Question

Solve the following systems of equations by the method of cross-multiplication:
$\frac{ax}{b}-\frac{by}{a}=a+b$
$ax-by=2ab$

Answer 1

Given equations can be written as

$\frac{ax}{b}-\frac{by}{a}=a+b\Rightarrow a^{2}x-b^{2}y-ab(a+b)=0$ is in the form of $a_{1}x+b_{1}y+c_1$

$ax-by=2ab\Rightarrow ax-by-2ab=0$ is in the form of $a_{2}x+b_{2}y+c_2$

$\Rightarrow a_1=a^2,b_1=-b^2,c_1=-ab(a+b)$

$\Rightarrow a_2=a,b_2=-b,c_2=-2ab$

Using cross-multiplication method,

$\frac{x}{\left(b_1\right)\left(c_2\right)-\left(b_2\right)\left(c_1\right)}=\frac{y}{\left(c_1\right)\left(a_2\right)-\left(c_2\right)\left(a_1\right)}=\frac{1}{\left(a_1\right)\left(b_2\right)-\left(a_2\right)(-b_1)}$

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

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

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

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

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

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

$\therefore x=bandy=-a$