A piece of twine of length t is cut into two pieces. The length of the

I Algebra

\(t=\) length of twine
\(S=\) length of shorter piece
\(L=3S + 2 =\) length of longer piece
\(L - 2 = 3S\)
\(\frac{L-2}{3}=S\)

\(L+S=t\)
\(L + \frac{(L-2)}{3}=t\)
\(3L +(L-2)=3t\)
\(4L=3t+2\)
\(L=\frac{(3t+2)}{4}\)

Answer E

II. Assign values
Let smaller length, S = 1 yard
Longer length, L = (3S + 2) = (3 + 2) = 5 yds
Twine length, t = (S + L) = (1 + 5) = 6 yds

Using t=6, find the answer that yields L = 5

(A) \(\frac{(t+3)}{3}=\frac{(5+3)}{3}=\frac{8}{3}\) NO

(B) \(\frac{(3t+2)}{3}=\frac{(3*6)+2}{3}=\frac{20}{3}=6.xx\) NO

(C) \(\frac{(t-2)}{4}=\frac{(6-2)}{4}=1\) NO

(D) \(\frac{(3t+4)}{4}=\frac{(3*6)+4}{4}=\frac{22}{4}=5.5\) NO

(E) \(\frac{(3t+2)}{4}=\frac{(3*6)+2}{4}=\frac{20}{4}=5\) YES

Answer E