Question
Solve the following simultaneous equations :$3(2u + v) = 7uv;3(u + 3v) = 11uv$
✓
Answer
$ 3(2 u+v)=7 u v$
$3(u+3 v)=11 u v $
Dividing by $uv,$ we get,
$ \frac{6}{v}+\frac{3}{u}=7 \ldots \ldots . . (1)$
$3(u+3 v)=11 u v$
$3 u+9 v=11 u v $
Dividing by $uv,$ we get
$ \frac{3}{v}+\frac{8}{u}=11 \dots...........(2) $
Multiplying $(1)$ by $3,$ we get,
$ \frac{18}{v}+\frac{9}{u}=21\dots ........(3) $
Subtracting $(2)$ from $(3),$ we get,
$ \frac{15}{v}=10$
$\Rightarrow v =\frac{15}{10}=\frac{3}{2}$
$\therefore \frac{3}{u} $
$ =7-6 \times \frac{2}{3}$
$=7-4$
$=3$
$\Rightarrow u=1 $
Thus, the solution set is $\left(1, \frac{3}{2}\right)$.
$3(u+3 v)=11 u v $
Dividing by $uv,$ we get,
$ \frac{6}{v}+\frac{3}{u}=7 \ldots \ldots . . (1)$
$3(u+3 v)=11 u v$
$3 u+9 v=11 u v $
Dividing by $uv,$ we get
$ \frac{3}{v}+\frac{8}{u}=11 \dots...........(2) $
Multiplying $(1)$ by $3,$ we get,
$ \frac{18}{v}+\frac{9}{u}=21\dots ........(3) $
Subtracting $(2)$ from $(3),$ we get,
$ \frac{15}{v}=10$
$\Rightarrow v =\frac{15}{10}=\frac{3}{2}$
$\therefore \frac{3}{u} $
$ =7-6 \times \frac{2}{3}$
$=7-4$
$=3$
$\Rightarrow u=1 $
Thus, the solution set is $\left(1, \frac{3}{2}\right)$.
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