Solve a + 2b = 4, 3a - b = 5 by Cramer's rule

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The question

Question: solve a + 2b = 4 and 3a − b = 5 by Cramer's rule.

We call them ① and ②.

We want the a and b that make both true.

Cramer's rule uses three determinants: D, D a and D b.

Then a is D a over D, and b is D b over D.

1 Find D

D: the numbers in front of a and b, in the same places as in the equations.

So D = 1 × (−1) − 3 × 2.

1 times (−1) is −1.

3 times 2 is 6.

−1 minus 6 is −7.

So D = −7.

2 Find Da

D a: the same, but the a column is replaced by the numbers on the right: 4 and 5.

Careful!

For D a, we replace the a column: the first one.

The b column stays.

So D a = 4 × (−1) − 5 × 2.

4 times (−1) is −4.

5 times 2 is 10.

−4 minus 10 is −14.

So D a = −14.

3 Find Db

D b: now the b column is replaced by 4 and 5.

So D b = 1 × 5 − 3 × 4.

1 times 5 is 5.

3 times 4 is 12.

5 minus 12 is −7.

So D b = −7.

4 Divide

a is D a over D: −14 over (−7).

−14 divided by (−7) is 2.

So a = 2.

b is D b over D: −7 over (−7).

−7 divided by (−7) is 1.

So b = 1.

So a = 2 and b = 1, both at the same time.

We write it as (2, 1).

5 Check

Let's check in every equation.

We put a = 2, b = 1 into ①.

We get 4, the same as the right side.

Now ②.

We get 5, the same as the right side.

Both work.

So the answer is right!

Let's look back.

First: find D.

Then: find Da.

Then: find Db.

Then: divide.

Last: check.