SME
Grade 9 Math · Problem type
Solve a system by elimination
Use this for “Solve 3x + 2y = 16 and 5x − 4y = 12 by elimination.”
Chart
First: y = (−3/2)x + 8
Second: y = (5/4)x − 3
They cross at (4, 2). −3/2 × 4 + 8 = 2
5/4 × 4 − 3 = 2
Sum: 11x = 44, so x = 4: y is gone, and the upright line goes through the crossing
This is an example. Type your own values to start a new problem.
a
x
+
b
y
=
c
d
x
+
(
e
)
y
=
f
(x,
y)
=
(
x
,
y
)
L
Least common multiple of the y terms
worked out (not typed)
k₁
Multiply the first by
worked out (not typed)
k₂
Multiply the second by
worked out (not typed)
p
x in the sum
worked out (not typed)
r
Right side of the sum
worked out (not typed)
Formulas
L
=
least
common
multiple
of
b
and
e
4 = least common multiple of 2 and −4
k₁
=
L
÷
|
b
|
2 = 4 ÷ |2|
k₂
=
−
k₁
×
b
e
1
=
−2
×
2
−4
p
=
k₁
×
a
+
k₂
×
d
11 = 2 × 3 + 1 × 5
r
=
k₁
×
c
+
k₂
×
f
44 = 2 × 16 + 1 × 12
x
=
r
p
4
=
44
11
y
=
c
−
a
×
x
b
2
=
16 − 3 × 4
2
Assumptions
•
Multiply one or both equations so the y terms are opposites.
•
Add the equations: y cancels, leaving one equation in x.
•
Put x back into the first equation to find y, then check in the second.
Step-by-step
WE KNOW
a = 3, b = 2, c = 16, d = 5, e = −4, f = 12
FIND
least common multiple of the y terms (L), multiply the first by (k₁), multiply the second by (k₂), x in the sum (p), right side of the sum (r), solution x (x), solution y (y)
Step 1 · Find least common multiple of the y terms (L)
Use L = least common multiple of b and e
The
smallest
number
both
y
coefficients
go
into
(their
sizes,
without
the
signs).
L
=
least
common
multiple
of
2
and
−4
L
=
4
Step 2 · Find multiply the first by (k₁)
Use k₁ = L ÷ |b|
Multiply
the
first
equation
so
its
y
term
is
L
or
−
L
.
k₁
=
4
÷
|2|
k₁
=
4
2
k₁
=
2
Step 3 · Find multiply the second by (k₂)
Use k₂ = −k₁ × b ÷ e
Multiply
the
second
so
its
y
term
is
the
opposite
of
the
first’s:
k₂
e
=
−k₁
b
.
k₂
=
−2
×
2
−4
k₂
=
−
4
−4
k₂
=
1
Step 4 · Find x in the sum (p)
Use p = k₁ × a + k₂ × d
Add
the
x
terms
of
the
two
multiplied
equations;
the
y
terms
cancel.
p
=
2
×
3
+
1
×
5
2(
3
x
+
2
y
)
=
2(16):
6
x
+
4
y
=
32
5
x
−
4
y
=
12
Add:
11
x
=
44
p
=
11
Step 5 · Find right side of the sum (r)
Use r = k₁ × c + k₂ × f
Add the right sides the same way.
r
=
2
×
16
+
1
×
12
r
=
32
+
12
r
=
44
Step 6 · Find solution x (x)
Use x = r ÷ p
The
sum
is
px
=
r
:
divide
both
sides
by
p
.
x
=
44
11
x
=
4
Step 7 · Find solution y (y)
Use y = (c − a × x) ÷ b
Put
x
into
the
first
equation,
take
ax
from
both
sides,
then
divide
by
b
.
y
=
16 − 3 × 4
2
y
=
16 − 12
2
y
=
4
2
y
=
2
Check
Put the numbers back in. Both sides should match.
4 = least common multiple of 2 and −4 ✓
2 = 4 ÷ |2| ✓
1
=
−2
×
2
−4
✓
11 = 2 × 3 + 1 × 5 ✓
44 = 2 × 16 + 1 × 12 ✓
4
=
44
11
✓
5 × 4 + (−4) × 2 = 12 ✓
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