SME
Grade 9 Math · Problem type
Compound inequality with or
Use this for “Solve 2x + 3 < −1 or 3x − 2 ≥ 7.”
Diagram
x < −2 or x ≥ 3: a number in either part is a solution.
The open circle leaves out −2; the closed circle takes in 3.
Test 4: 4 < −2 is false, 4 ≥ 3 is true: one part is enough, a solution.
This is an example. Type your own values to start a new problem.
a
x
+
b
<
c
or
d
x
+
(
e
)
≥
f
x
<
L
or
x
≥
U
t
Test number
example
h
Is the test true?
worked out (not typed)
Formulas
L
=
c
−
b
a
−2
=
−1 − 3
2
U
=
f
−
e
d
3
=
7 − (−2)
3
test
t
in
ax
+
b
<
c
or
dx
+
e
≥
f
:
h
test 4 in 2x + 3 < −1 or 3x + (−2) ≥ 7: true
Assumptions
•
Solve each part on its own.
•
The solutions are every number that makes at least one part true: two rays.
•
If the rays meet or overlap, every number is a solution.
Step-by-step
WE KNOW
a = 2, b = 3, c = −1, d = 3, e = −2, f = 7, t = 4
FIND
first bound (L), second bound (U), is the test true? (h)
Step 1 · Find first bound (L)
Use L = (c − b) ÷ a
First
part:
take
b
from
both
sides,
then
divide
by
a
.
L
=
−1 − 3
2
L
=
−
4
2
L
=
−2
Step 2 · Find second bound (U)
Use U = (f − e) ÷ d
Second
part:
take
e
from
both
sides,
then
divide
by
d
.
U
=
7 − (−2)
3
U
=
9
3
U
=
3
→
x
<
−2
or
x
≥
3
Step 3 · Find is the test true? (h)
Use test t in ax + b < c or dx + e ≥ f: h
Put the test number into both parts: is it true? One true part is enough.
2(4) + 3 = 11
3(4) − 2 = 10
11 < −1 or 10 ≥ 7 is true
h
=
true
Check
Put the numbers back in. Both sides should match.
−2
=
−1 − 3
2
✓
3
=
7 − (−2)
3
✓
3 × 4 − 2 = 10 ≥ 7, so 4 is a solution ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Refresh
Related lessons
Solving linear and compound inequalities
Main lesson
Compound inequality with and
Use this for “Solve −5 < 3x + 4 ≤ 13” (the x term positive).
Graph y (sign) mx + b
Use this for “Graph y ≥ 2x − 3 and test a point”, or with <, ≤ or >.
Whole-number answers
Use this for “$12 to join plus $4 a visit, and at most $50. How many visits?”