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Find the number of ordered pairs (a,b) of integers such that \(\frac{a + 2}{a + 5} = \frac{b}{3}\)
I don't know where to start. I only got two solutions (4, 2) and (-2, 0)

by just trying to find ordered pairs a and b where the expression (which is b) and are integers, but 2 isn't the correct answer. What approach am I supposed to take?

 Apr 21, 2020
 #1
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From what I can see, there are 6: (-14,4),(-8,6),(-6,12),(-4,-6),(-2,0),(4,2)

 

I looked at the graph \(b= \frac {3a+6} {a+5}\), since there is a horizontal asymptote at b=3, there are only a finite number of points to check

 Apr 21, 2020
 #2
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Find the number of ordered pairs \((a,b)\) of integers such that \(\dfrac{a + 2}{a + 5} = \dfrac{b}{3}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{a + 2}{a + 5}} &=& \mathbf{\dfrac{b}{3}} \\\\ 3(a+2) &=& b(a+5) \\ 3a+6 &=& ab+5b \\ -3a+5b+ab &=& 6 \\ -3a+5b+ab+(-3*5) &=& (-3*5)+6 \\ -3a+5b+ab-15 &=& -15+6 \\ \mathbf{(a+5)(b-3)} &=& \mathbf{-9} \\ \mathbf{m*n} &=& \mathbf{-9} \\ && \text{find all integer products }\ m*n = -9, \\ && \text{where}\ m=a+5,\ \text{ and }\ n=b-3 \\ \hline \end{array} \)

 

all integer products

\(\begin{array}{|ccl|r|r|r|r|} \hline && m*n & m & n & a=m-5 & b=n+3 \\ \hline -9 &=& 1\times -9 & 1 & -9 & -4 & -6 \\ -9 &=& -1\times 9 & -1 & 9 & -6 & 12 \\ -9 &=& 3\times -3 & 3 & -3 & -2 & 0 \\ -9 &=& -3\times 3 & -3 & 3 & -8 & 6 \\ -9 &=& 9\times -1 & 9 & -1 & 4 & 2 \\ -9 &=& -9\times 1 & -9 & 1 & -14 & 4 \\ \hline \end{array} \)

 

There are 6 integer products.

The number of ordered pairs (a,b) are 6.

\((-4,-6),\ (-6,12),\ (-2,0),\ (-8,6),\ (4,2),\ (-14,4) \)

 

laugh

 Apr 21, 2020

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