Solve the following systems of equations by using the method of c…

Question

Solve the following systems of equations by using the method of cross multiplication:
$\frac{\text{a}}{\text{x}}-\frac{\text{b}}{\text{y}}=0$
$\frac{\text{ab}^2}{\text{x}}+\frac{\text{a}^2\text{b}}{\text{y}}=\text{a}^2+\text{b}^2,$ where $\text{x}\neq0$ and $\text{y}\neq0$

✓

Answer

Substituting $\frac{1}{\text{x}}=\text{u}$ and $\frac{1}{\text{y}}=\text{v}$ in the given equations, we get
$au - bv + 0 = 0 ...(i)$
$ab^2u + a^2bv - (a^2 + b^2) = 0 ...(ii)$
Here, $ a_1 = a, b_1 = -b, c_1 = 0, a_2 = ab^2, b_2 = a^2b$ and $c_2 = -(a^2 + b^2).$
By cross multiplication, we have:
$\frac{\text{u}}{\text{b}_1\text{c}_2-\text{b}_2\text{c}_1}=\frac{\text{v}}{\text{c}_1\text{a}_2-\text{c}_2\text{a}_1}=\frac{1}{\text{a}_1\text{b}_2-\text{a}_2\text{b}_1}$
$\Rightarrow\frac{\text{u}}{(-\text{b})[-(\text{a}^2+\text{b}^2)]-(\text{a}^2\text{b})(0)}=\frac{\text{v}}{(0)(\text{a}^2\text{b})-(-\text{a}^2-\text{b}^2)(\text{a})}$
$=\frac{1}{(\text{a})(\text{a}^2\text{b})-(\text{ab}^2)(-\text{b})}$
$\Rightarrow\frac{\text{u}}{\text{b}(\text{a}^2-\text{b}^2)}=\frac{\text{v}}{\text{a}(\text{a}^2+\text{b}^2)}=\frac{1}{\text{ab}(\text{a}^2+\text{b}^2)}$
$\Rightarrow\text{u}=\frac{\text{b}(\text{a}^2+\text{b}^2)}{\text{ab}(\text{a}^2+\text{b}^2)},\ \text{v}=\frac{\text{a}(\text{a}^2+\text{b}^2)}{\text{ab}(\text{a}^2+\text{b}^2)}$
$\Rightarrow\text{u}=\frac{1}{\text{a}},\ \text{v}=\frac{1}{\text{b}}$
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{\text{a}},\ \frac{1}{\text{y}}=\frac{1}{\text{b}}$
$\Rightarrow\text{x}=\text{a},\ \text{y}=\text{b}$
Hence, $x = a$ and $y = b$ is the required solution.