Chapter 2, Problem 3CA

To determine

To Category: The given inequalities into two categories.

Answer to Problem 3CA

At least one integer solution	No integer solution
$\begin{array}{l}5x-6+x\ge 2x-8\\ 9x-3<12\text{ or }6x+2>-10\\ 5(x-1)\le 5x-3\end{array}$	$\begin{array}{l}2(3x+8)>3(2x+6)\\ x-8+4x\le 3(x-3)+2x\\ 17<4x+5<21\end{array}$

Explanation of Solution

At least one integer solution

  $\begin{array}{l}5x-6+x\ge 2x-8\\ \Rightarrow 6x-6\ge 2x-8\\ \Rightarrow 6x-2x\ge -8+6\\ \Rightarrow 4x\ge -2\\ \Rightarrow x\ge \frac{-2}{4}\\ \Rightarrow x\ge \frac{-1}{2}\end{array}$

Hence there are many integer solutions.

  $\begin{array}{l}9x-3<12\text{ or }6x+2>-10\\ \Rightarrow 9x<12+3\text{ or }6x>-10-2\\ \Rightarrow 9x<15\text{ or }6x>-12\\ \Rightarrow x<\frac{15}{9}\text{ or }x>\frac{-12}{6}\\ \Rightarrow x<\frac{5}{3}\text{ or }x>-2\\ \Rightarrow -2<x<\frac{5}{3}\end{array}$

There are three integer solutions, $x=-1,0,1$

  $\begin{array}{l}5(x-1)\le 5x-3\\ \Rightarrow 5x-5\le 5x-3\\ \Rightarrow -5\le -3\end{array}$

Possible for all integers, hence all integer solution.

No integer solutions

  $\begin{array}{l}2(3x+8)>3(2x+6)\\ \Rightarrow 6x+16>6x+18\\ \Rightarrow 16>18\end{array}$

This is not possible and hence there is no integer solution.

  $\begin{array}{l}x-8+4x\le 3(x-3)+2x\\ \Rightarrow 5x-8\le 3x-9+2x\\ \Rightarrow 5x-8\le 5x-9\\ \Rightarrow -8\le -9\end{array}$

This is not possible and hence there is no integer solution

  $\begin{array}{l}17<4x+5<21\\ \Rightarrow 12<4x<16\\ \Rightarrow \frac{16}{4}<x<\frac{16}{4}\\ \Rightarrow 3<x<4\end{array}$

No integer between 3 and 4 and hence no integer solution.