**Question:**

Solve each of the following systems of equations by the method of cross-multiplication :

$a^{2} x+b^{2} y=c^{2}$

$b^{2} x+a^{2} y=d^{2}$

**Solution:**

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 $y=\frac{a^{2} d^{2}-b^{2} c^{2}}{a^{4}-b^{4}}$