Question

a) Solve the inequation and represent the solution set on the number line.
$-3+x\le \frac{8x}{3}+2\le \frac{14}{3}+2x,wherexϵI$

b) Solve the given inequation and graph the solution on the number line.
$2y-3<y+1\le 4y+7;yϵR$

Answer 1

a) $-3+x\le \frac{8x}{3}+2\le \frac{14}{3}+2x,wherexϵI-3+x\le \frac{8x}{3}+2;\frac{8x}{3}+2\le \frac{14}{3}+2x\Rightarrow \frac{8x}{3}-x\ge -3-2;\frac{8x}{3}-2x\le \frac{14}{3}-2\Rightarrow \frac{5x}{3}\ge -5;\frac{2x}{3}\le \frac{8}{3}\Rightarrow 5x\ge -15;2x\le 8\Rightarrow x\ge -3;x\le 4\therefore -3\le x\le 4Solutionset=\{-3,-2,-1,-0,1,2,3,4\}.$

b) $Given:2y-3<y+1\le 4y+7;yϵR$
$\Rightarrow 2y-y<3+1;1-7\le 4y-y$
$\Rightarrow y<4;$ $-6\le 3y$
$\Rightarrow y<4;$ $-2\le y$
Hence, the solution set $=\{-2\le y<4,yϵR\}$
The graph of the solution is shown below.