Worked examples
The question
Question: solve x + y = 7 and x − y = 1 by Cramer's rule.
x + y = 7①
We call them ① and ②.
We want the x and y that make both true.
x − y = 1②
Cramer's rule uses three determinants: D, D x and D y.
Then x is D x over D, and y is D y over D.
x = DxD, y = DyD
1 Find D
D: the numbers in front of x and y, in the same places as in the equations.
D = |1, 1; 1, −1|
So D = 1 × (−1) − 1 × 1.
D = 1 × (−1) − 1 × 1
1 times (−1) is −1.
1 times 1 is 1.
D = −1 − 1
−1 minus 1 is −2.
So D = −2.
D = −2
2 Find Dx
D x: the same, but the x column is replaced by the numbers on the right: 7 and 1.
Dx = |7, 1; 1, −1|
Careful!
For D x, we replace the x column: the first one.
The y column stays.
Careful: not the y column replaced ✗
So D x = 7 × (−1) − 1 × 1.
Dx = 7 × (−1) − 1 × 1
7 times (−1) is −7.
1 times 1 is 1.
Dx = −7 − 1
−7 minus 1 is −8.
So D x = −8.
Dx = −8
3 Find Dy
D y: now the y column is replaced by 7 and 1.
Dy = |1, 7; 1, 1|
So D y = 1 × 1 − 1 × 7.
Dy = 1 × 1 − 1 × 7
1 times 1 is 1.
1 times 7 is 7.
Dy = 1 − 7
1 minus 7 is −6.
So D y = −6.
Dy = −6
4 Divide
x is D x over D: −8 over (−2).
x = DxD = −8−2
−8 divided by (−2) is 4.
So x = 4.
x = 4
y is D y over D: −6 over (−2).
y = DyD = −6−2
−6 divided by (−2) is 3.
So y = 3.
y = 3
So x = 4 and y = 3, both at the same time.
We write it as (4, 3).
(x, y) = (4, 3)
5 Check
Let's check in every equation.
We put x = 4, y = 3 into ①.
① 4 + 3 = 7
We get 7, the same as the right side.
Now ②.
② 4 − 3 = 1
We get 1, the same as the right side.
Both work.
So the answer is right!
Let's look back.
First: find D.
Then: find Dx.
Then: find Dy.
Then: divide.
Last: check.
▶ Watch it solved
The question
Question: solve 3x − 2y = 4 and 2x + y = 12 by Cramer's rule.
3x − 2y = 4①
We call them ① and ②.
We want the x and y that make both true.
2x + y = 12②
Cramer's rule uses three determinants: D, D x and D y.
Then x is D x over D, and y is D y over D.
x = DxD, y = DyD
1 Find D
D: the numbers in front of x and y, in the same places as in the equations.
D = |3, −2; 2, 1|
So D = 3 × 1 − 2 × (−2).
D = 3 × 1 − 2 × (−2)
3 times 1 is 3.
2 times (−2) is −4.
D = 3 − (−4)
Taking away a negative is adding: 3 + 4.
3 − (−4) = 3 + 4
3 minus (−4) is 7.
So D = 7.
D = 7
2 Find Dx
D x: the same, but the x column is replaced by the numbers on the right: 4 and 12.
Dx = |4, −2; 12, 1|
Careful!
For D x, we replace the x column: the first one.
The y column stays.
Careful: not the y column replaced ✗
So D x = 4 × 1 − 12 × (−2).
Dx = 4 × 1 − 12 × (−2)
4 times 1 is 4.
12 times (−2) is −24.
Dx = 4 − (−24)
Taking away a negative is adding: 4 + 24.
4 − (−24) = 4 + 24
4 minus (−24) is 28.
So D x = 28.
Dx = 28
3 Find Dy
D y: now the y column is replaced by 4 and 12.
Dy = |3, 4; 2, 12|
So D y = 3 × 12 − 2 × 4.
Dy = 3 × 12 − 2 × 4
3 times 12 is 36.
2 times 4 is 8.
Dy = 36 − 8
36 minus 8 is 28.
So D y = 28.
Dy = 28
4 Divide
x is D x over D: 28/7.
x = DxD = 287
28 divided by 7 is 4.
So x = 4.
x = 4
y is D y over D: 28/7.
y = DyD = 287
28 divided by 7 is 4.
So y = 4.
y = 4
So x = 4 and y = 4, both at the same time.
We write it as (4, 4).
(x, y) = (4, 4)
5 Check
Let's check in every equation.
We put x = 4, y = 4 into ①.
① 3(4) − 2(4) = 12 − 8 = 4
We get 4, the same as the right side.
Now ②.
② 2(4) + 4 = 8 + 4 = 12
We get 12, the same as the right side.
Both work.
So the answer is right!
Let's look back.
First: find D.
Then: find Dx.
Then: find Dy.
Then: divide.
Last: check.
▶ Watch it solved
The question
Question: solve 2x + 4y = 3 and x − 2y = 1 by Cramer's rule.
2x + 4y = 3①
We call them ① and ②.
We want the x and y that make both true.
x − 2y = 1②
Cramer's rule uses three determinants: D, D x and D y.
Then x is D x over D, and y is D y over D.
x = DxD, y = DyD
1 Find D
D: the numbers in front of x and y, in the same places as in the equations.
D = |2, 4; 1, −2|
So D = 2 × (−2) − 1 × 4.
D = 2 × (−2) − 1 × 4
2 times (−2) is −4.
1 times 4 is 4.
D = −4 − 4
−4 minus 4 is −8.
So D = −8.
D = −8
2 Find Dx
D x: the same, but the x column is replaced by the numbers on the right: 3 and 1.
Dx = |3, 4; 1, −2|
Careful!
For D x, we replace the x column: the first one.
The y column stays.
Careful: not the y column replaced ✗
So D x = 3 × (−2) − 1 × 4.
Dx = 3 × (−2) − 1 × 4
3 times (−2) is −6.
1 times 4 is 4.
Dx = −6 − 4
−6 minus 4 is −10.
So D x = −10.
Dx = −10
3 Find Dy
D y: now the y column is replaced by 3 and 1.
Dy = |2, 3; 1, 1|
So D y = 2 × 1 − 1 × 3.
Dy = 2 × 1 − 1 × 3
2 times 1 is 2.
1 times 3 is 3.
Dy = 2 − 3
2 minus 3 is −1.
So D y = −1.
Dy = −1
4 Divide
x is D x over D: −10 over (−8).
x = DxD = −10−8
Made simpler: x = 5/4.
x = 54
y is D y over D: −1 over (−8).
y = DyD = −1−8
Made simpler: y = 1/8.
y = 18
So x = 5/4 and y = 1/8, both at the same time.
We write it as (5/4, 1/8).
(x, y) = (54, 18)
5 Check
Let's check in every equation.
We put x = 5/4, y = 1/8 into ①.
① 2(54) + 4(18) = 52 + 12 = 3
We get 3, the same as the right side.
Now ②.
② (54) − 2(18) = 54 − 14 = 1
We get 1, the same as the right side.
Both work.
So the answer is right!
Let's look back.
First: find D.
Then: find Dx.
Then: find Dy.
Then: divide.
Last: check.
▶ Watch it solved
