How do you solve #| x - 3 | + 2 >7#?

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Answer 1 · 2016-04-17T01:33:38

#x in (-oo, -2) uu (8, + oo)#

The first thing to do here is isolate the modulus on one side of the inequality by subtracting #2# from both sides

#|x-3| + color(red)(cancel(color(black)(2))) - color(red)(cancel(color(black)(2))) > 7 - 2#

#|x-3| > 5#

Now, because you're dealing with an absolute value inequality, you must account for two possible cases

#color(white)(a)#

When this is the case, the inequality becomes

#x - 3 > 5#

#x > 8#

#color(white)(a)#

This time, you will have

#-(x-3) > 5#

#-x + 3 >5#

#-x > 2 implies x < -2#

So, you need to have #x < -2# and #x > 8#, which means that the solution interval for this inequality will be

#color(green)(|bar(ul(color(white)(a/a)color(black)(x in (-oo, -2) uu (8, + oo))color(white)(a/a)|)))#