# Inequality problemWatch

Thread starter 1 year ago
#1
Write the inequality -3 < x < 8 in the form |x - a| < b
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Thread starter 1 year ago
#2
Hello i have got very stuck trying to solve the question, any help would be appreciated.

Write the inequality -3 < x < 8 in the form |x-a| < b
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1 year ago
#3
(Original post by znx7)
Write the inequality -3 < x < 8 in the form |x - a| < b
From looking at the title of this thread I thought it was about geography/politics.
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1 year ago
#4
|x-5/2 |< 11/2
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1 year ago
#5
-3 < x < 8 in the form |x-a| < b

if x-a>0 then
x-a<b
x<a+b

if x-a<0 the
a-x<b
a-b<x

that's too much for a hint
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1 year ago
#6
(Original post by znx7)
Hello i have got very stuck trying to solve the question, any help would be appreciated.

Write the inequality -3 < x < 8 in the form |x-a| < b
|x-a|<b is equivalent to saying a-b<x<a+b
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Thread starter 1 year ago
#7
(Original post by Kotsiozz)
|x-5/2 |< 11/2
could you explain how you got that answer please?
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1 year ago
#8
|x-a|<b is equal to -b<x-a<b
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1 year ago
#9
|x-a|<b is equal to -b<x-a<b and then solve simultaneous eqs
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1 year ago
#10
(Original post by znx7)
Write the inequality -3 < x < 8 in the form |x - a| < b
This is easy if you understand that means the distance from a variable point to a fixed point .

So e.g. means that is always less than 3 units away from the +ve number 2 on the number line. That means it can go 3 less than 2, to -1, or 3 more than 2, to +5, i.e. . Here, can vary over an interval of length 6, but it can only go a distance of half that, 3, from the mid-point, 2.

The last inequality is not symmetrical, but note that an inequality like *is* symmetrical - we've lost symmetry somehow. Note that we can get back to a nice symmetrical form by subtracting 2 from all sides:

That is now symmetrical since 2 is the midpoint (or average) of -1 and 5 i.e. , and half the length of the interval is

So given we can rewrite it symmetrically as:

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