SME
Grade 9 Math · Algebra 1
Absolute value equations and inequalities
Use this for “Solve |2x − 3| = 7” and find both solutions.
Diagram
|x − 1.5| = 3.5: the numbers exactly 3.5 from 1.5
1.5 − 3.5 = −2
1.5 + 3.5 = 5
The solutions are −2 and 5; nothing between them is one.
This is an example. Type your own values to start a new problem.
|
a
x
+
(
b
)
|
=
c
x
=
x₁
or
x
=
x₂
h
Center
worked out (not typed)
d
Distance from the center
worked out (not typed)
Formulas
x₁
=
c
−
b
a
5
=
7 − (−3)
2
x₂
=
−
c
−
b
a
−2
=
−7 − (−3)
2
h
=
−
b
a
3
2
=
−
−3
2
d
=
c
÷
|
a
|
7
2
=
7
÷
|2|
Assumptions
•
|u| = c means u is c from 0: u = c or u = −c.
•
A negative c has no solution, and c = 0 has just one.
•
Check each answer in the first equation.
Step-by-step
WE KNOW
c = 7, a = 2, b = −3
FIND
solution from ax + b = c (x₁), solution from ax + b = −c (x₂), center (h), distance from the center (d)
Step 1 · Find solution from ax + b = c (x₁)
Use x₁ = (c − b) ÷ a
Set
ax
+
b
=
c
:
take
b
from
both
sides,
then
divide
by
a
.
x₁
=
7 − (−3)
2
x₁
=
10
2
x₁
=
5
Step 2 · Find solution from ax + b = −c (x₂)
Use x₂ = (−c − b) ÷ a
Set
ax
+
b
=
−
c
the
same
way.
When
c
=
0
the
two
are
one
solution.
x₂
=
−7 − (−3)
2
x₂
=
−
4
2
x₂
=
−2
Step 3 · Find center (h)
Use h = −b ÷ a
The
center,
where
ax
+
b
=
0.
h
=
−
−3
2
h
=
3
2
Step 4 · 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
Check
Put the numbers back in. Both sides should match.
5
=
7 − (−3)
2
✓
−2
=
−7 − (−3)
2
✓
3
2
=
−
−3
2
✓
7
2
=
7
÷
|2|
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.