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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