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  1. #1
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    inequality

    Solve the inequality |x+4|+|x-1|>5 .

    I know the strategy here is to get rid of the modulus by picking cases .

    my questino here is how to pick these cases ?
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  2. #2
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    Quote Originally Posted by hooke View Post
    Solve the inequality |x+4|+|x-1|>5 .

    I know the strategy here is to get rid of the modulus by picking cases .

    my questino here is how to pick these cases ?
    Use the definition of the absolute value:

    |x|=\left\{ \begin{array}{l}x,\ if \ x \geq 0 \\ -x,\ if \ x<0 \end{array} \right.

    With your example you have:

    |x+4|=\left\{ \begin{array}{l}x+4,\ if \ x \geq -4 \\ -(x+4),\ if \ x<-4 \end{array} \right.

    |x-1|=\left\{ \begin{array}{l}x-1,\ if \ x \geq 1 \\ -(x-1),\ if \ x<1 \end{array} \right.

    You now have 3 intervals with different inequalities:

    \begin{array}{ll}x\geq 1 :&  x+4+x-1>5 \\ -4\leq x<1 :& x+4+(-(x-1)) >5 \\  x<-4:  &-(x+4)+(-(x-1))>5 \end{array}

    Solve for x.

    Draw the graphs of

    f(x)=|x+4|+|x-1| and y = 5

    Determine those points on the graph of f where the y-coordinate is greater than 5.
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  3. #3
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    Quote Originally Posted by earboth View Post
    Use the definition of the absolute value:

    |x|=\left\{ \begin{array}{l}x,\ if \ x \geq 0 \\ -x,\ if \ x<0 \end{array} \right.

    With your example you have:

    |x+4|=\left\{ \begin{array}{l}x+4,\ if \ x \geq -4 \\ -(x+4),\ if \ x<-4 \end{array} \right.

    |x-1|=\left\{ \begin{array}{l}x-1,\ if \ x \geq 1 \\ -(x-1),\ if \ x<1 \end{array} \right.

    You now have 3 intervals with different inequalities:

    \begin{array}{ll}x\geq 1 :& x+4+x-1>5 \\ -4\leq x<1 :& x+4+(-(x-1)) >5 \\ x<-4: &-(x+4)+(-(x-1))>5 \end{array}

    Solve for x.

    Draw the graphs of

    f(x)=|x+4|+|x-1| and y = 5

    Determine those points on the graph of f where the y-coordinate is greater than 5.
    hey thank you for clarifying !
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