A 46​-inch piece of steel is cut into three pieces so that the second piece is twice as long as the first​ piece, and the third piece is one inch more than six times the length of the first piece. Find the lengths of the pieces.

Question

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Part 1
x
2x
6x+~x
6x~
6x+1
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Part 1
The lengths of the​ first, the​ second, and the third pieces are
  
enter your response here ​in,
  
enter your response here ​in, and
  
enter your response here in respectively.

Answer 1

To solve the problem, let's define the lengths of the three pieces of steel based on the given information.

1. Let \( x \) be the length of the first piece.
2. The second piece is twice as long as the first piece, so its length is \( 2x \).
3. The third piece is one inch more than six times the length of the first piece, so its length is \( 6x + 1 \).

Now, we know that the total length of the three pieces is 46 inches. Therefore, we can set up the following equation:

\[ x + 2x + (6x + 1) = 46 \]

Now, we simplify and solve for \( x \):

\[ x + 2x + 6x + 1 = 46 \]

Combine like terms:

\[ 9x + 1 = 46 \]

Subtract 1 from both sides:

\[ 9x = 45 \]

Now divide by 9:

\[ x = 5 \]

Now that we have the length of the first piece (\( x = 5 \)), we can find the lengths of the second and third pieces.

• Length of the second piece: \[ 2x = 2(5) = 10 \text{ inches} \]

• Length of the third piece: \[ 6x + 1 = 6(5) + 1 = 30 + 1 = 31 \text{ inches} \]

So, the lengths of the three pieces are:

• First piece: \( 5 \) inches
• Second piece: \( 10 \) inches
• Third piece: \( 31 \) inches

In conclusion, the final answer is:

• The lengths of the first, second, and third pieces are \( 5 \) in, \( 10 \) in, and \( 31 \) in respectively.