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I can solve simple type of inequalities by number line . For example, if I want to solve $x>6$ by number line , then I have to plot this solution into number line which is described in the following image .

enter image description here

How can I solve inequalities with one variable by number line ? Suppose that , I have the following inequalities:

$$x^2-5x+6<0$$

Remember I have to solve this inequality by number line . Suppose , I have applied the following procedure to solve this :

  1. the solution to the above inequality is $2<x<3$
  2. Then I have plotted the solution $2<x<3$ in number line .

This solution is not acceptable because I have to solve the above inequality by number line. Can you guys plz help me to solve this problem ?

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To start with I would talk to the teacher, asking why certain solutions are not accepted.

For the inequality $$x^2-5x+6 <0\tag{1}$$ it is easier to factorise $x^2-5x+6$ first. That is we look for two numbers $A$ and $B$ such that $x^2-5x+6=A\cdot B$, because solving $$A\cdot B <0\tag{2}$$ is easy: (2) can only happen when both $A<0$ and $B>0$, or both $A>0$ and $B<0$.

Now, to factorise $x^2-5x+6$ we first complete the square
\begin{eqnarray}x^2-5x+6& =& \left(x - \frac{5}{2}\right)^2 -\left(\frac{5}{2}\right)^2+6\\ &=& \left(x - \frac{5}{2}\right)^2 -\frac{25}{4}+6\\ &=& \left(x - \frac{5}{2}\right)^2 -\frac{25}{4}+\frac{6\cdot4}{4}\\ &=& \left(x - \frac{5}{2}\right)^2 -\frac{25}{4}+\frac{24}{4} &=& \left(x - \frac{5}{2}\right)^2 -\frac{1}{4} \end{eqnarray} and use the conjugation rule $a^2-b^2 = (a+b)(a-b)$, so that \begin{eqnarray} x^2-5x+6 &=& \left(x - \frac{5}{2}\right)^2 -\frac{1}{4} \\&=& \left(x - \frac{5}{2} +\frac12\right)\left(x - \frac{5}{2} -\frac12\right)\\ &=& \left(x - 2\right)\left(x - 3\right) \end{eqnarray} Now, for (1) we get the inequalities (or rather system of inequalities) $$\left(x - 2\right)<0 \text{ and }\left(x - 3\right)>0\tag{3}$$ and $$\left(x - 2\right)>0 \text{ and }\left(x - 3\right)<0\tag{4}$$ which are linear and can be "solved using number line" (or by the algebraic method of adding a number to both sides of the inequalities). Note that (3) has no solution, while (4) is the line segment you found.

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