Question
Solve: $\frac{\text{ax}}{\text{b}}-\frac{\text{by}}{\text{a}}=\text{a}+\text{b},$ $\text{ax}-\text{by}=\text{2ab}.$
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Answer
The given equation may be written as follows:
$\frac{\text{ax}}{\text{b}}-\frac{\text{by}}{\text{a}}-(\text{a+b})=0\ ...(\text{i})$
$ax - by - 2ab = 0 ...(ii)$
Here, $\text{a}_1=\frac{\text{a}}{\text{b}},\ \text{b}_1=\frac{\text{b}_1}{\text{a}},$ $\text{c}_1=-(\text{a+b}),\ \text{a}_2=\text{a},$ $\text{b}_2= -\text{b},\ \text{c}_2=-2\text{ab}$
By cross multiplying, we have:
$\therefore\frac{\text{x}}{\big(-\frac{\text{b}}{\text{a}}\big)\times(-2\text{ab})-(-\text{b})\times(-(\text{a+b}))}\\\ \ \ =\frac{\text{y}}{-(\text{a+b})\times\text{a}-(-2\text{ab})\times\frac{\text{a}}{\text{b}}}\\\ \ \ =\frac{1}{\frac{\text{a}}{\text{b}}\times(-\text{b})-\text{a}\times\big(-\frac{\text{b}}{\text{a}}\big)}$
$\Rightarrow\frac{\text{x}}{2\text{b}^2-\text{b}(\text{a+b})}=\frac{\text{y}}{-(\text{a+b})+2\text{a}^2}=\frac{1}{-\text{a+b}}$
$\Rightarrow\frac{\text{x}}{2\text{b}^2-\text{ab}-\text{b}^2}=\frac{\text{y}}{-\text{a}^2-\text{ab}+2\text{a}^2}=\frac{1}{-\text{a+b}}$
$\Rightarrow\frac{\text{x}}{\text{b}^2-\text{ab}}=\frac{\text{y}}{\text{a}^2-\text{ab}}=\frac{1}{-(\text{a}-\text{b})}$
$\Rightarrow\frac{\text{x}}{-\text{b}(-\text{a}-\text{b})}=\frac{\text{y}}{\text{a}(\text{a}-\text{b})}=\frac{1}{-(\text{a}-\text{b})}$
$\Rightarrow\text{x}=\frac{-\text{b}(\text{a}-\text{b})}{-(\text{a}-\text{b})}=\text{b},\ \text{y}=\frac{\text{a}(\text{a}-\text{b})}{-(\text{a}-\text{b})}=-\text{a}$
Hence, $x = b$ and $y = -a$ is the required solution.
$\frac{\text{ax}}{\text{b}}-\frac{\text{by}}{\text{a}}-(\text{a+b})=0\ ...(\text{i})$
$ax - by - 2ab = 0 ...(ii)$
Here, $\text{a}_1=\frac{\text{a}}{\text{b}},\ \text{b}_1=\frac{\text{b}_1}{\text{a}},$ $\text{c}_1=-(\text{a+b}),\ \text{a}_2=\text{a},$ $\text{b}_2= -\text{b},\ \text{c}_2=-2\text{ab}$
By cross multiplying, we have:
$\therefore\frac{\text{x}}{\big(-\frac{\text{b}}{\text{a}}\big)\times(-2\text{ab})-(-\text{b})\times(-(\text{a+b}))}\\\ \ \ =\frac{\text{y}}{-(\text{a+b})\times\text{a}-(-2\text{ab})\times\frac{\text{a}}{\text{b}}}\\\ \ \ =\frac{1}{\frac{\text{a}}{\text{b}}\times(-\text{b})-\text{a}\times\big(-\frac{\text{b}}{\text{a}}\big)}$
$\Rightarrow\frac{\text{x}}{2\text{b}^2-\text{b}(\text{a+b})}=\frac{\text{y}}{-(\text{a+b})+2\text{a}^2}=\frac{1}{-\text{a+b}}$
$\Rightarrow\frac{\text{x}}{2\text{b}^2-\text{ab}-\text{b}^2}=\frac{\text{y}}{-\text{a}^2-\text{ab}+2\text{a}^2}=\frac{1}{-\text{a+b}}$
$\Rightarrow\frac{\text{x}}{\text{b}^2-\text{ab}}=\frac{\text{y}}{\text{a}^2-\text{ab}}=\frac{1}{-(\text{a}-\text{b})}$
$\Rightarrow\frac{\text{x}}{-\text{b}(-\text{a}-\text{b})}=\frac{\text{y}}{\text{a}(\text{a}-\text{b})}=\frac{1}{-(\text{a}-\text{b})}$
$\Rightarrow\text{x}=\frac{-\text{b}(\text{a}-\text{b})}{-(\text{a}-\text{b})}=\text{b},\ \text{y}=\frac{\text{a}(\text{a}-\text{b})}{-(\text{a}-\text{b})}=-\text{a}$
Hence, $x = b$ and $y = -a$ is the required solution.
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