SME
Grade 9 Math · Problem type
Compound inequality with and
Use this for “Solve −5 < 3x + 4 ≤ 13” (the x term positive).
Diagram
−3 < x ≤ 3: −3 < x and x ≤ 3, both true at once: the numbers between.
The open circle leaves out −3; the closed circle takes in 3.
Test 0: 0 > −3 is true, 0 ≤ 3 is true: both true, a solution.
This is an example. Type your own values to start a new problem.
l
a
x
+
b
r
L
x
U
x₀
Test number
example
h
Is the test true?
worked out (not typed)
Formulas
L
=
l
−
b
a
−3
=
−5 − 4
3
U
=
r
−
b
a
3
=
13 − 4
3
test
x₀
in
l
(sign
s
)
ax
+
b
(sign
t
)
r
:
h
test 0 in −5 (sign <) 3x + 4 (sign ≤) 13: true
Assumptions
•
Do the same to all three parts: take b from each, then divide each by a.
•
Here a is positive, so the signs stay: < leaves a bound out, ≤ takes it in.
•
The solutions are between the two bounds: both parts must be true.
Step-by-step
WE KNOW
l = −5, s: <, a = 3, b = 4, t: ≤, r = 13, x₀ = 0
FIND
lower bound (L), upper bound (U), is the test true? (h)
Step 1 · Find lower bound (L)
Use L = (l − b) ÷ a
Take
b
from
the
left
part,
then
divide
it
by
a
.
L
=
−5 − 4
3
L
=
−
9
3
L
=
−3
Step 2 · Find upper bound (U)
Use U = (r − b) ÷ a
Do the same to the right part.
U
=
13 − 4
3
U
=
9
3
U
=
3
→
−3
<
x
≤
3
Step 3 · Find is the test true? (h)
Use test x₀ in l (sign s) ax + b (sign t) r: h
Put the test number in the middle: is it true? Both parts must be true.
3(0) + 4 = 4
−5 < 4 ≤ 13 is true
h
=
true
Check
Put the numbers back in. Both sides should match.
−3
=
−5 − 4
3
✓
3
=
13 − 4
3
✓
3 × 0 + 4 = 4, so 0 is a solution ✓
Topic check
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