Question
Solve the following pair of linear $($simultaneous$)$equation using method of elimination by substitution:$\frac{3 x}{2}-\frac{5 y}{3}+2=0,\frac{x}{3}+\frac{y}{2}=2 \frac{1}{6}$
✓
Answer
$ \frac{x}{3}+\frac{y}{2}=2 \frac{1}{6}$
$ \frac{x}{3}+\frac{y}{2}=\frac{13}{6}$
$ \Rightarrow \frac{2 x+3 y}{6}=\frac{13}{6}$
$ \Rightarrow 2 x+3 y=13$
$ \Rightarrow 2 x=13-3 y$
$\Rightarrow x=\frac{13-3 y}{2}\ldots .(1)$
And,
$\frac{3 x}{2}-\frac{5 y}{3}+2=0$
$ \Rightarrow \frac{3}{2}\left(\frac{13-3 y}{2}\right)-\frac{5 y}{3}=-2$
$ \Rightarrow \frac{39-9 y}{4}-\frac{5 y}{3}=-2$
$ \Rightarrow \frac{117-27 y-20 y}{12}=-2$
$ \Rightarrow \frac{117-47 y}{12}=-2$
$ \Rightarrow 117-47 y=-24$
$ \Rightarrow 47 y=141$
$ \Rightarrow y=3$
Substituting the value of $y$ in $(1),$ we have
$x=\frac{13-3 \times 3}{2}=\frac{13-9}{2}=\frac{4}{2}=2$
$\therefore$ Solution is $x=2$ and $y=3$.
$ \frac{x}{3}+\frac{y}{2}=\frac{13}{6}$
$ \Rightarrow \frac{2 x+3 y}{6}=\frac{13}{6}$
$ \Rightarrow 2 x+3 y=13$
$ \Rightarrow 2 x=13-3 y$
$\Rightarrow x=\frac{13-3 y}{2}\ldots .(1)$
And,
$\frac{3 x}{2}-\frac{5 y}{3}+2=0$
$ \Rightarrow \frac{3}{2}\left(\frac{13-3 y}{2}\right)-\frac{5 y}{3}=-2$
$ \Rightarrow \frac{39-9 y}{4}-\frac{5 y}{3}=-2$
$ \Rightarrow \frac{117-27 y-20 y}{12}=-2$
$ \Rightarrow \frac{117-47 y}{12}=-2$
$ \Rightarrow 117-47 y=-24$
$ \Rightarrow 47 y=141$
$ \Rightarrow y=3$
Substituting the value of $y$ in $(1),$ we have
$x=\frac{13-3 \times 3}{2}=\frac{13-9}{2}=\frac{4}{2}=2$
$\therefore$ Solution is $x=2$ and $y=3$.
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