3 6 absolute inequalities-geom

Absolute Value Inequalities
(Geometric Method)
Frank Ma © 2011

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

In this section we solve simple absolute–value inequalities by the geometric methods. That is, we interpret these inequalities as statements about distances and obtain the solutions by taking simple measurements.

|x| means “the distance between x and 0”, so the expression |x| < c means “the distance between x and 0 is smaller than c”

|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).

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

We are to draw all numbers which are within 7 units from the number 0.

x
x
0
-7
-7
7
-7 < x < 7

|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
x
0
-7
-7
7
-7 < x < 7
The open circles means the end points are not included in the solution.

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

Because |x – y| is “the distance between x and y” so the expression |x – y| < cmeans the “the distance between x and y is less than c”.

Because |x – y| is “the distance between x and y” so the expression |x – y| < cmeans the “the distance 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.

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

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

|x – 2| < 3
less than 3

|x – 2| < 3
less than 3
Draw
2

|x – 2| < 3
less than 3
x
x
Draw
right 3
left 3
2

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

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

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”.

The geometric meaning of the inequality is that “the distance between x and 0 is 7 or more”. In picture
end point
included
0
x
x
-7
-7
7

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

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

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

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

end point
included
0
x
x
-7
-7
7
x < –7 or 7 < x
Because |x – y| is “the distance between 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|.

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.

For example,
|2x – 4| = |2(x – 2)| = 2|x – 2|
|2x + 3| = |2(x + 3/2)| = 2|x + 3/2|
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|
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|
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|
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|
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
the distance between x and –3/2

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

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

For example,
|2x – 4| = |2(x – 2)| = 2|x – 2|
|2x + 3| = |2(x + 3/2)| = 2|x + 3/2|
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
more than 3/2
x
x
right 3/2
left 3/2
–3/2

For example,
|2x – 4| = |2(x – 2)| = 2|x – 2|
|2x + 3| = |2(x + 3/2)| = 2|x + 3/2|
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
more than 3/2
x
x
right 3/2
left 3/2
–3/2
–3
0

For example,
|2x – 4| = |2(x – 2)| = 2|x – 2|
|2x + 3| = |2(x + 3/2)| = 2|x + 3/2|
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
more than 3/2
x
x
right 3/2
left 3/2
–3/2
–3
0
Hence x < –3 or 0 < x.

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
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. Pull out the coefficient of the x–term, translate and solve the expressions geometrically. Draw the solution.
21. |5 – 2x| ≤ 3
22. |3 + 2x| < 7
23. |–2x + 3| > 5
24. |4 – 2x| ≤ –3
25. |3x + 3| < 5
26. |3x + 1| ≤ 7
27. |8 – 4x| ≤ 3
28. |4x – 5| < 9
29. |4x + 1| ≤ 9

3 6 absolute inequalities-geom

by math123b on Aug 28, 2011

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