Algebra 2 question! (Grade 10 honors alg 2)

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chocoholic42

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Nov 11, 2012
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Hi, we are learning absolute value inequalities and I am in grade 10 Alg 2 honors.

2/(lx+4l) greater than or equal to 1 (the "l" represents absolute value bars)

my work so far:

2 greater than or equal to lx+4l(1)

so: x+4 less than or equal to 2 OR x+4 greater than or equal to -2
then i got:
x less than or equal to 2 x greater than or equal to -2 as my final answers

BUT, my solution sheet says this is wrong... what is wrong?
 
chocoholic42 said:
Hi, we are learning absolute value inequalities and I am in grade 10 Alg 2 honors.

2/(lx+4l) greater than or equal to 1 (the "l" represents absolute value bars)

my work so far:

2 greater than or equal to lx+4l(1)

so: x+4 less than or equal to 2 OR x+4 greater than or equal to -2
then i got:
x less than or equal to 2 x greater than or equal to -2 as my final answers

BUT, my solution sheet says this is wrong... what is wrong?
Click to expand...

What's wrong is that the rule you are trying to use says |u|< a if and only if -a < u < a. This is an AND statement, not an OR. u<a AND -a < u. So the answer to your problem is an interval centered at 4 rather than what amounts to the entire real line (every x is either <=2 or >= -2)
 
girodd said:

What's wrong is that the rule you are trying to use says |u|< a if and only if -a < u < a. This is an AND statement, not an OR. u<a AND -a < u. So the answer to your problem is an interval centered at 4 rather than what amounts to the entire real line (every x is either <=2 or >= -2)
Click to expand...

How is it an "AND" statement if the original problem is greater than? I thought all >'s meant or statements
 
girodd said:
When you multiply through by |x-4| you get |x-4|<=2. That's a "less than" you have to deal with.
Click to expand...

This is correct.

Also, the answer [-6,-4)U(-4,-2)] is correct.

Do you see why -4 is excluded? Hint: look at the problem in its original form.
 
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