Question 168553: absolute value [2x+1]less than or equal to 11
Answer by gonzo(654)   (Show Source):
You can put this solution on YOUR website! the basic formula for solving an inequality is:
if |x| = d, then x = d, and x = -d
if |x| < d, then x < d, and x > -d
if |x| > d, then x > d, and x < -d
likewise,
if |x| <= d, then x <= d, and x >= -d
if |x| >= d, then x >= d, and x >= -d
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if |2x+1| <= 11, then
2x + 1 <= 11, and 2x + 1 >= -11
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working with 2x + 1 <= 11:
2x + 1 <= 11
2x <= 10
x <= 5
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working with 2x + 1 >= -11:
2x + 1 >= -11
2x >= -12
x >= -6
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answer is x <= 5 and x >= -6
this translates to:
-6 <= x <= 5
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to test:
let x = -7
|2x+1| = |-14+1| = |-13| = 13 NOT <= 11 = ok since -7 is not >= -6 which violates one of the conditions.
let x = 6
|2x+1| = |12+1| = |13| = 13 NOT <= 11 = ok since 6 is not <= 5 which violates one of the conditions.
let x = -6
|2x+1| = |-12+1| = |-11| = 11 <= 11 = ok since -6 = -6 which is one of the conditions.
let x = 5
|2x+1| = |10+1| = |11| = 11 <= 11 = ok since 5 = 5 which is one of the conditions.
let x = 0
|2x+1| = |-+1| = |1| = 1 <= 11 = ok since 1 > -6 and < 5 which is one of the conditions.
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answer proves to be good.
|2x-1| <= 11 is a true equation when:
-6 <= x <= 5
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