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Thread: Solve the inequality

  1. #1
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    Solve the inequality

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  2. #2
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    You have $\displaystyle \displaystyle \frac{3}{a + 1} \geq 3$.

    First, note that $\displaystyle \displaystyle a + 1 \neq 0 \implies a \neq -1$.

    In order to solve for $\displaystyle \displaystyle a$, you need to consider two cases, since multiplying or dividing by a negative number reverses the inequality sign...


    Case 1: $\displaystyle \displaystyle a + 1 < 0 \implies a < -1$.

    Then $\displaystyle \displaystyle \frac{3}{a + 1} \geq 3$

    $\displaystyle \displaystyle 3 \leq 3(a + 1)$

    $\displaystyle \displaystyle 1 \leq a + 1$

    $\displaystyle \displaystyle 0 \leq a$.

    So in this case, you are told $\displaystyle \displaystyle a < -1$ and $\displaystyle \displaystyle a \geq 0$. Since there are not any values of $\displaystyle \displaystyle a$ that satisfy both conditions, there are not any solutions in this case.


    Case 2: $\displaystyle \displaystyle a + 1 > 0 \implies a > -1$.

    $\displaystyle \displaystyle \frac{3}{a + 1} \geq 3$

    $\displaystyle \displaystyle 3 \geq 3(a + 1)$

    $\displaystyle \displaystyle 1 \geq a + 1$

    $\displaystyle \displaystyle 0 \geq a$.

    So in this case, the conditions are $\displaystyle \displaystyle a >-1$ and $\displaystyle \displaystyle a \leq 0$.

    Therefore, the solution in this case is $\displaystyle \displaystyle -1 < a \leq 0$.


    So the solution of your inequality is $\displaystyle \displaystyle -1 < a \leq 0$.
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  3. #3
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    Wow, that is a big post, what have you tried?
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  4. #4
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    Quote Originally Posted by pickslides View Post
    Wow, that is a big post, what have you tried?
    Shouting it from the roof tops!
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