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

The question

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

We want the x and y that make both true.

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.

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.

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.

2 Get 0s under the 1 in column 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.

3 Get a 1 in row 2

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.

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

We know y = 1.

We put it in.

Work it out: x + 2 = 4.

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

So x = 2.

We write it as (2, 1).

5 Check

Let's check in every equation.

We put x = 2, y = 1 into ①.

We get 4, the same as the right side.

Now ②.

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.

▶ Watch it solved

Solve x + y + z = 6, 2x - y + z = 3, x + 2y - z = 2 by gaussian… Medium

The question

Question: solve x + y + z = 6, 2x − y + z = 3 and x + 2y − z = 2 by Gaussian elimination.

We want the x, y and z that make all three true.

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.

1 Write the augmented matrix

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

The bar stands for the = signs.

This is called an augmented matrix.

We call the rows R 1, R 2 and R 3.

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.

2 Get 0s under the 1 in column 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.

3 Get a 1 in row 2

4 Get 0s under the 1 in column 2

5 Get a 1 in row 3

6 Back substitution

Start at the bottom.

Each row is an equation again.

Row 3 is 1z = 3, with 0 for the other letters.

So z = 3.

Row 2 is the equation y − 2z = −4.

We know z = 3.

We put it in.

Work it out: y − 6 = −4.

To remove the − 6, we add 6 on both sides.

So y = 2.

Row 1 is the equation x + y + z = 6.

We know y = 2 and z = 3.

We put them in.

Work it out: x + 5 = 6.

To remove the + 5, we take away 5 on both sides.

So x = 1.

We write it as (1, 2, 3).

7 Check

Let's check in every equation.

We put x = 1, y = 2, z = 3 into ①.

We get 6, the same as the right side.

Now ②.

We get 3, the same as the right side.

Now ③.

We get 2, the same as the right side.

All of them 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: get 0s under the 1 in column 2.

Then: get a 1 in row 3.

Then: back substitution.

Last: check.

▶ Watch it solved

Solve 2x + y + z = 7, x + y + z = 6, x - y + 2z = 5 by gaussian… Harder

The question

Question: solve 2x + y + z = 7, x + y + z = 6 and x − y + 2z = 5 by Gaussian elimination.

We want the x, y and z that make all three true.

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.

1 Write the augmented matrix

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

The bar stands for the = signs.

This is called an augmented matrix.

We call the rows R 1, R 2 and R 3.

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.

2 Get a 1 in row 1

3 Get 0s under the 1 in column 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.

4 Get a 1 in row 2

5 Get 0s under the 1 in column 2

6 Get a 1 in row 3

7 Back substitution

Start at the bottom.

Each row is an equation again.

Row 3 is 1z = 3, with 0 for the other letters.

So z = 3.

Row 2 is the equation y + z = 5.

We know z = 3.

We put it in.

Work it out: y + 3 = 5.

To remove the + 3, we take away 3 on both sides.

So y = 2.

Row 1 is the equation x + y + z = 6.

We know y = 2 and z = 3.

We put them in.

Work it out: x + 5 = 6.

To remove the + 5, we take away 5 on both sides.

So x = 1.

We write it as (1, 2, 3).

8 Check

Let's check in every equation.

We put x = 1, y = 2, z = 3 into ①.

We get 7, the same as the right side.

Now ②.

We get 6, the same as the right side.

Now ③.

We get 5, the same as the right side.

All of them work.

So the answer is right!

Let's look back.

First: write the augmented matrix.

Then: get a 1 in row 1.

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

Then: get a 1 in row 2.

Then: get 0s under the 1 in column 2.

Then: get a 1 in row 3.

Then: back substitution.

Last: check.

▶ Watch it solved