Question 1178983
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Solve the following inequality  {{{sqrt(7-x)}}} > x-1
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            This inequality is  SPECIAL  and its solution requires  ACCURACY - so trace attentively each my step.



<pre>
1.  The domain of this inequality is the set  7-x >= 0,  or   x <= 7.



2.  All the set  x-1 < 0,  or  x < 1,  is the solution, BECAUSE the right side is negative there, 

                                                         while the left side is not negative.


    Therefore, we need analyze for the solution only the remaining part  1 <= x <= 7.

    Let's call this set { 1 <= x <= 7 } as A17.



3.  Next, analyzing for the set A17, square both sides of the original inequality.  You will get


        7 - x > x^2 - 2x + 1

        x^2 - x - 6 < 0

        (x-3)*(x+2) < 0


    The solution set to the last inequality  is   -2 < x < 3.      

    Taking the intersection with the set A17, we have the solution set  { 1 <= x <3 }.



4.  Thus we have two parts of the entire solution set  (a)  { x < 1 }  from n.2,  and  (b) { 1 <= x < 3 } from n.3.


     The final solution set is the union  sets (a) and (b), i.e.  { x < 1 } U { 1 <= x < 3},  which is,  OBVIOUSLY,  (-oo < x < 3 }.



<U>ANSWER</U>.  The solution to the original inequality is the set  { -oo < x < 3 },  or  (-oo,3).



                          <U>Visual CHECK</U>



    {{{graph( 400, 400, -5, 5, -5, 5,        
              sqrt(7-x), x-1
)}}}


    Plot  y = {{{sqrt(7-x)}}} (red line)  and  y = x-1  (green line)



The solution set is the set of all points of the x-axis, where red line is above the green line.    
</pre>


Solved, &nbsp;answered, &nbsp;explained, &nbsp;visualized, &nbsp;checked and completed.




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<H3><U>The post-solution note</U></H3>

    &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The plot is indispensable helper in such an analysis.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Honestly, &nbsp;it is THE &nbsp;PLOT, &nbsp;who directs and guides the flow of your mental reasonings.



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Do not forget to post your &nbsp;"THANKS" &nbsp;to me for my teaching.