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# HELP ME PLS

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Find the number of ordered pairs $$(a,b)$$ of integers such that $$\frac{a + 2}{a + 5} = \frac{b}{3}$$

May 1, 2020

#2
+30086
+3

Rewrite the equation as:   b = 3(a + 2)/(a + 5)

Let z = a+5    z is also an integer

Then  b = 3(z-3)/z  or   b = 3 - 9/z

For this to result in an integer, we must have 9/z as an integer.  This means the following values are the only ones allowed for z:

z = -9, -3, -1, 1, 3, 9  which means  a = -14, -8, -6, -4, -2, 4  with corresponding values for b of:  b = 4, 6, 12, -6, 0, 2

I'll leave you to order them in pairs.

May 2, 2020
edited by Alan  May 2, 2020

#1
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Well, correct me if I'm wrong, but a+2=b, and a+5=3.  We can change the equation, to make it _+5=3, then we can revere it so it says 3-5=_, or rather a.  That equation simplified would be a=(-2).  Now that we know a=(-2), we can see that b=0.  I'm not sure if this is what you're looking for, but I hope that this helps!

May 2, 2020
#2
+30086
+3

Rewrite the equation as:   b = 3(a + 2)/(a + 5)

Let z = a+5    z is also an integer

Then  b = 3(z-3)/z  or   b = 3 - 9/z

For this to result in an integer, we must have 9/z as an integer.  This means the following values are the only ones allowed for z:

z = -9, -3, -1, 1, 3, 9  which means  a = -14, -8, -6, -4, -2, 4  with corresponding values for b of:  b = 4, 6, 12, -6, 0, 2

I'll leave you to order them in pairs.

Alan May 2, 2020
edited by Alan  May 2, 2020
#3
+111329
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Thanks, alan......that's a very crafty solution   !!!!!

CPhill  May 2, 2020