Question
Solve:$\frac{a}{x}-\frac{b}{y}=0, \frac{a b^2}{x}+\frac{a^2 b}{y}=a^2+b^2$
✓
Answer
Given equation are $\frac{a}{x}-\frac{b}{y}=0$ and $\frac{a b^2}{x}+\frac{a^2 b}{y}=a^2+b^2$
Taking $\frac{1}{x}=u$ and $\frac{1}{y}=v$, the above system of equations become
$a u-b v+0=0$
$ a b^2 u+a^2 b v-\left(a^2+b^2\right)=0$
By cross$-$multiplication, we have
$\frac{u}{-b \times\left[-\left(a^2+b^2\right)\right]-a^2 b \times 0}=\frac{-v}{a \times\left[-\left(a^2+b^2\right)\right]-a b^2 \times 0}=\frac{1}{a \times a^2 b-a b^2 \times(-b)}$
$ \Rightarrow \frac{u}{b\left(a^2+b^2\right)}=\frac{-v}{-a\left(a^2+b^2\right)}=\frac{1}{a^3 b+a b^3}$
$ \Rightarrow \frac{u}{b\left(a^2+b^2\right)}=\frac{v}{a\left(a^2+b^2\right)}=\frac{1}{a b\left(a^2+b^2\right)}$
$ \Rightarrow u=\frac{b\left(a^2+b^2\right)}{a b\left(a^2+b^2\right)}$ and $v=\frac{a\left(a^2+b^2\right)}{a b\left(a^2+b^2\right)}$
$ \Rightarrow \mathrm{u}=\frac{1}{a}$ and $\mathrm{v}=\frac{1}{b}$
$ \Rightarrow \frac{1}{x}=\frac{1}{a}$ and $\frac{1}{y}=\frac{1}{b}$
$ \Rightarrow \mathrm{x}=\mathrm{a}$ and $\mathrm{y}=\mathrm{b}$
Taking $\frac{1}{x}=u$ and $\frac{1}{y}=v$, the above system of equations become
$a u-b v+0=0$
$ a b^2 u+a^2 b v-\left(a^2+b^2\right)=0$
By cross$-$multiplication, we have
$\frac{u}{-b \times\left[-\left(a^2+b^2\right)\right]-a^2 b \times 0}=\frac{-v}{a \times\left[-\left(a^2+b^2\right)\right]-a b^2 \times 0}=\frac{1}{a \times a^2 b-a b^2 \times(-b)}$
$ \Rightarrow \frac{u}{b\left(a^2+b^2\right)}=\frac{-v}{-a\left(a^2+b^2\right)}=\frac{1}{a^3 b+a b^3}$
$ \Rightarrow \frac{u}{b\left(a^2+b^2\right)}=\frac{v}{a\left(a^2+b^2\right)}=\frac{1}{a b\left(a^2+b^2\right)}$
$ \Rightarrow u=\frac{b\left(a^2+b^2\right)}{a b\left(a^2+b^2\right)}$ and $v=\frac{a\left(a^2+b^2\right)}{a b\left(a^2+b^2\right)}$
$ \Rightarrow \mathrm{u}=\frac{1}{a}$ and $\mathrm{v}=\frac{1}{b}$
$ \Rightarrow \frac{1}{x}=\frac{1}{a}$ and $\frac{1}{y}=\frac{1}{b}$
$ \Rightarrow \mathrm{x}=\mathrm{a}$ and $\mathrm{y}=\mathrm{b}$
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