Solve 3x + 4y = 1, 2x + 3y = 0 by Cramer's rule Step by Step

Step by step

The question

Question: solve 3x + 4y = 1 and 2x + 3y = 0 by Cramer's rule.

3x + 4y = 1①

We call them ① and ②.

We want the x and y that make both true.

2x + 3y = 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 = |3, 4; 2, 3|

So D = 3 × 3 − 2 × 4.

D = 3 × 3 − 2 × 4

3 times 3 is 9.

2 times 4 is 8.

D = 9 − 8

9 minus 8 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, 4; 0, 3|

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 × 3 − 0 × 4.

D_x = 1 × 3 − 0 × 4

1 times 3 is 3.

0 times 4 is 0.

D_x = 3 − 0

3 minus 0 is 3.

So D x = 3.

D_x = 3

3 Find Dy

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

D_y = |3, 1; 2, 0|

So D y = 3 × 0 − 2 × 1.

D_y = 3 × 0 − 2 × 1

3 times 0 is 0.

2 times 1 is 2.

D_y = 0 − 2

0 minus 2 is −2.

So D y = −2.

D_y = −2

4 Divide

x is D x over D: 3/1.

x = D_xD = 31

3 divided by 1 is 3.

So x = 3.

x = 3

y is D y over D: −2/1.

y = D_yD = −21

−2 divided by 1 is −2.

So y = −2.

y = −2

So x = 3 and y = −2, both at the same time.

We write it as (3, −2).

(x, y) = (3, −2)

5 Check

Let's check in every equation.

We put x = 3, y = −2 into ①.

① 3(3) + 4(−2) = 9 − 8 = 1

We get 1, the same as the right side.

Now ②.

② 2(3) + 3(−2) = 6 − 6 = 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