Solve:||x-3|-2|≤1
Please show work!
Thanks

Step-by-step explanation:

Hello,

We know that

So we need to take into account two cases

Case 1 -

Then, |x-3|=x-3

||x-3|-2|=|x-3-2|=|x-5|

Either x-5 is positive and then |x-5|=x-5 and

Or x-5 is negative and then, |x-5|=-x+5

So the solution is [4;6]

Case 2 -

Then, |x-3|=-x+3

||x-3|-2|=|-x+3-2|=|-x+1|

Either -x+1 is positive and then |-x+1|=-x+1 and

Or -x+1 is negative and then, |-x+1|=x-1

So the solution is [0;2]

Conclusion

The solution is [0;2]∪[4;6]

Thanks

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

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step-by-step explanation: