Question
Solve the following pair of linear $($Simultaneous $)$ equation using method of elimination by substitution$\frac{2 x+1}{7}+\frac{5 y-3}{3}=12,\frac{3 x+2}{2}-\frac{4 y+3}{9}=13$
✓
Answer
$\frac{2 x+1}{7}+\frac{5 y-3}{3}=12 \quad ($given$)$
$ \Rightarrow \frac{3(2 x+1)+7(5 y-3)}{21}=12$
$ \Rightarrow 6 x+3+35 y-21=252$
$ \Rightarrow 6 x+35 y-18=252$
$ \Rightarrow 6 x+35 y=270$
$ \Rightarrow 6 x=270-35 y$
$ \Rightarrow x=\frac{270-35 y}{6}$
$\frac{3 x+2}{2}-\frac{4 y+3}{9}=13 \quad($given$)$
$ \Rightarrow \frac{9(3 x+2)-2(4 y+3)}{18}=13$
$ \Rightarrow 27 x+18-8 y-6=234$
$ \Rightarrow 27 x-8 y+12=234$
$\Rightarrow 27 x-8 y=222\ldots .(1)$
Substituting $x=\frac{270-35 y}{6}$ in $(1)$, we get
$27\left(\frac{270-35 y}{6}\right)-8 y=222$
$ \Rightarrow 7290-945 y-48 y=1332$
$ \Rightarrow-993 y=-5958$
$ \Rightarrow y=6$
Substituting $y=6$ in $x=\frac{270-35 y}{6}$, we get
$x=\frac{270-35 \times 6}{6}=\frac{270-210}{6}=\frac{60}{6}=10$
$\therefore$ Solution is $x=10$ and $y=6$.
$ \Rightarrow \frac{3(2 x+1)+7(5 y-3)}{21}=12$
$ \Rightarrow 6 x+3+35 y-21=252$
$ \Rightarrow 6 x+35 y-18=252$
$ \Rightarrow 6 x+35 y=270$
$ \Rightarrow 6 x=270-35 y$
$ \Rightarrow x=\frac{270-35 y}{6}$
$\frac{3 x+2}{2}-\frac{4 y+3}{9}=13 \quad($given$)$
$ \Rightarrow \frac{9(3 x+2)-2(4 y+3)}{18}=13$
$ \Rightarrow 27 x+18-8 y-6=234$
$ \Rightarrow 27 x-8 y+12=234$
$\Rightarrow 27 x-8 y=222\ldots .(1)$
Substituting $x=\frac{270-35 y}{6}$ in $(1)$, we get
$27\left(\frac{270-35 y}{6}\right)-8 y=222$
$ \Rightarrow 7290-945 y-48 y=1332$
$ \Rightarrow-993 y=-5958$
$ \Rightarrow y=6$
Substituting $y=6$ in $x=\frac{270-35 y}{6}$, we get
$x=\frac{270-35 \times 6}{6}=\frac{270-210}{6}=\frac{60}{6}=10$
$\therefore$ Solution is $x=10$ and $y=6$.
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