Lodz697 wrote:
If xy = 3(x + 1) + y; and, x and y are integers, x could be any of the following values except:
A. 7
B. 5
C. 4
D. 3
E. 2
What's the best way to solve a problem likethis?
You can use the method shownhere . Or do the following:
There are several ways to solve this question through algebraic manipulations. You can use the method shownhere . Or do the following:
Essentially, our goal is to arrive at an equation of the form\( x = expression \ with \ y \only\) or\( y = expression \ with \ x \only\).
\( xy = 3(x + 1) +y\)
\( xy -y= 3(x +1)\)
\( y(x-1) = 3x +3\)
\( y=\frac{3x +3}{x-1}\)
Next, let's attempt to split the fraction to isolate\(x\) in a singleterm:
\( y=\frac{3x -3 + 3 +3}{x-1}\)
\( y=\frac{3x -3 }{x-1}+ \frac{3+3}{x-1}\)
\( x=3+\frac{6}{x-1}\)
Since\(y\) is an integer, for\( 3+\frac{6}{x-1}\) to be an integer,\(\frac{6}{x-1}\) must also be an integer. This means that\(x-1\) must be a divisor of 6. From the options, only if x = 5, x - 1 is not a factor of 6, making it the correct answer.
Answer: B.
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Statistics : Posted by Bunuel • on 06 Feb 2024, 03:06 • Replies 6 • Views 1004
