Question 61887
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Could someone help please?
Solve the following inequalities. Write the answers in interval notation.
(a) x^2 + 7x - 18 => 0
(b) (3x-4)/(2x+1) <= 6
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The solutions by @jai_kos are incorrect for both inequalities.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;I came to bring correct solutions for both problems.



<pre>
(a)  Your starting inequality is

         x^2 + 7x - 18 => 0.    (1)


     Factor left side quadratic polynomial

         (x+9)*(x-2) >= 0.


     Critical points are  x = -9  and  x = 2.

     On the left of the critical point  x = -9,  both factors (x+9) and (x-2) are negative, so their product is positive.

     Between the critical points -9 < x < 2, factor (x+9) is positive, while factor (x-2) is negative,
             so their product is negative.

     On the right of the critical point  x = 2,  both factors (x+9) and (x-2) are positive, so their product is positive.


     Thus the solution set for inequality (1) is  x < -9  or  x > 2,  

     or, in interval notation, the union ({{{-infinity}}},{{{-9}}}) U ({{{2}}},{{{infinity}}}).
</pre>

Part (a) is solved correctly.


<pre>
(b)  Your starting inequality is

     {{{(3x-4)/(2x+1)}}} <= 6.    (1)


Transform it equivalently this way

     {{{(3x-4)/(2x+1)}}} - 6 <= 0              <<<---===  moving 6 from right side to left side with changing the sign

     {{{(3x-4)/(2x+1)}}} - {{{(6(2x+1))/(2x+1)}}} <= 0    <<<---=== writing '6' with the common denominator

     {{{((3x-4) - 6(2x+1))/(2x+1)}}} <= 0       <<<---===  simplifying

     {{{(-9x -10)/(2x+1)}}} <= 0                <<<---===  simplifying further



Now, the left side rational function can be non-positive if and only if

    EITHER the numerator is non-negative and denominator is negative

        -9x - 10 >= 0  and  2x + 1 < 0    (2)

    OR     the numerator is non-positive and denominator is positive

        -9x - 10 <= 0  and  2x + 1 > 0.   (3)



In case (2),  9x <= -10  and  x < -1/2,  which is the same as  

              x <= -10/9 and  x < -1/2.

              These both inequalities, taken together, have the solution set  x <= -10/9.



In case (3),  9x >= -10  and  2x > -1,  which is the same as  

              x >= -10/9 and  x > -1/2.

              These both inequalities, taken together, have the solution set  x >= -1/2.


Thus the final solution to the given inequality is this set of real numbers  { x <= -10/9 } OR { x >= -1/2 }.
In the interval notation, the solution set is the union of two sets ({{{-infinity}}},{{{-10/9}}}] U ({{{-1/2)}}},{{{infinity}}}).
</pre>

Both problems/questions are solved correctly.