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Compare the process of solving |x – 1| + 1 < 15 to that of solving |x – 1| + 1 > 15. Check all of the following you included in your response. Both absolute values would need to be isolated first. You would need to write a compound inequality for each. Both compound inequalities would compare x – 1 to –15 and 15. The inequality with “<” would use an “and” statement, while the “>” would use an “or” statement.

Respuesta :

okpalawalter8

The required Comparison of the inequalities are

  • The |x – 1| + 1 > 15 represents the value of x lies between 13<x<15.

The range of values encompassing the region's junction is (-13, 15).

  • If x is more than or equal to 15, then x-11+1>15 indicates the value of x is greater than or equal to 13. None of the regions in the intersection are empty.

What is inequality?

When comparing two numbers, an inequality indicates whether one is less than, larger than, or not equal to the other.

We take into account the various variables of the inequality

|x – 1| + 1 > 15

Therefore

|-x-1|+1-1<15-1

|-x-1|-1 <14

13<x<15

The required region lies between the inequality -13 <x< 15.

Simplify the inequality Ix-11+1 > 15 we get,

|x-1|+1 > 15

|x+1| +1-1 >15-1

|x-1| > 14

x> 15  

x<-13

  • If x has a value between -13 and x + 15, then the expression "|x-1|+1+115" is true. The range "(-13, 15)" contains the intersection of the region.
  • If "|x-1|+1>15" then either "x >15" or "x-13" applies to the value of x. This region's intersection is unoccupied.

Read more about inequalities

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