Inequalities are solved much like equalities. The only thing you have to watch out for is multiplying or dividing by negative numbers. When you do that the direction of the sign must change:
EXAMPLE 1:
6 + 11x > -60
Subtract 6 from both sides:
11x > -60 - 6
11x > -66
Divide both sides by 11:
x > -66/11
x > -6
That one didn't involve a sign change.
EXAMPLE 2:
-26 < 4 - 5x
Subtract 4 from both sides:
-26 - 4 < -5x
-30 < -5x
Now divide both sides by -5 (remember to change the sign!)
-30/-5 > x
6 > x
Now swap the whole equation (reverse the sign too):
x < 6
Does that help?
2007-12-05 12:31:19
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answer #1
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answered by Puzzling 7
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Subtract 6 from both sides. 11x > -66. Divide
x > -11
Subtract 4 from both sides. -30 < -5x. Divide
6 > x since inequality is reversed when dividing by a negative. 6 > x means x < 6.
2007-12-05 20:30:23
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answer #2
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answered by richardwptljc 6
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6 + 11x > - 60
11x > - 66 {subtract 6 from both sides}
x > - 6 {divide both sides by 6}
-26 < 4 - 5x
5x - 26 < 4 {subtract 5x from both sides}
5x < 30 {add 26 to both sides}
x < 6 {divide both sides by 5}
2007-12-05 20:34:48
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answer #3
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answered by kindricko 7
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Inqeualities are simple, just make sure you get a number on one side of the ineuqlity sign, and a variable(x) on the other side. Then if you get a coefficient like 2x on one side of the inequality sign and a number like 24 on the other side, divide both sides by 2 and x will be 12. here is an example.
2x<24, divide both sides by 2 and get 12
or like 3x+8<26, subtract 8 from both sides, then divide each side by 3 and get 6 as x.
2007-12-05 20:32:25
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answer #4
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answered by Mr. 3.14™ 7
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solve just like equation, unless you divide by negative, then flip sign
6+11x >-60
11x>-66
x>-6
-26 < 4 - 5x
-30<-5x
6>x
could be written as x<6
2007-12-05 20:30:37
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answer #5
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answered by RickSus R 5
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6 + 11x > -60 11x > -66 (SUBTRACT -6 FROM BOTH SIDES) x > -6 -30 < -5x x< 6
2007-12-05 20:30:41
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answer #6
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answered by APPLES 2
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6+11x>-60
11x>-60--6
11x>-66
x>-6
2007-12-05 20:37:57
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answer #7
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answered by des 1
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2007-12-05 20:28:46
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answer #8
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answered by Anonymous
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