SME
Grade 6 Math · Problem type
Equivalent algebraic expressions
Use this for “Are 3(2 + x) and 6 + 3x equivalent?”
Diagram
6 + 12 = 18.
This is an example. Type your own values to start a new problem.
(
+
x
)
=
+
L
Value with parentheses
worked out (not typed)
R
Value multiplied out
worked out (not typed)
Formulas
L
=
n
(
m
+
x
)
(value
with
parentheses
=
number
outside
×
(number
inside
+
the
number
for
x))
18 = 3(2 + 4)
u
=
n
×
m
(first
part
=
number
outside
×
number
inside)
6 = 3 × 2
w
=
nx
(second
part
=
number
outside
×
the
number
for
x)
12 = 3 × 4
R
=
u
+
w
(value
multiplied
out
=
first
part
+
second
part)
18 = 6 + 12
Assumptions
•
Equivalent expressions are equal for every value of x.
•
3(2 + x) = 3 × 2 + 3 × x = 6 + 3x: multiply each part inside by the number outside.
•
Like terms add: 2x + 3x = 5x.
•
Numbers to 12 outside and 20 inside; x from 0 to 20.
Step-by-step
WE KNOW
n = 3 (number outside), m = 2 (number inside), x = 4 (number put in for x)
FIND
value with parentheses (L), first part (u), second part (w), value multiplied out (R)
Step 1 · Find value with parentheses (L)
Use L = n(m + x) (value with parentheses = number outside × (number inside + the number for x))
Put
4
in
for
x
.
Add
inside
the
parentheses
first.
3 × (2 + 4) = 3 × 6
L
=
18
Step 2 · Find first part (u)
Use u = n × m (first part = number outside × number inside)
Multiply the number inside by the number outside.
u
=
3
×
2
u
=
6
Step 3 · Find second part (w)
Use w = nx (second part = number outside × the number for x)
Multiply
x
by
the
number
outside:
3
x
with
4
put
in
for
x
.
w
=
3
×
4
w
=
12
Step 4 · Find value multiplied out (R)
Use R = u + w (value multiplied out = first part + second part)
Multiply
each
part
inside
by
3:
3(2
+
x
)
=
3
×
2
+
3
×
x
=
6
+
3
x
.
Add
the
two
parts.
R
=
6
+
12
R
=
18
(the
value
with
parentheses
is
18
too:
the
two
match
for
every
x
)
Check
Put the numbers back in. Both sides should match.
18 = 3 × (2 + 4) ✓
6 = 3 × 2 ✓
12 = 3 × 4 ✓
18 = 6 + 12 ✓
Topic check
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