Heart of Algebra Problem with Solution

from 320 SAT Math Problems

Today I’d like to provide you with a solution to the last Heart of Algebra problem I gave you. Here is the question one more time, followed by a complete explanation.

Level 4 Heart of Algebra

2x + 5 + k = 7x
2y + 5 + t = 7y

In the equations above, k and t are constants. If t is k minus 1/3, which of the following is true?

(A) x is y minus 1/3
(B) x is y minus 1/15
(C) x is y plus 1/3
(D) x is y plus 1/15

* Solution using the elimination method: We begin by replacing t by k – 1/3, and interchanging the order of the two equations.

2y + 5 + k – 1/3 = 7y

2x + 5 + k          = 7x

We then subtract the bottom equation from the top equation.

2y – 2x – 1/3 = 7y – 7x

We solve this equation for x by adding 7x, subtracting 2y, and adding  1/3  to each side of the equation,

5x = 5y + 1/3

Finally, we multiply each side of this last equation by  1/5  to get

x = 1/5 (5y + 1/3) = 1/5 ⋅ 5y + 1/5 ⋅ 1/3 = y + 1/15

So x is y plus 1/15, choice D.

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