PDA

View Full Version : Solving compound linear inequalities


kayla_coop
Sep 29, 2007, 11:47 AM
2p-2 is lssthan/equalto 4p-8is lessthan/equalto 3p-3


Could you help me:confused:

galactus
Sep 29, 2007, 01:25 PM
There are different approaches.


2p-2\leq{4p-8}\leq{3p-3}

Factor:

2(p-1)\leq{4(p-2)}\leq{3(p-1)}

Divide by p-1:

2\leq{4(\frac{p-2}{p-1})}\leq{3}

2\leq{4-\frac{4}{p-1}}\leq{3}

2\geq{\frac{4}{p-1}}\geq{1}

Now, can you finish.