Bunuel wrote:

Which of the following is equivalent to \(\frac{x+y}{3}−\frac{x−y}{4}\)?

A. \(\frac{x+y}{12}\)

B. \(\frac{7x+y}{12}\)

C. \(\frac{7x−y}{12}\)

D. \(\frac{7y+x}{12}\)

E. \(\frac{7y−x}{12}\)

\(\frac{x+y}{3}−\frac{x−y}{4}\)=

\((\frac{x+y}{3}*\frac{4}{4}) −(\frac{x−y}{4}*\frac{3}{3})\)

\(\frac{4(x+y)}{3*4} −\frac{3(x−y)}{4*3}=\)

\((\frac{4x+4y}{12})-(\frac{3x-3y}{12})=\)

(Change the sign of terms being subtracted!)

\(\frac{4x+4y-3x+3y}{12}=\)

=\(\frac{x + 7y}{12}\)

Answer D

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In the depths of winter, I finally learned

that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"