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

1 Answer
Apr 17, 2016

Answer:

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

Explanation:

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)#

  • #color(purple)(ul(color(black)(x - 3 >= 0 implies |x- 3| = x-3))#

When this is the case, the inequality becomes

#x - 3 > 5#

#x > 8#

#color(white)(a)#

  • #color(purple)(ul(color(black)(x - 3 <0 implies | x -3| = -(x-3)))#

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)|)))#