How do you solve # 3 > 2|x-2|+5|x-2|#?

1 Answer
Geoff K.
Aug 8, 2017

Combine like terms; isolate the absolute value; solve for it; rewrite the inequality without the absolute value signs; isolate #x#.

#11/7"  "<"  "x"  "<"  "17/7#.

Explanation:

Given:

#3>2abs(x-2)+5abs(x-2)#

Step 1: Since the value inside the absolute brackets is the same for both terms on the RHS, we can add them together:

#3>7abs(x-2)#

Step 2: We can now isolate the absolute value brackets by dividing both sides by 7, like this:

#3/7>abs(x-2)#

Step 3: When the absolute value of something (#x-2#) is smaller than a specified value (#3/7#), that means its magnitude must be less than that value. In this case, that means #x-2# can be positive or negative, but it must be within #3/7# units of 0 either way. We can rewrite this as a new inequality, thereby eliminating the absolute value brackets:

#–3/7"  "<"  "x-2"  "<"  "3/7#

Step 4: Isolate #x#; add 2 to all three "sides".

#–3/7+2"  "<"  "x"  "<"  "3/7+2#

which reduces to

#11/7"  "<"  "x"  "<"  "17/7#.

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