I have a slightly different answer .... |x - 2| + |x + 4| + |x + 3| = 10
Remember that if the quantity within the abolute value bars is positive, just remove the
bars; but, if what is inside, you need to rewrite the expression as: - ( expression ).
There are 8 cases:
1) (+,+,+) x - 2 >= 0 ---> x >= 2
x - 4 >= 0 ---> x >= 4
x + 3 >= 0 ---> x >= -3
< This is possible for x >= 4 >
|x - 2| + |x + 4| + |x + 3| = 10 ---> (x - 2) + (x - 4) + (x + 3) = 10
3x - 3 = 10
3x = 13
This answer works because x >= 4: x = 13/3
2) (+, +, -) x - 2 >= 0 ---> x >= 2
x - 4 >= 0 ---> x >= 4
x + 3 <= 0 ---> x <= -3
< This is impossible because x can't both be greater than 4 and less that -3 >
3) (+,-,+) x - 2 >= 0 ---> x >= 2
x - 4 <= 0 ---> x <= 4
x + 3 >= 0 ---> x >= -3
< This is possible only for x >= 2 and x <= 4 >
|x - 2| + |x + 4| + |x + 3| = 10 ---> (x - 2) - (x - 4) + (x + 3) = 10
x + 5 = 10
This answer doesn't work because it isn't between 2 and 4: x = 5
4) (+,-,-) x - 2 >= 0 ---> x >= 2
x - 4 <= 0 ---> x <= 4
x + 3 <= 0 ---> x <= -3
< This is impossible because x can't be both >= 2 and <= -3 >
5) (-, +, +) x - 2 <= 0 ---> x <= 2
x - 4 >= 0 ---> x >= 4
x + 3 >= 0 ---> x >= -3
< This is impossible because x can't be both <= 2 and >= 4 >
6) (-,+,-) x - 2 <= 0 ---> x <= 2
x - 4 >= 0 ---> x >= 4
x + 3 <= 0 ---> x ><-3
< This is impossible because x can't be both >= 4 and <= 2 >
7) (-,-,+) x - 2 <= 0 ---> x <= 2
x - 4 <= 0 ---> x <= 4
x + 3 >= 0 ---> x >= -3
< This is possible only for x >= -3 and x <= 2 >
|x - 2| + |x + 4| + |x + 3| = 10 ---> - (x - 2) - (x - 4) + (x + 3) = 10
-x + 9 = 10
- x = 1
This answer works because it is between -3 and 2 x = -1
8) (-,-,-) x - 2 <= 0 ---> x <= 2
x - 4 <= 0 ---> x <= 4
x + 3 <= 0 ---> x <= -3
< This is possible only for x <= -3>
|x - 2| + |x + 4| + |x + 3| = 10 ---> - (x - 2) - (x - 4) - (x + 3) = 10
-3x + 3 = 10
- 3x = 7
This answer doesn't work because it isn't <= -3 x = -7/3
There are two answers:: -1 and 13/3