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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:
$\frac{1}{\text{x}}+\frac{1}{\text{y}}=7,$
$\frac{2}{\text{x}}-\frac{3}{\text{y}}=17 $ $(\text{x}\neq0,\ \text{y}\neq0).$
✓

Answer

Taking $\frac{1}{ x }= u$ and $\frac{1}{ y }= v$, the given equations become: $u + v =72 u +3 v =17$ The given equations may be written as: $u+v-7=0 \ldots$ (i) $2 u+3 v-17=0 \ldots$ (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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