A 31-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 three inches more than four times the length of the first piece. Find the lengths of the pieces.

The length of the first piece is 4 inches, that of the second is 8 inches and the third 19 inches.

Step-by-step explanation:

Total length of the steel = 31 inches

Let the length of the first piece = x

let the length of the second piece = y

Let the length of the third piece = z

So,    

  y = 2x

  z = 4x + 3

x + 2x + 4x +3  = total length of the steel = 31

   7x + 3 + 31

    7x = 31 - 3

    7x = 28

    x= 4 inches

Thus; y = 2x = 2 × 4 = 8 inches

         z = 4x +3 = 4×4 +3 =19 inches

∴The length of the first piece is 4 inches, that of the second is 8 inches and the third 19 inches.

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!