Bunuel wrote:
Given the two equations 3r + s = 17 and r + 2s = 9, by how much does r exceed s?
A. 3
B. 4
C. 5
D. 6
E. 7
We can multiply the second equation by -3 and we have:
-3r - 6s = -27
When we add this equation to the first equation, the terms containing r cancel out, and we now have:
-5s = -10
s = 2
Thus:
r + 2(2) = 9
r = 5
So, r is 3 greater than s.
Answer: A
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