# Thread:

1. ## 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 ?

2. Originally Posted by hooke
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:

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

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

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

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. Originally Posted by earboth
Use the definition of the absolute value:

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

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

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

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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