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1.) |2x| >= -64

2.) |x + 9| >= 17

3.) |3x +7| =< 26

4.) |6x + 25| + 14 < 6

2006-08-09 20:02:46 · 3 answers · asked by No One! 2 in Science & Mathematics Mathematics

3 answers

remeber the definition: |z| is z if z>0 and -z if z<0

so you solve such equation by considering two cases, (z>=0 and z<0), replacing |z| with z or -z, solve like a regular equation and intersecting with condition for z>0 or z<0, then combining the solutions for two cases

by the way, 1) has every x as solution and 4) has no solutions

Example for 2)

case 1: x+9>=0 , i.e. x>=-9
equation becomes x+9>=17 i.e. x>=8
Intersect x>=-9 and x>=8, get x>=8

case 2: x+9<0 , i.e. x<-9
equation becomes -(x+9)>=17 i.e. x<=-26
intersect x<-9 and x<=-26, get x<=-26

Combine the cases: x>=8 or x<=-26
Plot - draw the real line, highlight to the right of 8 and to the left of -26

2006-08-09 20:06:48 · answer #1 · answered by Anonymous · 0 0

i have already answered 1-3 so i will answer #4.

|6x + 25| + 14 < 6
|6x + 25| < -8

since you can't get a negative value for absolute values, this problem is unsolvable.

2006-08-10 02:15:48 · answer #2 · answered by Sherman81 6 · 0 0

I do not if vertical bar are absolute values. let think it is.
So
1) All x because absolute values are always positive
2) x>=8 or x<=-26
3)-11<=x <=19/3
4)No solution as the absolute is always positive

2006-08-09 20:18:31 · answer #3 · answered by fred 055 4 · 0 0

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