Solve 5x + 2y = 1, 3x + y = 0 by Cramer's rule Step by Step

Step by step

The question

Question: solve 5x + 2y = 1 and 3x + y = 0 by Cramer's rule.

5x + 2y = 1①

We call them ① and ②.

We want the x and y that make both true.

3x + y = 0②

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 = D_xD, y = D_yD

1 Find D

D: the numbers in front of x and y, in the same places as in the equations.

D = |5, 2; 3, 1|

So D = 5 × 1 − 3 × 2.

D = 5 × 1 − 3 × 2

5 times 1 is 5.

3 times 2 is 6.

D = 5 − 6

5 minus 6 is −1.

So D = −1.

D = −1

2 Find Dx

D x: the same, but the x column is replaced by the numbers on the right: 1 and 0.

D_x = |1, 2; 0, 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 = 1 × 1 − 0 × 2.

D_x = 1 × 1 − 0 × 2

1 times 1 is 1.

0 times 2 is 0.

D_x = 1 − 0

1 minus 0 is 1.

So D x = 1.

D_x = 1

3 Find Dy

D y: now the y column is replaced by 1 and 0.

D_y = |5, 1; 3, 0|

So D y = 5 × 0 − 3 × 1.

D_y = 5 × 0 − 3 × 1

5 times 0 is 0.

3 times 1 is 3.

D_y = 0 − 3

0 minus 3 is −3.

So D y = −3.

D_y = −3

4 Divide

x is D x over D: 1 over (−1).

x = D_xD = 1−1

1 divided by (−1) is −1.

So x = −1.

x = −1

y is D y over D: −3 over (−1).

y = D_yD = −3−1

−3 divided by (−1) is 3.

So y = 3.

y = 3

So x = −1 and y = 3, both at the same time.

We write it as (−1, 3).

(x, y) = (−1, 3)

5 Check

Let's check in every equation.

We put x = −1, y = 3 into ①.

① 5(−1) + 2(3) = −5 + 6 = 1

We get 1, the same as the right side.

Now ②.

② 3(−1) + 3 = −3 + 3 = 0

We get 0, 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.

Answer