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Linear Equation In Two Variable — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsLinear Equation In Two Variable4 Marks
Question
Solve the following systems of equations by using the method of cross multiplication:
$\frac{1}{\text{x}}+\frac{1}{\text{y}}=7,$
$\frac{2}{\text{x}}-\frac{3}{\text{y}}=17 $ $(\text{x}\neq0,\ \text{y}\neq0).$
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Answer

Taking $\frac{1}{\text{x}}=\text{u}$ and $\frac{1}{\text{y}}=\text{v},$
the given equations become: $u + v = 7 2u + 3v = 17$
The given equations may be written as:
$u + v - 7 = 0 ...(i) $
$2u + 3v - 17 = 0 ...(ii)$
Here, $a_1 = 1, b_1 = 1, c_1 = -7,$
$ a_2 = 2, b_2 = 3$ and $c_2 = -17$
By cross multiplication, we have:

$\therefore\frac{\text{u}}{[1\times(-17)-3\times(-7)]}=\frac{\text{v}}{[(-7)\times2-1\times(-17)]}=\frac{1}{[3-2]}$
$\Rightarrow\frac{\text{u}}{-17+21}=\frac{\text{v}}{-14+17}=\frac{1}{1}$
$\Rightarrow\frac{\text{u}}{4}=\frac{\text{v}}3{}=\frac{1}{1}$
$\Rightarrow\text{u}=\frac{4}{1}=4,\ \text{v}=\frac{3}1{}=3$
$\Rightarrow\frac{1}{\text{x}}=4,\ \frac{1}{\text{y}}=3$
$\Rightarrow\text{x}=\frac{1}4{},\ \text{y}=\frac{1}3{}$
Hence, $\text{x}=\frac{1}4{}$ and $\text{y}=\frac{1}{3}$ is the required solution.

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