Question:

Can someone show me how to solve this inequality?

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Does anyone know the step by step answer to this inequality. Its not making sense to me because it is an or statement and when I graph it, the x is sandwiched between the two values. Please solve in iterval and set notation using -inf and inf for infinity. Thank you so much!

3x + 1 < 0 or 5x + 5 > −8

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  1. Okay, for &quot;3x + 1&lt;0&quot;.

    3x +1&lt;0

         -1  -1

    -------------

    (the 1 cancel each other out)

    it becomes:

    3x &lt; -1

    divide the 3 from both sides.

    3/3 &lt;0/-1

    3 and 3 cancel each other out.

    it then becones -1/3(fraction)

    the answer is x = -1/3

    to check your answer you would,

    3(-1/3) + 1 &lt;0

    which changes to,

    -1 + 1 &lt;0


  2. Solve each equation separately

    3x + 1 &lt; 0

    3x &lt; -1

    x &lt; -1/3

    5x + 5 &gt; -8

    5x &gt; -13

    x &gt; -13/5

    x&gt; -2 3/5

    This give us x &lt; -1/3 OR x &gt; -2 3/5

    The first is all numbers to the left of -1/3 on the number line

    The second is all numbers to the right of -2 3/5 on the number line

    OR means to combine the 2 sets and since -2 3/5 is to the left of -1/3 the answer is all numbers

    If the problem had said AND instead of OR then you would have gotten your answer values between -1/3 and -2 3/5

  3. 3x+1 &lt; 0 ==&gt;  x&lt; -1/3    or   5x+5&gt; -8 ==&gt;  x&gt;-13/5

    As the conjunction is or  you must take all elements of both set.

    That is: all x less than -1/3 and all x greater than -13/5

    In this case, as -1/3 is at right of -13/5 the set is all real numbers... that is  the interval (-infinity, + infinity)

    ..............................-13/5 @@@@@@@@@@@@@@@

    @@@@@@@@@@@@@@@-1/3......................

    The union is:

    @@@@@@@@@@@@@@@@@@@@@@@@

  4. 3x+1&lt;0

    3x+1-1&lt;0-1

    3x&lt;-1

    3x/3&lt;-1/3

    x&lt;-1/3

    and

    5x+5&gt;-8

    5x+5-5&gt;-8-5

    5x&gt;-13

    5x/5&gt;-13/5

    x&gt;-13/5 sooooo;

    -13/5&lt;x&lt;-1/3

    ok?

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