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First algebraically, SP= shorter piece, LP=longer piece.
3(SP)=LP-2yards
LP+SP=t (the variable t is given by the author).
We need to show LP in terms of t variable.
If LP+SP=t, therefore SP=t-LP. So now we can plug this into our above equation.
3(t-LP)=LP-2yards --> 3t-3LP=LP-2 --> 3t=4LP-2 --> 4LP=3t+2 --> LP= 3t+2/4
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Bunuel
A piece of twine of length t is cut into two pieces. The length of the longer piece is 2 yards greater than 3 times the length of the shorter piece which of the following of the length, in yards, of the longer piece


(A) \(\frac{(t+3)}{3}\)

(B) \(\frac{(3t+2)}{3}\)

(C) \(\frac{(t-2)}{4}\)

(D) \(\frac{(3t+4)}{4}\)

(E) \(\frac{(3t+2)}{4}\)

A piece of twine of length t is cut into two pieces.
Let x = the length of the LONGER piece in yards
So, t - x = the length of the SHORTER piece in yards

The length of the longer piece is 2 yards greater than 3 times the length of the shorter piece.
In other words: (longer piece) = 3(shorter piece) + 2
In other words: x = 3(t - x) + 2

Which of the following is the length, in yards, of the longer piece?
So, we must solve for x
Take: x = 3(t - x) + 2
Expand right side: x = 3t - 3x + 2
Add 3x to both sides: 4x = 3t + 2
Divide both sides by 4 to get: \(x = \frac{3t + 2}{4}\)

Answer: E
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Bunuel
A piece of twine of length t is cut into two pieces. The length of the longer piece is 2 yards greater than 3 times the length of the shorter piece which of the following of the length, in yards, of the longer piece


(A) \(\frac{(t+3)}{3}\)

(B) \(\frac{(3t+2)}{3}\)

(C) \(\frac{(t-2)}{4}\)

(D) \(\frac{(3t+4)}{4}\)

(E) \(\frac{(3t+2)}{4}\)

We can create the equation:

L = 2 + 3S

and

L + S = t

S = t - L

Substituting, we have:

L = 2 + 3(t - L)

L = 2 + 3t - 3L

4L = 2 + 3t

L = (2 + 3t)/4

Answer: E
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Bunuel
A piece of twine of length t is cut into two pieces. The length of the longer piece is 2 yards greater than 3 times the length of the shorter piece which of the following of the length, in yards, of the longer piece


(A) \(\frac{(t+3)}{3}\)

(B) \(\frac{(3t+2)}{3}\)

(C) \(\frac{(t-2)}{4}\)

(D) \(\frac{(3t+4)}{4}\)

(E) \(\frac{(3t+2)}{4}\)

Whenever you see variables in options, try to plug in.

Say length of shorter piece is 1. Then length of longer piece is 3*1 + 2 = 5
And total length t = 1 + 5 = 6

Now put t = 6 in the options and see which one gives you 5.

Only (E) does.
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