Question

 Solve the following system of equations by the method of cross-multiplication. `\frac{a}{x}-\frac{b}{y}=0;\text{}\frac{ab^{2}}{x}+\frac{a^{2}b}{y}=a^{2}+b^{2};` Where x ≠ 0, y ≠ 0

Answer 1

The given system of equations is

`\frac{a}{x}-\frac{b}{y}=0 ………(1)`

`\frac{ab^{2}}{x}+\frac{a^{2}b}{y}-( a^{2}+b^{2})=0 ………(2)`

Putting `\frac { a }{ x }=u` and `\frac { b }{ y }=v` in equatinos (1) and (2) the system of equations reduces to

u – v + 0 = 0

`b^2 u + a^2 v – (a^2 + b^2 ) = 0`

By the method of cross-multiplication, we have

`\Rightarrow \frac{u}{a^{2}+b^{2}-a^{2}\times0}=\frac{-v}{-(a^{2}+b^{2})-b^{2}\times0}=\frac{1}{a^{2}-(-b^{2})}`

`\Rightarrow\frac{u}{a^{2}+b^{2}}=\frac{-v}{-(a^{2}+b^{2})}=\frac{1}{a^{2}+b^{2}}`

`\Rightarrow\frac{u}{a^{2}+b^{2}}=\frac{1}{a^{2}+b^{2}}\Rightarrow u=1`

`and\text{}\frac{-v}{-(a^{2}+b^{2})}=\frac{1}{a^{2}+b^{2}}\Rightarrowv=1and\text{ u}=\frac{a}{x}=1\Rightarrow x=a`

`v=\frac{b}{y}=1\Rightarrow y=b`

Hence, the solution of the given system of equations is x = a, y = b.