SME
Grade 9 Math · Problem type
Distribute first
Use this for “2(3x − 4) = 4x + 2” (numbers after distributing up to 10).
Diagram
6x − 8 = 4x + 2.
x = 5: each side is 22.
This is an example. Type your own values to start a new problem.
p
(
a
x
+
(
b
)
)
=
c
x
+
d
x
=
A
x after distributing
worked out (not typed)
B
Number after distributing
worked out (not typed)
Formulas
A
=
p
×
a
6 = 2 × 3
B
=
p
×
b
−8 = 2 × (−4)
p
×
(
a
×
x
+
b
)
=
c
×
x
+
d
2 × (3 × 5 + (−4)) = 4 × 5 + 2
A
×
x
+
B
=
c
×
x
+
d
6 × 5 + (−8) = 4 × 5 + 2
Assumptions
•
Multiply every term in the brackets by the number outside: p(ax + b) = pax + pb.
•
Then collect the x terms on one side and the numbers on the other.
•
Check by putting x back into both sides of the first equation.
Step-by-step
WE KNOW
d = 2, p = 2, a = 3, b = −4, c = 4
FIND
x after distributing (A), number after distributing (B), solution (x)
Step 1 · Find x after distributing (A)
Use A = p × a
Multiply
the
x
term
in
the
brackets
by
the
number
outside.
A
=
2
×
3
A
=
6
Step 2 · Find number after distributing (B)
Use B = p × b
Multiply the number in the brackets too: every term inside.
B
=
2
×
(−4)
B
=
−8
→
2(
3
x
−
4)
=
4
x
+
2
becomes
6
x
−
8
=
4
x
+
2
Step 3 · Find solution (x)
Use A × x + B = c × x + d
Take
cx
from
both
sides
so
the
x
terms
are
on
the
left,
take
B
away,
then
divide.
x
=
d
−
B
A
−
c
x
=
2 − (−8)
6 − 4
Take
4
x
from
both
sides:
2
x
−
8
=
2
Add
8
to
both
sides:
2
x
=
10
Divide
both
sides
by
2:
x
=
5
x
=
5
Check
Put the numbers back in. Both sides should match.
6 = 2 × 3 ✓
−8 = 2 × (−4) ✓
2 × (3 × 5 − 4) = 4 × 5 + 2 ✓
2 × (3 × 5 + (−4)) = 4 × 5 + 2 ✓
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