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Math Help - Solving compound inequalities problem

  1. #1
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    Solving compound inequalities problem

    I have read through the solving inequalities pdf and am still struggling with this problem.

    -7<\frac{1}{x}\leq1

    I've tried taking the -7<\frac{1}{x} and the \frac{1}{x}\leq1 on their own and combining the solutions but my answers are obviously wrong. Could someone show me how to solve these kinds of inequalities?
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  2. #2
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    Re: Solving compound inequalities problem

    Hello, BobRoss!

    This one is tricky and dangerous.
    There are several approaches.
    I'll show you mine . . .


    \text{Solve: }\:\text{-}7\:<\:\frac{1}{x}\:\le\:1

    Like you, I split it into two inequalities: . -7 \:<\:\frac{1}{x}\:\text{ and }\:\frac{1}{x} \:\le\:1
    Note the "and"; we want both to be true.

    We'd like to multiply through by , but we must be very careful!
    What we get depends on whether x is positive or negative.


    Suppose x is positive.
    Multiply both inequalities by x.

    We have: . \text{-}7x \:<\:1 \quad\Rightarrow\quad x \:>\:\text{-}\tfrac {1}{7}
    . . . and: . 1 \:\le\:x \quad\Rightarrow\quad x \:\ge\:1

    To make both statements true, we use: . \boxed{x\:\ge\:1}


    Suppose x is negative.
    Multiply both inequalities by negative x.

    We have: . \text{-}7x \:>\:1 \quad\Rightarrow\quad x \:<\:\text{-}\tfrac{1}{7}
    . . . and: . 1 \:\ge\:x \quad\Rightarrow\quad x \:\le\:1

    To make both statements true, we use: . \boxed{x \:<\:\text{-}\tfrac{1}{7}}


    Solution: . \left(x\:<\:\text{-}\tfrac{1}{7}\right)\:\cup\:\left(x\:\ge\:1\right)

    n . . . . . . . \left(\text{-}\infty,\,\text{-}\tfrac{1}{7}\right)\:\cup\:\left[1,\,\infty\right)


    . . . . . \begin{array}{ccccc} === & \circ & --- & \bullet & === \\ & \text{-}\frac{1}{7} && 1 \end{array}
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  3. #3
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    Re: Solving compound inequalities problem

    Quote Originally Posted by Soroban View Post
    Hello, BobRoss!

    This one is tricky and dangerous.
    There are several approaches.
    I'll show you mine . . .



    Like you, I split it into two inequalities: . -7 \:<\:\frac{1}{x}\:\text{ and }\:\frac{1}{x} \:\le\:1
    Note the "and"; we want both to be true.

    We'd like to multiply through by , but we must be very careful!
    What we get depends on whether x is positive or negative.


    Suppose x is positive.
    Multiply both inequalities by x.

    We have: . \text{-}7x \:<\:1 \quad\Rightarrow\quad x \:>\:\text{-}\tfrac {1}{7}
    . . . and: . 1 \:\le\:x \quad\Rightarrow\quad x \:\ge\:1

    To make both statements true, we use: . \boxed{x\:\ge\:1}


    Suppose x is negative.
    Multiply both inequalities by negative x.

    We have: . \text{-}7x \:>\:1 \quad\Rightarrow\quad x \:<\:\text{-}\tfrac{1}{7}
    . . . and: . 1 \:\ge\:x \quad\Rightarrow\quad x \:\le\:1

    To make both statements true, we use: . \boxed{x \:<\:\text{-}\tfrac{1}{7}}


    Solution: . \left(x\:<\:\text{-}\tfrac{1}{7}\right)\:\cup\:\left(x\:\ge\:1\right)

    n . . . . . . . \left(\text{-}\infty,\,\text{-}\tfrac{1}{7}\right)\:\cup\:\left[1,\,\infty\right)


    . . . . . \begin{array}{ccccc} === & \circ & --- & \bullet & === \\ & \text{-}\frac{1}{7} && 1 \end{array}
    Even though there is nothing wrong with your solution, we should note that when we checked what happens with positive x and negative x, we are bringing in ANOTHER restriction for x, that also needs to be satisfied. In this case, it doesn't make any difference, but there might be a time in future where it does.
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    Re: Solving compound inequalities problem

    Thanks so much, that helps a lot! Although I don't quite understand this part:

    Quote Originally Posted by Prove It View Post
    Even though there is nothing wrong with your solution, we should note that when we checked what happens with positive x and negative x, we are bringing in ANOTHER restriction for x, that also needs to be satisfied. In this case, it doesn't make any difference, but there might be a time in future where it does.
    What is the new restriction and could you possibly give me an example of when I might see that?
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    Re: Solving compound inequalities problem

    Quote Originally Posted by BobRoss View Post
    Thanks so much, that helps a lot! Although I don't quite understand this part:



    What is the new restriction and could you possibly give me an example of when I might see that?
    The new restriction comes from saying "if we think about positive values for x", we are immediately making the restriction x > 0, and if we say "if we think about negative values for x", we are immediately making the restriction x < 0.

    Take for example, the inequality \displaystyle \begin{align*} \frac{x^2 + 3x -2}{x} < 3 \end{align*}.

    In order to solve this, we need to think of two cases, the first where x < 0, and the second where x > 0.

    In case 1, with x < 0, we have

    \displaystyle \begin{align*} \frac{x^2 + 3x - 2}{x} &< 3 \\ x^2 + 3x - 2 &> 3x \\ x^2 - 2 &> 0 \\ x^2 &> 2 \\ |x| &> \sqrt{2} \\ x < -\sqrt{2} \textrm{ or } x &> \sqrt{2} \end{align*}

    So by solving the inequality we have \displaystyle \begin{align*} x < -\sqrt{2} \end{align*} or \displaystyle \begin{align*} x > \sqrt{2} \end{align*}, BUT to get this, we originally had to restrict \displaystyle \begin{align*} x < 0 \end{align*}. So that means that the solution for this case is just \displaystyle \begin{align*} x < -\sqrt{2} \end{align*}.
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