Inequality problem Watch

znx7
Badges: 3
Rep:
?
#1
Report Thread starter 1 year ago
#1
Write the inequality -3 < x < 8 in the form |x - a| < b
0
quote
reply
znx7
Badges: 3
Rep:
?
#2
Report 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
0
quote
reply
silversocks012
Badges: 17
Rep:
?
#3
Report 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.
0
quote
reply
Kotsiozz
Badges: 8
Rep:
?
#4
Report 1 year ago
#4
|x-5/2 |< 11/2
Posted on the TSR App. Download from Apple or Google Play
0
quote
reply
OregonHooligan
Badges: 14
Rep:
?
#5
Report 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
0
quote
reply
B_9710
Badges: 15
Rep:
?
#6
Report 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
0
quote
reply
znx7
Badges: 3
Rep:
?
#7
Report Thread starter 1 year ago
#7
(Original post by Kotsiozz)
|x-5/2 |< 11/2
could you explain how you got that answer please?
0
quote
reply
Kotsiozz
Badges: 8
Rep:
?
#8
Report 1 year ago
#8
|x-a|<b is equal to -b<x-a<b
Posted on the TSR App. Download from Apple or Google Play
0
quote
reply
Kotsiozz
Badges: 8
Rep:
?
#9
Report 1 year ago
#9
|x-a|<b is equal to -b<x-a<b and then solve simultaneous eqs
Posted on the TSR App. Download from Apple or Google Play
0
quote
reply
atsruser
Badges: 11
Rep:
?
#10
Report 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 |x-a| means the distance from a variable point x to a fixed point a.

So e.g. |x-2| &lt; 3 means that x 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. -1 &lt; x &lt; 5. Here, x 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 |x-a| &lt; b *is* symmetrical - we've lost symmetry somehow. Note that we can get back to a nice symmetrical form by subtracting 2 from all sides:

-1 &lt; x &lt; 5 \Leftrightarrow -1-2 &lt; x-2 &lt; 5-2 \Leftrightarrow -3 &lt; x-2 &lt; 3 \Leftrightarrow |x-2| &lt; 3

That is now symmetrical since 2 is the midpoint (or average) of -1 and 5 i.e. 2 = \frac{1}{2}(5+(-1)), and half the length of the interval is 3 = \frac{1}{2}(5-(-1))

So given a &lt; x &lt; b we can rewrite it symmetrically as:

|x-\frac{b+a}{2}| &lt; \frac{b-a}{2}
0
quote
reply
X

Reply to thread

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of Lincoln
    Brayford Campus Undergraduate
    Wed, 12 Dec '18
  • Bournemouth University
    Midwifery Open Day at Portsmouth Campus Undergraduate
    Wed, 12 Dec '18
  • Buckinghamshire New University
    All undergraduate Undergraduate
    Wed, 12 Dec '18

Do you like exams?

Yes (169)
18.61%
No (553)
60.9%
Not really bothered about them (186)
20.48%

Watched Threads

View All