SME
Grade 9 Math · Problem type
Absolute value inequality: within
Use this for “Solve |2x + 1| ≤ 7.”
Diagram
|x + 0.5| ≤ 3.5: every number at most 3.5 from −0.5
−0.5 − 3.5 = −4
−0.5 + 3.5 = 3
So −4 ≤ x ≤ 3.
Both circles are closed: −4 and 3 are taken in.
Test 1: |1 + 0.5| = 1.5, and 1.5 ≤ 3.5 is true.
This is an example. Type your own values to start a new problem.
|
a
x
+
b
|
≤
c
L
≤
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
|
7
2
=
7
÷
|2|
L
=
h
−
d
−4
=
−
1
2
−
7
2
U
=
h
+
d
3
=
−
1
2
+
7
2
test
t
in
|ax
+
b
|
≤
c
:
k
test 1 in |2x + 1| ≤ 7: true
Assumptions
•
|u| ≤ c means u is within c of 0: −c ≤ u ≤ c, one “and” inequality.
•
The solutions are between two bounds, centered where ax + b = 0.
•
A negative c has no solution: a distance is never negative.
Step-by-step
WE KNOW
a = 2, b = 1, c = 7, 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
=
7
÷
|2|
d
=
7
2
d
=
7
2
Step 3 · Find lower bound (L)
Use L = h − d
Go the distance to the left of the center.
L
=
−
1
2
−
7
2
L
=
−4
Step 4 · Find upper bound (U)
Use U = h + d
And the same distance to the right.
U
=
−
1
2
+
7
2
U
=
3
→
−4
≤
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 = 3
|3| = 3
3 ≤ 7 is true
k
=
true
Check
Put the numbers back in. Both sides should match.
−
1
2
=
−
1
2
✓
7
2
=
7
÷
|2|
✓
−4
=
−
1
2
−
7
2
✓
3
=
−
1
2
+
7
2
✓
3 < 7, so the test number is a solution ✓
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Absolute value inequality: beyond
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