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One of the examples to solve in the book is: $$ -3<4-x<2 \quad\text{and}\quad -1 \leq x-5 \leq 2 $$ What I've done so far:

I've solved each inequality getting to $7>x>2$ and $4 \leq x \leq 7$

All I need to do now is to make it into one equality, I can do it when there is only 2 parts to the inequality but not when there is 3

Thanks!!

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  • $\begingroup$ Sorry guys there is something wrong with this the original question is: $\endgroup$ – Mark Zakhem Oct 9 '13 at 7:12
  • $\begingroup$ One of the examples to solve in the book is: -3<4-x<2 and -1 </= x-5 </= 2 What I've done so far: I've solved each inequality getting to 7>x>2 and 4<x< 7 All I need to do now is to make it into one equality, I can do it when there is only 2 parts to the inequality but not when there is 3 Thanks!! Also can someone please change the </= to the normal less/ more than or equal to sign $\endgroup$ – Mark Zakhem Oct 9 '13 at 7:12
  • $\begingroup$ can someone edit it into the question itself. Would be greatly appreciated! $\endgroup$ – Mark Zakhem Oct 9 '13 at 7:12
  • $\begingroup$ try to not read $1\lt x\lt 5$ as "1 is less than x is less than 5". Instead read it as "x is between $1$ and $5$ exclusively". Similarly, read $1\le x\le 5$ as "x is between $1$ and $5$ inclusively". $\endgroup$ – John Joy Apr 19 '14 at 19:59
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Once you get the two separate inequalities $$ \begin{cases} \color{red}{2 < x < 7} \\[5pt] \color{blue}{4 \leq x \leq 7} \end{cases} $$ it is convenient to represent them graphically. Just represent the real numbers as a straight line and mark the numbers that appear in you inequalities in the correct order from the lowest to the highest (their position need not be properly scaled though) as in the picture ($2$, $4$ and $7$ in your case).

enter image description here

Then represent each inequality as a rectangle as shown in the picture. Different inequalities should correspond to rectangles of different heights. It is convenient to (for instance) mark the top corners with either filled or hollow circles: filled circles tell you that the inequality includes the corresponding point, while hollow circles tell you they do not. In your example

  • in the equation $\color{red}{2<x<7}$, since neither $2$ and $7$ are part of the solution, then you mark both the top vertices with a hollow circle;
  • in the equation $\color{blue}{4\leq x\leq 7}$, both $4$ and $7$ are part of the solution, therefore the corresponding vertices are to be marked with a filled circle.

Since you are looking for the intersection of the two equations, you are interested in the region (of the real line) that belongs to both the two, i.e. the one that goes from $4$ to $7$.

Now, how can we tell whether $4$ and $7$ are part of the solution or not? That's where hollow and filled circles help us out: it should be easy to understand that the solution turns out to be $$ 4 ~\leq~ x ~<~ 7 $$ Can you tell why?

(edit) Further details:

Consider the following system of inequalities: $$ \begin{cases}\tag{S} \color{red}{2 < x < 7 \quad\vee\quad 9\leq x<11} \\[5pt] \color{blue}{4 \leq x < 10} \end{cases} $$ The '$\vee$' sign (logical or, set union) can be graphically translated as "draw another rectangle of the same height". Therefore the system (S) corresponds to the following graphical representation:

enter image description here

The solution, again, is the area spanned by both $\color{red}{red}$ and $\color{blue}{blue}$ rectangles, that is $$ 4 \leq x < 7 \quad\vee\quad 9\leq x <10 $$

It should be straightforward to understand how all these reasonings apply to systems of more than two inequalities as well.

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  • $\begingroup$ Wow! Great answer! The reason why is because the more than or equal to sign shows that 4 COULD be part of the answer and the less than sign shows that 7 is NOT part of the answer $\endgroup$ – Mark Zakhem Oct 9 '13 at 8:14
  • $\begingroup$ Yep! Note that point $x=7$ has both a hollow and a filled circle, which tells you that one inequality includes it but the other does not. Since you want the points that satisfy both the inequalities, $7$ is not part of the solution. In this sense, wherever you have hollow circles, those points do not belong to the solution ( hollow beats filled :) ). $\endgroup$ – AndreasT Oct 9 '13 at 8:27
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    $\begingroup$ @MarkZakhem Note that it would be the other way around if you instead of $x$ that statisfy all the inequalities, wanted $x$ that statify at least one of the inequalities (that would be the case if there's an or between them instead of an and). Then filled circles would trumph hollow, and the solution would've been $2<x\leq 7$ You can still use the same figure to help you visualize the multiple inequalities, you would just have to interpret it slightly differently. $\endgroup$ – Arthur Oct 9 '13 at 8:49
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You did first part right

$7>x>2$

$(2)$ $-1 \le x-5\le2$

$4 \le x\le7$

combining both parts we ,get

$4 \le x<7$ (you can check it ,by representing in number line)

Represent solution of both parts in number line

Your new solution is intersection of both parts

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  • $\begingroup$ I know it's just making the 2 inequalities into one $\endgroup$ – Mark Zakhem Oct 9 '13 at 7:19
  • $\begingroup$ How did you combine it $\endgroup$ – Mark Zakhem Oct 9 '13 at 7:40
  • $\begingroup$ @MarkZakhem you can't combine these inequalities to an equality. They are not too restrictive to form an equality when combined. $\endgroup$ – Ruslan Oct 9 '13 at 7:41
  • $\begingroup$ to form the INequality $\endgroup$ – Mark Zakhem Oct 9 '13 at 7:42
  • $\begingroup$ it helps if you change the $sense$ of the inequality $$7\ge x\ge 4$$into$$4\le x\le 7$$ $\endgroup$ – John Joy Apr 19 '14 at 19:52

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