42 inch board is cut into 3 pieces. Piece 2 is twice as long as the 1st & 3rd piece. Piece 3 is three times as long as piece 1. Find inches of all three pieces

Is piece 2 twice as long as 1 & 3 combined? Or separately?

separately

If piece 2 is twice piece 1, then

x = piece 1
3x = piece 3 (3 times piece 1)
2x = piece 2 (twice piece 1)
x + 3x + 2x = 42
6x = 42
x = 7
Piece 1 - 7
Piece 2 - 14
Piece 3 - 21

If piece 2 is twice piece 3, then

x = piece 1
3x = piece 3 (3 times piece 1)
2(3x) = piece 2 (twice piece 3)
x + 3x + 2(3x) = 42
10x = 42
x = 4.2
Piece 1 - 4.2
Piece 2 - 12.6
Piece 3 - 25.2

If piece 2 is twice piece 1 & 3, then

x = piece 1
3x = piece 3 (3 times piece 1)
2(3x + x) = piece 2 (twice piece 1 & 3)
x + 3x + 2(4x) = 42
12x = 42
x = 3.5
Piece 1 - 3.5
Piece 2 - 10.5
Piece 3 - 28

I'm confused. Shouldn't pc 2 be the largest since it says that pc2 is twice as long as pc 1&3, and pc 3/three times as long as pc1

To find the length of the three pieces, let's assign variables to represent the length of each piece. Let's call the length of the first piece "x".

According to the question, the second piece is twice as long as the first piece, so its length would be 2x.

The third piece is three times as long as the first piece, so its length would be 3x.

We also know that the sum of the lengths of the three pieces should equal 42 inches.

So, we can write an equation to express this as:

x + 2x + 3x = 42

Simplifying the equation, we get:
6x = 42

Dividing both sides of the equation by 6, we find:
x = 7

Now that we know the value of x, we can substitute it back into the expressions for the lengths of the three pieces:

First piece: x = 7 inches
Second piece: 2x = 2 * 7 = 14 inches
Third piece: 3x = 3 * 7 = 21 inches

Therefore, the lengths of the three pieces are:
First piece: 7 inches
Second piece: 14 inches
Third piece: 21 inches