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24 Sep 2023, 03:07
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D
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Difficulty:
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Question Stats:
28% (02:41) correct
72%
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wrong
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If x and y are integers such that 4x + 3y = 3xy, how many distinct pairs of (x, y) are there?
A. 0
B. 1
C. 2
D. 3
E. 6
A. 0
B. 1
C. 2
D. 3
E. 6
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20 Mar 2024, 02:54
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Official Solution:
If \(x\) and \(y\) are integers such that \(4x + 3y = 3xy\), how many distinct pairs of \((x, y)\) are there?
A. 0
B. 1
C. 2
D. 3
E. 6
There are several ways to solve this question through algebraic manipulations. Essentially, our goal is to arrive at an equation of the form \(x = expression \ with \ y \ only\) or \(y = expression \ with \ x \ only\).
\(4x + 3y = 3xy\)
\(3y = 3xy-4x\)
\(3y = x(3y-4)\)
\(x=\frac{3y}{3y-4}\)
Next, let's attempt to split the fraction to isolate \(y\) in a single term:
\(x=\frac{3y-4+4}{3y-4}\)
\(x=\frac{3y-4}{3y-4} + \frac{4}{3y-4}\)
\(x=1+ \frac{4}{3y-4}\)
Since \(x\) is an integer, for \(1+ \frac{4}{3y-4}\) to be an integer, \(\frac{4}{3y-4}\) must also be an integer. This means that \(3y-4\) must be a divisor of 4. The number 4 has six factors: -4, -2, -1, 1, 2, and 4. However, for -2, 1, and 4, \(y\) does not result in an integer. Therefore, \(3y-4\) can only be -4, -1, or 2. Accordingly, \(y\) can only be 0, 1, or 2. For each of these \(y\) values, we find a corresponding \(x\) value, leading to a total of three distinct pairs of \((x, y)\).
Answer: D
If \(x\) and \(y\) are integers such that \(4x + 3y = 3xy\), how many distinct pairs of \((x, y)\) are there?
A. 0
B. 1
C. 2
D. 3
E. 6
There are several ways to solve this question through algebraic manipulations. Essentially, our goal is to arrive at an equation of the form \(x = expression \ with \ y \ only\) or \(y = expression \ with \ x \ only\).
\(4x + 3y = 3xy\)
\(3y = 3xy-4x\)
\(3y = x(3y-4)\)
\(x=\frac{3y}{3y-4}\)
\(x=\frac{3y-4+4}{3y-4}\)
\(x=\frac{3y-4}{3y-4} + \frac{4}{3y-4}\)
\(x=1+ \frac{4}{3y-4}\)
Answer: D
General Discussion
24 Sep 2023, 04:11
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Bunuel
\(4x + 3y = 3xy\)
\(4*x = 3*y*(x-1)\)
One of the pairs that we can infer relatively quickly is (x,y) = (0,0). So we can eliminate A.
For other pairs, we need to understand how the multiplications can be grouped
4x = 3 ; y(x-1) = 1 → Eliminate, as x = 4/3. The question premise states that x is an integer.
4x = (x-1) ; 3y = 1 → Eliminate, as y = 1/3. The question premise states that y is an integer.
4x = 3(x-1) ; y = 1 → Valid, as x = -3. (x,y) = (-3,1). We can eliminate B.
x = 3; y*(x-1) = 4 → Valid, as y = 2. (x,y) = (3,2). We can eliminate C.
y = 4 ; 3(x-1) = x → Eliminate, as x = 3/2. The question premise states that x is an integer.
4 = (x-1) ; 3y = x → Eliminate, as y = 5/3. The question premise states that y is an integer.
I don't suppose any other pairing is possible.
The pairs are (0,0), (-3,1),(3,2)
Option D
