Solution set of x + 2y + 3z = 6, 2x + 4y + 6z = 12 Step by Step

Step by step

The question

Question: find all the solutions of x + 2y + 3z = 6 and 2x + 4y + 6z = 12.

x + 2y + 3z = 6①

We want every x, y and z that make both true.

There are 3 letters, but only 2 equations.

2x + 4y + 6z = 12②

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, after the bar.

= [1, 2, 3, 6; 2, 4, 6, 12]

We call the rows R 1 and R 2.

2 Row reduce

We use the three row moves: swap two rows, times a row by a number that is not 0, add a multiple of one row to another.

They never change the answers.

Our goal: a leading 1 in each row, with 0s above and below it.

swap two rows

× a row by a number (not 0)

add a multiple of one row to another

→ [1, 2, 3, 6; 0, 0, 0, 0]

3 Find the free letters

The first number that is not 0 in each row is a 1.

We call it a leading 1, and ring it.

The letter x has a leading 1 in its column.

The letters y and z have none: they are free.

A free letter can be any number.

A row of 0s says 0 = 0: always true, so it tells us nothing.

leading 1: x

free: y, z

4 Write the leading letters with the free ones

Row 1 as an equation: x + 2y + 3z = 6.

x + 2y + 3z = 6

We want x alone.

To remove + 2y and + 3z, we take away 2y and take away 3z on both sides.

x ~~+ 2y + 3z − 2y − 3z~~ = 6 − 2y − 3z

They cancel on the left.

So x = 6 − 2y − 3z.

x = 6 − 2y − 3z

Careful!

The + 2y moves to the other side, so its sign changes.

So x is not 6 + 2y.

Careful: not x = 6 + 2y ✗

5 Write it as vectors

We stack the letters in a column.

The free letters stay as they are.

[x; y; z] = [6 − 2y − 3z; y; z]

We split it up.

The numbers alone make one vector: (6, 0, 0).

What goes with y makes another: (−2, 1, 0), and the same for z.

[x; y; z] = [6; 0; 0] + y[−2; 1; 0] + z[−3; 0; 1], y and z any numbers

So every solution is one point, plus any amount of each direction.

Two free letters: the solutions make a flat plane.

one point + any amount of each direction

6 Check

Let's check.

First the point: all free letters 0.

We put x = 6, y = 0 and z = 0 into every equation.

① 6 + 2(0) + 3(0) = 6 + 0 + 0 = 6

② 2(6) + 4(0) + 6(0) = 12 + 0 + 0 = 12

The point works in every equation.

Now the direction (−2, 1, 0).

Its left sides must give 0, so adding it changes nothing.

① (−2) + 2(1) + 3(0) = −2 + 2 + 0 = 0

② 2(−2) + 4(1) + 6(0) = −4 + 4 + 0 = 0

Now the direction (−3, 0, 1).

Its left sides must give 0, so adding it changes nothing.

① (−3) + 2(0) + 3(1) = −3 + 0 + 3 = 0

② 2(−3) + 4(0) + 6(1) = −6 + 0 + 6 = 0

Every left side gives 0.

So adding any amount of it keeps every equation true.

The answer is right!

Let's look back.

First: write the augmented matrix.

Then: row reduce.

Then: find the free letters.

Then: write the leading letters with the free ones.

Then: write it as vectors.

Last: check.

Answer