Okay, so I understand absolute value.
And I understand solving inequalities.
But I don't understand how to solve absolute value inequalites when one side is a negative...for example....
[3-x] > -5
^^the parentheses are the absolute value signs.
pleasee help me!
2007-09-11 13:04:55 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I know what you're asking, and from what you said, I assume you know how to break these things into two parts (positive case and negative case)
But this problem seems to violate the rules you've learned. Think about the results you've gotten and look at the question again. think about absolute value. Whatever I put in for x, won't it be > -5????
So all values work! In other words, there is no value of x that will not make it be true! The whole number line is shaded in (if your teacher makes you draw number lines to show your answers, that is.)
Hope this helps you see the answer!
Doug
2007-09-11 13:11:58 · answer #1 · answered by douglas 2 · 2⤊ 0⤋
x>8
2007-09-11 20:10:22 · answer #2 · answered by aldrin 2 · 1⤊ 0⤋
3 subtract what is greater than -5 x>that value
or
-5-3=-8
reverse signs x<8
2007-09-11 20:11:18 · answer #3 · answered by SCIENCE_MAN_88@YAHOO.COM 2 · 1⤊ 0⤋
you have to get the value of the 3-x first before applying the absolute
2007-09-11 20:10:49 · answer #4 · answered by kenn 2 · 1⤊ 0⤋
[3-x] > -5
inequality is true for all x because the absolute value of anything is positive and therefore > -5
2007-09-11 20:27:17 · answer #5 · answered by ironduke8159 7 · 1⤊ 0⤋
switch the -5 and 3-x and change sign. -5<3-x
hope this helps peace. by the way i've always made a A in math.
2007-09-11 20:11:20 · answer #6 · answered by Shorttie 2 · 1⤊ 0⤋
3 -x > -5
-x > -8
x > 8
2007-09-11 20:09:00 · answer #7 · answered by Runa 7 · 0⤊ 0⤋