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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

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Topic review (newest first)

mikau
2006-01-20 18:26:19

incase your not sure where I got the requirements from. If you have an expression such as |x + 1| if the value of x + 1 is less then zero, it will be negative. Therefore, if x + 1 is less then 1, the expression will haves its sign changed. Thus |x + 1| equals -(x + 1) when the expression is less then 1. To find when x + 1 is less then zero: x + 1 < 0  solved x < -1. Lets try a value less then -1 as a test. |-2 + 1| = 1. But -2 is less then 1 so instead of | x + 1| we can write -( x + 1) x = -2 so

-(-2 + 1)

-(-1)

1.

This gave us the same answer as the absolute value expression. Why? Because thats what the absolute value symbols mean. When the value is negative (less then 0) the sign is reversed to make if positve.

Thus   if (x < 0) |x| = -x

mikau
2006-01-20 18:17:35

doesn't look easy at all. Absolute value algebra is a nightmare!

Lets see, if x < 1 then the expression |x -1| will be - (x-1). If x > 4, then |8 - x| will be -(8 - 2x). Fortunatly, x can never be both less then 1 and greater then 4 so its either one or the other. Not both. Of course, if x is neither less then 1, nor greater then 4, then both the abolute value symbols can be eliminated. So lets check all three situations.

if x < 1

-(x - 1) - ( 8 - 2x) = 3  solved x = 10. 10 is not less then 1 so we discard this solution.


if (x > 4)

(x - 1) + (8 - 2x) = 3   solved  x = 4. 4 is not less then 4 so we discard this solution.


if (1 <= x <=4)

(x - 1) - (8 - 2x) = 3.   solved x = 4. 4 is greater then 1 and less then 4 so this is an acceptable solution.

So if my thinking is correct, x = 4 should be the solution.

RauLiTo
2006-01-20 17:23:49

X = ?

| X - 1 | - | 8 - 2x | = 3

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