# Solve the following system of equations by the method of cross-multiplication. a/x−b/y=0;(ab^2)/x+(a^2b)/y=a^2+b^2; Where x ≠ 0, y ≠ 0 - Mathematics

Sum

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

#### Solution

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.

Is there an error in this question or solution?

Share