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Linear Equations In Two Variables — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsLinear Equations In Two Variables4 Marks
Question
Solve the following systems of equations:
$\frac{2}{3\text{x}+2\text{y}}+\frac{3}{3\text{x}-2\text{y}}=\frac{17}{5},$
$\frac{5}{3\text{x}+2\text{y}}+\frac{1}{3\text{x}-2\text{y}}=2.$
✓

Answer

The given equations are
$\frac{2}{3\text{x}+2\text{y}}+\frac{3}{3\text{x}-2\text{y}}=\frac{17}{5}$
$\frac{5}{3\text{x}+2\text{y}}+\frac{1}{3\text{x}-2\text{y}}=2$
Let $\frac{1}{3\text{x}-2\text{y}}=\text{u}$ and $\frac{1}{3\text{x}-2\text{y}}=\text{v}$ then equations are
$2\text{u}+3\text{v}=\frac{17}{5}\ ......(\text{i})$
$5\text{u}+\text{v}=2\ ......(\text{ii})$
Multiply equation (ii) by 3 and subtract (ii) from (i) we get
$2\text{u}\ \ +3\text{v}=\frac{17}{5}\\15\text{u}+3\text{v}=6\\\overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\-13\text{u}=-\frac{13}{5}$
$\Rightarrow\text{u}=\frac{1}{5}$
Put the value of u in equation (i) we get
$2\times\frac{1}{5}+3\text{v}=\frac{17}{5}$
$\Rightarrow3\text{v}=3$
$\Rightarrow\text{v}=1$
Then
$\frac{1}{3\text{x}+2\text{y}}=\frac{1}{5}$
$\Rightarrow3\text{x}+2\text{y}=5\ ......(\text{iii})$
$\frac{1}{3\text{x}-2\text{y}}=1$
$\Rightarrow{3\text{x}-2\text{y}}=1\ ......(\text{iv})$
Add both equations we get
${3\text{x}\ +\ 2\text{y}}\ =\ 5\\{3\text{x}\ -\ 2\text{y}}\ =\ 1\over6\text{x}\ \ \ \ \ \ \ \ \ \ =\ 6$
$\Rightarrow\text{x}=1$
Put the value of x in equation (iii) we get
$3\times1+2\text{y}=5$
$\Rightarrow2\text{y}=2$
$\Rightarrow\text{y}=1$
Hence the value of x = 1 and y = 1.

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