Question
Solve the following pairs of linear $($simultaneous$)$ equation using method of elimination by substitution:$\frac{x}{6}+\frac{y}{15}=4;\frac{x}{3}-\frac{y}{12}=4 \frac{3}{4}$
✓
Answer
$ \frac{x}{6}+\frac{y}{15}=4$
$ \Rightarrow \frac{5 x+2 y}{30}=4$
$ \Rightarrow 5 x+2 y=120$
$ \Rightarrow 5 x=120-2 y$
$\Rightarrow x=\frac{120-2 y}{5}$$\ldots .(1)$
And,
$\frac{x}{3}-\frac{y}{12}=4 \frac{3}{4}$
$\Rightarrow \frac{1}{3}\left(x-\frac{y}{4}\right)=\frac{19}{4}$
$\Rightarrow \frac{1}{3}\left(\frac{120-2 y}{5}-\frac{y}{4}\right)=\frac{19}{4}$
$\Rightarrow \frac{480-8 y-5 y}{20}=\frac{57}{4}$
$\Rightarrow \frac{480-13 y}{20}=\frac{57}{4}$
$\Rightarrow 480-13 y=285$
$\Rightarrow 13 y=195$
$ \Rightarrow y=15$
Substituting the value of $y$ in $(1)$, we have
$x=\frac{120-2 \times 15}{5}=\frac{120-30}{5}=\frac{90}{5}=18$
$\therefore$ Solution is $x=18$ and $y=15$
$ \Rightarrow \frac{5 x+2 y}{30}=4$
$ \Rightarrow 5 x+2 y=120$
$ \Rightarrow 5 x=120-2 y$
$\Rightarrow x=\frac{120-2 y}{5}$$\ldots .(1)$
And,
$\frac{x}{3}-\frac{y}{12}=4 \frac{3}{4}$
$\Rightarrow \frac{1}{3}\left(x-\frac{y}{4}\right)=\frac{19}{4}$
$\Rightarrow \frac{1}{3}\left(\frac{120-2 y}{5}-\frac{y}{4}\right)=\frac{19}{4}$
$\Rightarrow \frac{480-8 y-5 y}{20}=\frac{57}{4}$
$\Rightarrow \frac{480-13 y}{20}=\frac{57}{4}$
$\Rightarrow 480-13 y=285$
$\Rightarrow 13 y=195$
$ \Rightarrow y=15$
Substituting the value of $y$ in $(1)$, we have
$x=\frac{120-2 \times 15}{5}=\frac{120-30}{5}=\frac{90}{5}=18$
$\therefore$ Solution is $x=18$ and $y=15$
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