^{2}=0$
$b^{2} x+a^{2} y-d^{2}=0$
By cross multiplication method we get
$\frac{x}{\left(-d^{2} b^{2}\right)-\left(-c^{2} a^{2}\right)}=\frac{-y}{\left(-d^{2} a^{2}\right)-\left(-c^{2} b^{2}\right)}=\frac{1}{a^{4}-b^{4}}$
$\frac{x}{\left(c^{2} a^{2}-d^{2} b^{2}\right)}=\frac{y}{\left(d^{2} a^{2}-c^{2} b^{2}\right)}=\frac{1}{a^{4}-b^{4}}$
Consider the following for x
$\frac{x}{\left(c^{2} a^{2}-d^{2} b^{2}\right)}=\frac{1}{a^{4}-b^{4}}$
$x=\frac{a^{2} c^{2}-b^{2} d^{2}}{a^{4}-b^{4}}$
Now consider the following for y
$\frac{-y}{\left(-d^{2} a^{2}\right)-\left(-c^{2} b^{2}\right)}=\frac{1}{a^{4}-b^{4}}$
$\frac{y}{\left(d^{2} a^{2}-c^{2} b^{2}\right)}=\frac{1}{a^{4}-b^{4}}$
$y=\frac{a^{2} d^{2}-b^{2} c^{2}}{a^{4}-b^{4}}$
Hence we get the value of $x=\frac{a^{2} c^{2}-b^{2} d^{2}}{a^{4}-b^{4}}$ and $y=\frac{a^{2} d^{2}-b^{2} c^{2}}{a^{4}-b^{4}}$
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