Question

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

$\frac{b}{a}x+\frac{a}{b}y=a^2+b^2$

$x+y=2ab$

Answer 1

Given: $\frac{b}{a}x+\frac{a}{b}y=a^2+b^2$...(i) and $x+y=2ab$...(ii)

Now, (i) can be transformed as,

$\Rightarrow b^{2}x+a^{2}y-ab(a^2+b^2)=0$...(iii) and

$\Rightarrow x+y-2ab=0$...(ii)

Now, using cross multiplication method we get,

$\Rightarrow \frac{x}{a^2\times (-2ab)-1\times [-ab(a^2+b^2]}=\frac{-y}{b^2\times (-2ab)-1\times [-ab(a^2+b^2]}=\frac{1}{b^2-a^2}$

$\Rightarrow \frac{x}{-2a^{3}b+a^{3}b+a{b}^3}=\frac{-y}{-2a{b}^3+a^{3}b+a{b}^3}=\frac{1}{b^2-a^2}$

$\Rightarrow \frac{x}{-a^{3}b+a{b}^3}=\frac{-y}{a^{3}b-a{b}^3}=\frac{1}{b^2-a^2}$

$\Rightarrow \frac{x}{ab(b^2-a^2)}=\frac{y}{ab(b^2-a^2)}=\frac{1}{b^2-a^2}$

$\Rightarrow \frac{x}{ab}=\frac{y}{ab}=\frac{1}{1}$

On comparing, we get,

$\Rightarrow x=ab\times \frac{1}{1}=ab$ and $y=ab\times \frac{1}{1}=ab$

Hence, $x=ab$ and $y=ab$