SME
Grade 12 Math · Problem type
Cramer’s rule for two equations
Use this for “Solve 2x + 3y = 13 and x − y = −1 by Cramer’s rule.”
Diagram
D = 2 × (−1) − 3 × 1 = −5
x
=
Dₓ
D
=
−
10
−5
=
2
y
=
Dᵧ
D
=
−
15
−5
=
3
Dₓ and Dᵧ put the right sides in that unknown’s column (shaded).
This is an example. Type your own values to start a new problem.
a
x
+
b
y
=
p
c
x
+
(
d
)
y
=
q
x
=
x
,
y
=
y
D
Determinant ad − bc
worked out (not typed)
Formulas
D
=
a
×
d
−
b
×
c
−5 = 2 × (−1) − 3 × 1
x
=
p
×
d
−
b
×
q
D
2
=
13 × (−1) − 3 × (−1)
−5
y
=
a
×
q
−
p
×
c
D
3
=
2 × (−1) − 13 × 1
−5
Assumptions
•
D is the determinant of the coefficients, ad − bc; it must not be 0.
•
For x, put the right sides in the x column: x = (pd − bq) ÷ D. For y, in the y column: y = (aq − pc) ÷ D.
•
The point (x, y) is where the two lines cross.
Step-by-step
WE KNOW
a = 2, b = 3, p = 13, c = 1, d = −1, q = −1
FIND
determinant ad − bc (D), solution x (x), solution y (y)
Step 1 · Find determinant ad − bc (D)
Use D = a × d − b × c
The determinant of the coefficients: down the main diagonal, take away the other.
D
=
2
×
(−1)
−
3
×
1
D
=
−2
−
3
D
=
−5
Step 2 · Find solution x (x)
Use x = (p × d − b × q) ÷ D
Dₓ
puts
the
right
sides
in
the
x
column;
divide
it
by
D
.
x
=
13 × (−1) − 3 × (−1)
−5
Dₓ = 13 × (−1) − 3 × (−1) = −10
−
10
−5
=
2
x
=
2
Step 3 · Find solution y (y)
Use y = (a × q − p × c) ÷ D
Dᵧ
puts
the
right
sides
in
the
y
column;
divide
it
by
D
.
y
=
2 × (−1) − 13 × 1
−5
Dᵧ = 2 × (−1) − 13 × 1 = −15
−
15
−5
=
3
y
=
3
Check
Put the numbers back in. Both sides should match.
−5 = 2 × (−1) − 3 × 1 ✓
2
=
13 × (−1) − 3 × (−1)
−5
✓
3
=
2 × (−1) − 13 × 1
−5
✓
Topic check
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