# Solving this equation of absolute value (with the step, please) :1. |x-3|+|x-2|+|x+1|<7 2. |x-3|=x-3 3. show it that |x|≤2 -->

beckden | High School Teacher | (Level 1) Educator

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1.   You divide the range up into ranges.   x>=3, x>=2 x>=-1 and x<=3.

in x>=3 all equations are positive so the equation becomes x-3+x-2+x+1=3x-4
3x-4<7 when 3x<11 or x<11/3 so the equation is true for x<11/3.

in 2<=x<=3 in this interval x-3 is negative so its absolute value can be calculated by 3-x. x-2 and x+1 are positive in this range so the equation becomes
3-x+x-2+x+1=x+2<7  x+2<7 when x<5, so the equation is true in this range and our solution now becomes 2<=x<11/3.

in -1<=x<=2 x-2 is negative and can be replaced with 2-x, x-3 is still negative and can be replaced with 3-x, but x+1 is still positive so our equation becomes
3-x+2-x+x+1=-x+6<7 when -x<1 or x>-1 so our range is now -1<x<11/3.  Note that x=-1 the equation is equal to 7 so -1 is not in our solution.

For x<-1 x-3, x-2 and x+1 are negative and can be replaced with their opposites, so the equation becomes 3-x+2-x+-1-x=-3x+4 and -3x+4<7 when
-3x<3 or x>-1 so the equation is never <7 in the interval x<=-1.  So we have our solution.

-1<x<11/7  is the solution to |x-3|+|x-2|+|x+1|<7

2.  |x-3|=x-3   since if x<3, x-3 is negative, and you cannot have an absolute value that is negative, so this is not true when x<3 (check a value out like 0, |0-3|=0-3 ==> |-3|=-3 ==> 3=-3 which is false.  You can also see that if x>=3, |x-3| = x-3 by definition. so this equation is true when x>=3.

3.  |x|<=2 --> |(2x^2+3x+2)/(x^2+2)|<=8

First look at (x^2+2) this is always >= 0, since x^2 is always >=0.
Now look at 2x^2+3x+2.  This is a parabola, where the vertex is at x = -b/2a = -3/(2(2)) = -3/4.   The y value of this parabola is (-3/4)^2 + 3(-3/4) + 2 = 9/16+ -9/4 + 2 = 9/16 + -36/16 + 32/16 = 5/16.  so this equation's minimum is positive.  So since both (2x^2+3x+2) and (x^2+2) are always positive then (2x^2+3x+2)/(x^2+2) is always positive and we can remove the absolute value signs around it.   (2x^2+3x+2)/(x^2+2) <=8.   so we want to solve when

Note that we can multiply both sides by (x^2+2) without worring about the <= because (x^2+2) is always positive.
(2x^2+3x+2)<= 8(x^2+2) we get
2x^2+3x+2 <= 8x^2+16
-6x^2 + 3x - 14 <= 0   We can find the maximum of this equation by the same method, find the vertex x = -b/2a = -3/(2(-6)) = 3/12 = 1/4.  We have
-6(1/4) + 3(1/4) - 14 = -6/4 + 3/4 - 56/4 = -59/4.  So since the maximum is < 0 this inequality -6x^2 + 3x - 14 is always true.   So for any value of x, |(2x^2+3x+2)/(x^2+2)| <= 8.

abdah12418 | Student, Undergraduate | (Level 2) eNoter

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for number 3 this is the equation

|x|≤2 --> |(2x^2 + 3x + 2) / (x^2+2)|≤8

Sory, cause i write the equation (with word) in that question was gone.