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P-1 Linear Equations in Two Variables — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsP-1 Linear Equations in Two Variables3 Marks
Question
Solve the following simultaneous equations.
$\frac{1}{2(3 x+4 y)}+\frac{1}{5(2 x-3 y)}=\frac{1}{4} ; \frac{5}{(3 x+4 y)}-\frac{2}{(2 x-3 y)}=-\frac{3}{2}$
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Answer

$\text { Let } \frac{1}{3 x+4 y}=m \text { and } \frac{1}{2 x-3 y}=n $
$\frac{1}{2} m+\frac{1}{5} n=\frac{1}{4} \Rightarrow 5 m+2 n=\frac{10}{4} \Rightarrow 20 m+8 n=10 \Rightarrow 10 m+4 n=5 \dots(I)$
$5 m-2 n=-\frac{3}{2} \Rightarrow 10 m-4 n=-3 \ldots \text { (II) }$
Equating Eq. I and II
$10 m+4 n=5$
$\underline {10 m-4 n=-3} $
$20 m=2 $
$m=\frac{2}{20} $
$m=\frac{1}{10}$
Substituting $m =\frac{1}{10}$ in Eq. I
$10 \times \frac{1}{10}+4 n =5 $
$1+4 n =5 $
$4 n =5-1$
$4 n =4 $
$n =\frac{4}{4} $
$n =1 $
$\therefore \frac{1}{3 x+4 y}= m \Rightarrow \frac{1}{3 x+4 y}=\frac{1}{10} \Rightarrow 3 x +4 y =10 \ldots (III)$
$\therefore \frac{1}{2 x -3 y }= n \Rightarrow \frac{1}{2 x -3 y }=1 \Rightarrow 2 x -3 y =1 \ldots {IV}$
Multiply Eq. III by 3 and Eq. IV by 4 and Equate
$9 x+12 y=30 \ldots(V) $
$\underline{8 x-12 y=4 \ldots(V I)} $
$17 x=34 $
$x=\frac{34}{17}$
$x=2$
Substituting $x =2$ in Eq. $V$
$9 \times 2+12 y =30 $
$18+12 y =30 $
$12 y =30-18 $
$12 y =12$
$y =\frac{12}{12}$
$y =1$
Hence, $(x, y)=(2,1)$

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