Compare the process of solving |x – 1| + 1 < 15 to that of solving |x – 1| + 1 > 15. Check all of the following you includ

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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 includ

ed 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.

Mathematics

1 answer:

zepelin [54] · Answer 1 · 2023-03-03T13:53:46+03:00

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.

<h3>What is inequality?</h3>

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.