Question
Solve $\frac{2}{3}(x+1)+4<10$ and represent its solution on a number line. Given replacement set is $\{-8,-6,-4,3,6,8,12\}$.
✓
Answer
$\frac{2}{3}(x+1)+4<10$
$\Rightarrow \frac{2}{3}(\mathrm{x}-1)<10-4$
$\Rightarrow \frac{2}{3}(\mathrm{x}-1)<6$
$\Rightarrow 2(x-1)<18$
$\Rightarrow x-1<9$
$\Rightarrow x-1+1<9+1 \ldots ($Adding $1$ to both sides$)$
$\Rightarrow x<10$
Thus $x<10$
Since, replacement set $=\{-8,-6,-4,3,6,8,12\}$
$\Rightarrow$ Solution set $=\{-8,-6,-4,3,6,8\}$
$\Rightarrow \frac{2}{3}(\mathrm{x}-1)<10-4$
$\Rightarrow \frac{2}{3}(\mathrm{x}-1)<6$
$\Rightarrow 2(x-1)<18$
$\Rightarrow x-1<9$
$\Rightarrow x-1+1<9+1 \ldots ($Adding $1$ to both sides$)$
$\Rightarrow x<10$
Thus $x<10$
Since, replacement set $=\{-8,-6,-4,3,6,8,12\}$
$\Rightarrow$ Solution set $=\{-8,-6,-4,3,6,8\}$
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