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Linear Equation In Two Variable — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsLinear Equation In Two Variable3 Marks
Question
Very-Short and Short-Answer Questions:
Solve $\frac{3}{\text{x}+\text{y}}+\frac{2}{\text{x}-\text{y}}=2$ and $\frac{9}{\text{x}+\text{y}}-\frac{4}{\text{x}-\text{y}}=1$
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Answer

The given system of equations is:
$\frac{3}{\text{x+y}}+\frac{2}{\text{x}-\text{y}}=2\ ....(\text{i})$
$\frac{9}{\text{x+y}}-\frac{4}{\text{x}-\text{y}}=1\ ...(\text{ii})$
Substituting $\frac{1}{\text{x+y}}=\text{u}$ and $\frac{1}{\text{x}-\text{y}}=\text{v}$ in $(i)$ and $(ii),$ the given equations are changed to:
$3v + 2v = 2 ...(iii)$
$9u - 4v = 1 ...(iv)$
Multiplying $(i)$ by $2$ and adding it with $(ii),$ we get
$15\text{u}=4+1$
$\Rightarrow\text{u}=\frac{1}{3}$
Multiplying $(i)$ by $3$ and subtracting $(ii)$ from it, we get
$6\text{v}+4\text{v}=6-1$
$\Rightarrow\text{u}=\frac{5}{10}=\frac{1}{2}$
Therefore,
$x + y = 3 ...(v)$
$x - y = 2 ....(vi)$
Now, adding $(v)$ and $(vi)$ we have
$2\text{x}=5$
$\Rightarrow\text{x}=\frac{5}{2}$
Substituting $\text{x}=\frac{5}{2}$ in $(v),$ we have
$\frac{5}{2}+\text{y}=3$
$\Rightarrow\text{y}=3-\frac{5}{2}=\frac{1}{2}$
Hence, $\text{x}=\frac{5}{2}$ and $\text{y}=\frac{1}{2}.$

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