3 3 absolute inequalities-geom-x

Absolute Value Inequalities
(Geometric Method)

In this section we solve simple absolute–value inequalities
by the geometric method.

by the geometric method.
Example A. Translate the meaning of |x| < 7 and draw x.

by the geometric method. We interpret these absolute-value
inequalities as statements about distances then obtain the
solutions by taking measurements on the real line.

|x| means “the distance between x and 0”,

so the expression |x| < c means “the distance between x
and 0 is smaller than c” (provided that c is not negative in
which case no such x exists).

|x| < 7
the distance between x and 0

|x| < 7
the distance between x and 0 is less than 7.

These are all the numbers which
are within 7 units from 0, from –7 to 7.
|x| < 7

–7 < x < 7-7-7 70
x
x
|x| < 7

–7 < x < 7-7-7 70
x
x
The open circles means the end points are not included in
the solution.
|x| < 7

I. (One Piece | |–Inequalities)
If |x| < c then –c < x < c.

Recall that |x – y| translates into “the distance between
x and y” so the expression |x – y| < c means the “the distance
between x and y is less than c”.

between x and y is less than c”. We use this geometric
meaning to solve these problems by simple drawings.

Example B. Translate the meaning of |x – 2| < 3 and solve.

|x – 2| < 3
Translate the symbols to a geometric description.

|x – 2| < 3
the distance between x and 2

|x – 2| < 3
the distance between x and 2 less than 3

|x – 2| < 3
2
Draw

|x – 2| < 3
2
x x
right 3left 3
Draw

|x – 2| < 3
–1 5
2
x x
right 3left 3
Draw

|x – 2| < 3
Hence –1 < x < 5.
–1 5
2
x x
right 3left 3
Draw

I. (Two Piece | |–Inequalities)
If |x| > c then x < –c or c < x.

Example C. Translate the meaning of |x| ≥ 7 and draw.

The geometric meaning of the inequality is that “the distance
between x and 0 is 7 or more”.

between x and 0 is 7 or more”. In picture
-7-7 7
0
x x
end point
included

x < –7 or 7 < x
-7-7 7
0
x x
end point
included

Note that the answer uses the word “or” for expressing the
choices of two sections–not the word “and”.
x < –7 or 7 < x
-7-7 7
0
x x
end point
included

choices of two sections–not the word “and”. The word “and”
is used when there are multiple conditions to be met.
x < –7 or 7 < x
-7-7 7
0
x x
end point
included

x < –7 or 7 < x
-7-7 7
0
x x
x and y” so the expression |x – y| > c means the “the distance
between x and y is more than c”.
end point
included

x < –7 or 7 < x
-7-7 7
0
x x
x and y” so the expression |x – y| > c means the “the distance
between x and y is more than c”.
We recall that we may break up | | for multiplication i.e.
|x * y| = |x| * |y|.
end point
included

We use this fact to pull out the coefficient of the x.

For example,
|2x – 4| =

For example,
|2x – 4| = |2(x – 2)|

For example,
|2x – 4| = |2(x – 2)| = 2|x – 2|

For example,
|2x – 4| = |2(x – 2)| = 2|x – 2|
|2x + 3| = |2(x + 3/2)|

For example,
|2x – 4| = |2(x – 2)| = 2|x – 2|
|2x + 3| = |2(x + 3/2)| = 2|x + 3/2|

For example,
|2x – 4| = |2(x – 2)| = 2|x – 2|
|2x + 3| = |2(x + 3/2)| = 2|x + 3/2|
We use this step to help us to extract the geometric
information.

For example,
|2x – 4| = |2(x – 2)| = 2|x – 2|
|2x + 3| = |2(x + 3/2)| = 2|x + 3/2|
Example D. Solve geometrically |2x + 3| > 3.
information.

For example,
|2x – 4| = |2(x – 2)| = 2|x – 2|
|2x + 3| = |2(x + 3/2)| = 2|x + 3/2|
information.
We want |2x + 3| = |2(x + 3/2)|

For example,
|2x – 4| = |2(x – 2)| = 2|x – 2|
|2x + 3| = |2(x + 3/2)| = 2|x + 3/2|
information.
We want |2x + 3| = |2(x + 3/2)| = 2|x + 3/2| > 3

For example,
|2x – 4| = |2(x – 2)| = 2|x – 2|
|2x + 3| = |2(x + 3/2)| = 2|x + 3/2|
information.
We want |2x + 3| = |2(x + 3/2)| = 2|x + 3/2| > 3 div. by 2
|x + 3/2| > 3/2

For example,
|2x – 4| = |2(x – 2)| = 2|x – 2|
|2x + 3| = |2(x + 3/2)| = 2|x + 3/2|
information.
We want |2x + 3| = |2(x + 3/2)| = 2|x + 3/2| > 3 div. by 2
|x + 3/2| > 3/2 or
|x – (–3/2)| > 3/2

For example,
|2x – 4| = |2(x – 2)| = 2|x – 2|
|2x + 3| = |2(x + 3/2)| = 2|x + 3/2|
the distance between x and –3/2
information.
We want |2x + 3| = |2(x + 3/2)| = 2|x + 3/2| > 3 div. by 2
|x + 3/2| > 3/2 or
|x – (–3/2)| > 3/2

For example,
|2x – 4| = |2(x – 2)| = 2|x – 2|
|2x + 3| = |2(x + 3/2)| = 2|x + 3/2|
the distance between x and –3/2 more than 3/2
information.
We want |2x + 3| = |2(x + 3/2)| = 2|x + 3/2| > 3 div. by 2
|x + 3/2| > 3/2 or
|x – (–3/2)| > 3/2

For example,
|2x – 4| = |2(x – 2)| = 2|x – 2|
|2x + 3| = |2(x + 3/2)| = 2|x + 3/2|
–3/2
information.
We want |2x + 3| = |2(x + 3/2)| = 2|x + 3/2| > 3 div. by 2
|x + 3/2| > 3/2 or
|x – (–3/2)| > 3/2

For example,
|2x – 4| = |2(x – 2)| = 2|x – 2|
|2x + 3| = |2(x + 3/2)| = 2|x + 3/2|
–3/2
x x
right 3/2left 3/2
information.
We want |2x + 3| = |2(x + 3/2)| = 2|x + 3/2| > 3 div. by 2
|x + 3/2| > 3/2 or
|x – (–3/2)| > 3/2

For example,
|2x – 4| = |2(x – 2)| = 2|x – 2|
|2x + 3| = |2(x + 3/2)| = 2|x + 3/2|
–3 0
–3/2
x x
right 3/2left 3/2
information.
We want |2x + 3| = |2(x + 3/2)| = 2|x + 3/2| > 3 div. by 2
|x + 3/2| > 3/2 or
|x – (–3/2)| > 3/2

For example,
|2x – 4| = |2(x – 2)| = 2|x – 2|
|2x + 3| = |2(x + 3/2)| = 2|x + 3/2|
Hence x < –3 or 0 < x.
–3 0
–3/2
x x
right 3/2left 3/2
information.
We want |2x + 3| = |2(x + 3/2)| = 2|x + 3/2| > 3 div. by 2
|x + 3/2| > 3/2 or
|x – (–3/2)| > 3/2

Let’s express intervals as absolute value inequalities.

If –c < x < c then |x| < c, i.e.

–c 0
right cleft c
+c
|x| < c

–c 0
right cleft c
x and y”
+c
|x| < c

–c 0
right cleft c
x and y” so the expression |x – y| < c means the
“the distance between x and y is less than c”.
+c
|x| < c

–c 0
right cleft c
We use this geometric meaning to write intervals into
inequalities. Specifically, we need to locate the midpoint of
the given interval first.
+c
|x| < c

–c 0
right cleft c
+c
|x| < c
The mid-point m between two numbers x and y
In picture:
o bm

–c 0
right cleft c
+c
|x| < c
The mid-point m between two numbers x and y is their average
that is m = .a + b
2 In picture:
o b
(a+b)/2
m

–c 0
right cleft c
+c
|x| < c
The mid-point m between two numbers x and y is their average
that is m = .a + b
2
For example, the mid-point of
2 and 4 is (2 + 4)/2 = 3.
In picture:
o b
(a+b)/2
m
2 4
(2+4)/2
m = 30

Example E. Express [2, 4] as an absolute value inequality in x.
2 40

To write an interval [a, b] into an absolute value inequality,
a b
2 40

view [a, b] as a “circle”
a b
2 40

view [a, b] as a “circle” with center at its midpoint m = (a + b)/2,
a bm = (b + a )/2
2 40

with radius r = (b – a)/2 which is half of the distance from a to b.
2 40
a b
r = (b – a) /2
m = (b + a )/2

2 40
a b
r = (b – a) /2
m = (b + a )/2
l x – m l ≤ r
An abs-value inequality

2 4m=30
a b
r = (b – a) /2
m = (b + a )/2
l x – m l ≤ r
Viewing [2, 4] as a circle, its center is at m = (2 + 4) / 2 = 3,

2 4m=30
a b
r = (b – a) /2
m = (b + a )/2
l x – m l ≤ r
with radius |4 – 2| / 2 = 1

2 4m=30
a b
r = (b – a) /2
m = (b + a )/2
l x – m l ≤ r
with radius |4 – 2| / 2 = 1 so the interval is lx – 3l ≤ 1.

Il. (Two–Piece | |–Inequalities)

If x < –c or c < x then c < |x| (c > 0), i.e.

–c 0
right cleft c
+c

–c 0
right cleft c
+c
c < |x|
An abs-value
inequality

–c 0
right cleft c
+c
c < |x|
An abs-value
inequality
Likewise the expression c < |x – y| means the
“the distance between x and y is more than c”.

–c 0
right cleft c
+c
c < |x|
An abs-value
inequality
Example F. Express the following two intervals
as an absolute value inequality in x.
2 40

–c 0
right cleft c
+c
c < |x|
An abs-value
inequality
2 40
Let’s find the abs-value inequality of the gap (2, 4).

–c 0
right cleft c
+c
c < |x|
An abs-value
inequality
2 4m=30
The interval (2, 4) has its center at m = (2 + 4) / 2 = 3,
with radius (4 – 2)/ 2 = 1

–c 0
right cleft c
+c
c < |x|
An abs-value
inequality
2 4m=30
with radius (4 – 2)/ 2 = 1 so (2, 4) is lx – 3l < 1.
lx – 3l < 1

–c 0
right cleft c
+c
c < |x|
An abs-value
inequality
2 4m=30
with radius (4 – 2)/ 2 = 1 so (2, 4) is lx – 3l < 1.
So the intervals to the two sides is lx – 3l >1.
lx – 3l < 1

Ex. A. Translate and solve the expressions geometrically.
Draw the solution.
1. |x| < 2 2. |x| < 5 3. |–x| < 2 4. |–x| ≤ 5
5. |x| ≥ –2 6. |–2x| < 6 7. |–3x| ≥ 6 8. |–x| ≥ –5
9. |3 – x| ≥ –5 10. |3 + x| ≤ 7 11. |x – 9| < 5
12. |5 – x| < 5 13. |4 + x| ≥ 9 14. |2x + 1| ≥ 3
21. |5 – 2x| ≤ 3 22. |3 + 2x| < 7 23. |–2x + 3| > 5
24. |4 – 2x| ≤ –3 25. |3x + 3| < 5 26. |3x + 1| ≤ 7
15. |x – 2| < 1 16. |3 – x| ≤ 5 17. |x – 5| < 5
18. |7 – x| < 3 19. |8 + x| < 9 20. |x + 1| < 3
Ex. B. Divide by the coefficient of the x–term, translate and
solve the expressions geometrically. Draw the solution.
27. |8 – 4x| ≤ 3 28. |4x – 5| < 9 29. |4x + 1| ≤ 9

Ex. C. Express the following intervals as absolute value
inequalities in x.
30. [–5, 5]
31. (–5, 5) 32. (–5, 2) 33. [–2, 5]
34. [7, 17] 35. (–49, 84) 36. (–11.8, –1.6) 37. [–1.2, 5.6]
–2 4
–15 –2
8 38
0 a
–a –a/2
a – b a + b
39.
42.
40.
41.
38.
43.

3 3 absolute inequalities-geom-x

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