A 38-inch piece of steel is cut into 3 pieces so that the second piece is twice as long as the first, and the third piece is three inches more than four times the length of the first piece. What is the length of the second piece?

66/7

Step-by-step explanation:

There are three pieces. Lets call the 1st piece A, the 2nd B and the 3rd C. As all of them sum 38:

A + B + C = 36 (eq 1)

As the second is twice as long as the first:

2A = B (eq. 2) the first multiplied by 2 is equal to the second

As the third is three inches more than the length of four times the 1st:

C = 3 + 4A (eq. 3) The 3rd is equal to four times the first plus 3 inches

Now we can replace eqs. 2 and 3 in eq. 1:

A + B + C = A + 2A + 3 + 4A = 36

7A + 3 = 36

Subtracting 3 in both sides:

7A = 33

Dividing by 7:

A = 33/7

Now, we can replace A in the 2nd eq. to get B:

2A = B

2 * (33/7) = B

66 / 7 = B

We can verify the results by finding C:

C = 3 + 4A = 3 + 4 (33/7) = 3 + 132/7 = 21/7 + 132/7 = 153/7

Then:

A+B+C = 33/7 + 66/7 + 153/7 = 252/7 = 36 - -> Verifies!