Question
Solve the following pair of linear $($simultaneous$)$ equation using method of elimination by substitution $:2x - 3y + 6 = 0,2x + 3y - 18 = 0$
✓
Answer
$ 2 x-3 y+6=0$
$ \Rightarrow 2 x=3 y-6$
$\Rightarrow x=\frac{3 y-6}{2}\ldots(1)$
And,
$2 x+3 y-18=0$
$ \Rightarrow 2\left(\frac{3 y-6}{2}\right)+3 y=18 \ldots[$ From$(1)] $
$\Rightarrow 3 y-6+3 y=18$
$\Rightarrow 6 y=24$
$ \Rightarrow y=4$
Substituting the value of $y$ in $(1),$
we have
$x=\frac{3 \times 4-6}{2}=\frac{12-6}{2}=\frac{6}{2}=3$
$\therefore$ Solution is $x=3$ and $y=4$.
$ \Rightarrow 2 x=3 y-6$
$\Rightarrow x=\frac{3 y-6}{2}\ldots(1)$
And,
$2 x+3 y-18=0$
$ \Rightarrow 2\left(\frac{3 y-6}{2}\right)+3 y=18 \ldots[$ From$(1)] $
$\Rightarrow 3 y-6+3 y=18$
$\Rightarrow 6 y=24$
$ \Rightarrow y=4$
Substituting the value of $y$ in $(1),$
we have
$x=\frac{3 \times 4-6}{2}=\frac{12-6}{2}=\frac{6}{2}=3$
$\therefore$ Solution is $x=3$ and $y=4$.
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