Question
Solve the following pair of linear $($Simultaneous $)$ equation using method of elimination by substitution $:2( x - 3 ) + 3( y - 5 ) = 0,5( x - 1 ) + 4( y - 4 ) = 0$
✓
Answer
Given equations are
$2(x-3)+3(y-5)=0\ldots(1)$
$5(x-1)+4(y-4)=0\ldots(2)$
From $(1)$, we get
$2 x-6+3 y-15=0$
$ \Rightarrow 2 x-21+3 y=0$
$ \Rightarrow 2 x=21-3 y$
$\Rightarrow \mathrm{x}=\frac{21-3 y}{2}$
From $(2),$ we get
$5(x-1)+4(y-4)=0$
$ \Rightarrow 5 x-5+4 y-16=0$
$\Rightarrow 5 x+4 y-21=0\ldots .(3)$
Substituting $\mathrm{x}=\frac{21-3 y}{2}$ in $(3),$ we get
$5\left(\frac{21-3 y}{2}\right)+4 y=21$
$ \Rightarrow \frac{105-3 y}{2}+4 y=21$
$ \Rightarrow \frac{105-15 y+8y}{2} =21$
$ \Rightarrow-7 y+63=0$
$ \Rightarrow 7 y=63$
$ \Rightarrow y=9$
Substituting $y=9$ in $x=\frac{21-3 y}{2}$, we get
$x=\frac{21-3(9)}{2}=\frac{21-27}{2}=-\frac{6}{2}=-3$
$\therefore$ Solution is $x=-3$ and $y=9$.
$2(x-3)+3(y-5)=0\ldots(1)$
$5(x-1)+4(y-4)=0\ldots(2)$
From $(1)$, we get
$2 x-6+3 y-15=0$
$ \Rightarrow 2 x-21+3 y=0$
$ \Rightarrow 2 x=21-3 y$
$\Rightarrow \mathrm{x}=\frac{21-3 y}{2}$
From $(2),$ we get
$5(x-1)+4(y-4)=0$
$ \Rightarrow 5 x-5+4 y-16=0$
$\Rightarrow 5 x+4 y-21=0\ldots .(3)$
Substituting $\mathrm{x}=\frac{21-3 y}{2}$ in $(3),$ we get
$5\left(\frac{21-3 y}{2}\right)+4 y=21$
$ \Rightarrow \frac{105-3 y}{2}+4 y=21$
$ \Rightarrow \frac{105-15 y+8y}{2} =21$
$ \Rightarrow-7 y+63=0$
$ \Rightarrow 7 y=63$
$ \Rightarrow y=9$
Substituting $y=9$ in $x=\frac{21-3 y}{2}$, we get
$x=\frac{21-3(9)}{2}=\frac{21-27}{2}=-\frac{6}{2}=-3$
$\therefore$ Solution is $x=-3$ and $y=9$.
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