Question
Solve the following simultaneous equations by the substitution method$:\ 2x + y = 8;3y = 3 + 4x$
✓
Answer
The given equations are
$ 2 x+y=8 ....(i)$
$3 y=3+4 x ....(ii) $
Now, consider equation $2 x+y=8$
$ \Rightarrow y =8 \text { - } 2 x ....(iii)$
Substituting the value of $y$ in eqn. $(ii),$ we get
$ 3(8-2 x)=3+4 x$
$\Rightarrow 24-6 x=3+4 x$
$\Rightarrow 6 x-4 x=3-24$
$\Rightarrow-10 x=-21$
$\Rightarrow x=\frac{21}{10} $
Puutting the value of $x$ in eqn. $(iii),$ we get
$ y=8-2\left(\frac{21}{10}\right)$
$=8-\frac{21}{5}$
$=\frac{40-21}{5}$
$=\frac{19}{5} $
Thus, the solution set is $\left(\frac{21}{10}, \frac{19}{5}\right)$.
$ 2 x+y=8 ....(i)$
$3 y=3+4 x ....(ii) $
Now, consider equation $2 x+y=8$
$ \Rightarrow y =8 \text { - } 2 x ....(iii)$
Substituting the value of $y$ in eqn. $(ii),$ we get
$ 3(8-2 x)=3+4 x$
$\Rightarrow 24-6 x=3+4 x$
$\Rightarrow 6 x-4 x=3-24$
$\Rightarrow-10 x=-21$
$\Rightarrow x=\frac{21}{10} $
Puutting the value of $x$ in eqn. $(iii),$ we get
$ y=8-2\left(\frac{21}{10}\right)$
$=8-\frac{21}{5}$
$=\frac{40-21}{5}$
$=\frac{19}{5} $
Thus, the solution set is $\left(\frac{21}{10}, \frac{19}{5}\right)$.
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