Solve x + 2y = 4, x + y = 3 by gaussian elimination Step by Step

Step by step

The question

Question: solve x + 2y = 4 and x + y = 3 by Gaussian elimination.

x + 2y = 4①

We want the x and y that make both true.

x + y = 3②

We will use three moves on the rows.

Swap two rows.

Times a row by a number that is not 0.

Add a multiple of one row to another.

Each move keeps the equations true, like in elimination.

swap two rows

× a row by a number (not 0)

add a multiple of one row to another

1 Write the augmented matrix

Each row is one equation: the numbers in front of x and y, then the number on the right.

The bar stands for the = signs.

This is called an augmented matrix.

= [1, 2, 4; 1, 1, 3]

We call the rows R 1 and R 2.

Our goal: 1s down the diagonal, and 0s under them.

This shape is called row-echelon form.

Then we go back up, one letter at a time.

goal: 1s down the diagonal, 0s under them

2 Get 0s under the 1 in column 1

→ [1, 2, 4; 0, −1, −1]

Careful!

A row move works on the whole row, the number after the bar too.

That number is the right side of the equation.

Careful: the whole row, after the bar too

3 Get a 1 in row 2

→ [1, 2, 4; 0, 1, 1]

4 Back substitution

Start at the bottom.

Each row is an equation again.

Row 2 is 1y = 1, with 0 for the other letters.

So y = 1.

y = 1

Row 1 is the equation x + 2y = 4.

x + 2y = 4

We know y = 1.

We put it in.

x + 2 × 1 = 4

Work it out: x + 2 = 4.

x + 2 = 4

To remove the + 2, we take away 2 on both sides.

x + 2 − 2 = 4 − 2

So x = 2.

x = 2

We write it as (2, 1).

(x, y) = (2, 1)

5 Check

Let's check in every equation.

We put x = 2, y = 1 into ①.

① 2 + 2(1) = 2 + 2 = 4

We get 4, the same as the right side.

Now ②.

② 2 + 1 = 3

We get 3, the same as the right side.

Both work.

So the answer is right!

Let's look back.

First: write the augmented matrix.

Then: get 0s under the 1 in column 1.

Then: get a 1 in row 2.

Then: back substitution.

Last: check.

Answer