algebra

A 100-foot rope is cut into three pieces. The first piece is three times as long as the second, and the third piece is 30 feet shorter than the second. How many feet long is the shortest piece?

 

Let First Piece = $p_1$

Let Second Piece = $p_2$

Let Third Piece = $p_3$

 

$p_1 + p_2 + p_3 = 100$

$p_1 = 3p_2$

$p_3 = p_2 - 30$

$3p_2 + p_2 + p_2 - 30 = 100$

$5p_2 = 130$

$p_2 = 26$

$p_1 = 78$

$p_3 = -4$

 

Where did I go wrong???

Say the second piece is length x. The first piece is 3 times that, so it is 3x.

The 3rd piece is 20 feet shorter than it, so it is x-20.

Adding them we get 5x-20, and that equals 100. 

Solving for x we get 24.

So the lengths are 72, 24 and 4.

f = 3s     and   t = s-30        and the three pieces  sum to 100 ft

3s      +      s     + s-30  = 100

5s = 130

s =  26        then f = 3s = 78      and t = 12         Which tells me this question is incorrect !