SME
Grade 9 Math · Problem type
Absolute value inequality: beyond
Use this for “Solve |2x − 1| ≥ 5.”
Diagram
|x − 0.5| ≥ 2.5: every number at least 2.5 from 0.5
0.5 − 2.5 = −2
0.5 + 2.5 = 3
So x ≤ −2 or x ≥ 3.
Both circles are closed: −2 and 3 are taken in.
Test 1: |1 − 0.5| = 0.5, and 0.5 ≥ 2.5 is false.
This is an example. Type your own values to start a new problem.
|
a
x
+
(
b
)
|
≥
c
x
≤
L
or
x
≥
U
h
Center
worked out (not typed)
d
Distance from the center
worked out (not typed)
t
Test number
example
k
Is the test true?
worked out (not typed)
Formulas
h
=
−
b
a
1
2
=
−
−1
2
d
=
c
÷
|
a
|
5
2
=
5
÷
|2|
L
=
h
−
d
−2
=
1
2
−
5
2
U
=
h
+
d
3
=
1
2
+
5
2
test
t
in
|ax
+
b
|
≥
c
:
k
test 1 in |2x + (−1)| ≥ 5: false
Assumptions
•
|u| ≥ c means u is at least c from 0: u ≤ −c or u ≥ c.
•
The solutions are two rays pointing away from the center.
•
A negative c makes every number a solution.
Step-by-step
WE KNOW
a = 2, b = −1, c = 5, t = 1
FIND
center (h), distance from the center (d), lower bound (L), upper bound (U), is the test true? (k)
Step 1 · Find center (h)
Use h = −b ÷ a
The
center,
where
ax
+
b
=
0.
h
=
−
−1
2
h
=
1
2
Step 2 · Find distance from the center (d)
Use d = c ÷ |a|
How
far
each
end
is
from
the
center:
|ax
+
b
|
=
|
a
|
×
|x
−
h
|.
d
=
5
÷
|2|
d
=
5
2
d
=
5
2
Step 3 · Find lower bound (L)
Use L = h − d
Go the distance to the left of the center.
L
=
1
2
−
5
2
L
=
−2
Step 4 · Find upper bound (U)
Use U = h + d
And the same distance to the right.
U
=
1
2
+
5
2
U
=
3
→
x
≤
−2
or
x
≥
3
Step 5 · Find is the test true? (k)
Use test t in |ax + b| ≥ c: k
Put
the
test
number
inside
the
bars,
then
compare
its
distance
from
0
with
c
.
2(1) − 1 = 1
|1| = 1
1 ≥ 5 is false
k
=
false
Check
Put the numbers back in. Both sides should match.
1
2
=
−
−1
2
✓
5
2
=
5
÷
|2|
✓
−2
=
1
2
−
5
2
✓
3
=
1
2
+
5
2
✓
1 < 5, so the test number is not a solution ✓
Topic check
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