[tex]2+\left(x+1\right)<5\\2+x+1<5\\x+3<5\\x+3-3<5-3\\\boxed{x<2}[/tex]
[tex]\frac{1}{2}\left(x+2\right)\le \:10\\\mathrm{mnozymy\:obustronnie\:przez\:}2\\2\cdot \frac{1}{2}\left(x+2\right)\le \:10\cdot \:2\\x+2\le \:20\\x+2-2\le \:20-2\\\boxed{x\le \:18}[/tex]
[tex]\frac{3}{4}\left(1-x\right)<2\\4\cdot \frac{3}{4}\left(1-x\right)<2\cdot \:4\\3\left(1-x\right)<8\\\frac{3\left(1-x\right)}{3}<\frac{8}{3}\\1-x<\frac{8}{3}\\1-x-1<\frac{8}{3}-1\\-x<\frac{5}{3}\\\left(-x\right)\left(-1\right)>\frac{5\left(-1\right)}{3}\\\boxed{x>-\frac{5}{3}}[/tex]
[tex]-4\left(2x+1\right)\ge \:-7\\\left(-4\left(2x+1\right)\right)\left(-1\right)\le \left(-7\right)\left(-1\right)\\4\left(2x+1\right)\le \:7\\\frac{4\left(2x+1\right)}{4}\le \frac{7}{4}\\2x+1\le \frac{7}{4}\\2x+1-1\le \frac{7}{4}-1\\2x\le \frac{3}{4}\\\frac{2x}{2}\le \frac{\frac{3}{4}}{2}\\\boxed{x\le \frac{3}{8}}[/tex]
