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

Guest May 1, 2020

#2**+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

#1**0 **

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!

Guest May 2, 2020

#2**+3 **

Best Answer

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