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Linear Equations In Two Variables — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsLinear Equations In Two Variables5 Marks
Question
Solve the following systems of equations by using the method of cross multiplication:
$\frac{\text{ax}}{\text{b}}-\frac{\text{by}}{\text{a}}=\text{a}+\text{b},$
$\text{ax}-\text{by}=\text{2ab}$
✓

Answer

The given equations may be written as: $\frac{\text{ax}}{\text{b}}-\frac{\text{by}}{\text{a}}-(\text{a}+\text{b})=0\ \dots(\text{i})$ $\text{ax}-\text{by}-\text{2ab}=0\ \dots(\text{ii})$
Here, $\text{a}_1=\frac{\text{a}}{\text{b}},\ \text{b}_1=\frac{-\text{b}}{\text{a}},$ $c_1 = -(a + b), a_2 = a, b_2 = -b and c_2 = -2$ab By cross multiplication, we have:

$\therefore\frac{\text{x}}{\big(-\frac{\text{b}}{\text{a}}\big)\times(-\text{2ab})-(-\text{b})\times(-(\text{a}+\text{b}))}=\frac{\text{y}}{-(\text{a}+\text{b})\times\text{a}-(-\text{2ab})\times\frac{\text{a}}{\text{b}}}=\frac{1}{\frac{\text{a}}{\text{b}}\times(-\text{b})-\text{a}\times\big(-\frac{\text{b}}{\text{a}}\big)}$
$\Rightarrow\frac{\text{x}}{\text{2b}^2-\text{b}(\text{a}+\text{b})}=\frac{\text{y}}{-\text{a}(\text{a}+\text{b})+\text{2a}^2}=\frac{1}{-\text{a}+\text{b}}$
$\Rightarrow\frac{\text{x}}{\text{2b}^2-\text{ab}+\text{b}^2}=\frac{\text{y}}{-\text{a}^2-\text{ab}+\text{2a}^2}=\frac{1}{-\text{a}+\text{b}}$ $\Rightarrow\frac{\text{x}}{(\text{b}^2-\text{ab})}=\frac{\text{y}}{(\text{a}^2-\text{ab})}=\frac{1}{-(\text{a}-\text{b})}$
$\Rightarrow\frac{\text{x}}{-\text{b}(\text{a}-\text{b})}=\frac{\text{y}}{\text{a}(\text{a}-\text{b})}=\frac{1}{-(\text{a}-\text{b})}$
$\Rightarrow\text{x}=\frac{-\text{b}(\text{a}-\text{b})}{-(\text{a}-\text{b})}=\text{b},\ \text{y}=\frac{\text{a}(\text{a}-\text{b})}{-(\text{a}-\text{b})}=-\text{a}$
Hence, x = b and y = -a is the required solution.

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