A pair of positive integers (x, y) satisfies the equation 31x + 29y = 1125. What is x + y?

I asked this question before, but now that I review that answers I received--it doesn't make much sense to me. Sorry.

I know that

if a×m + b×n = c

then (a - n)×m + (b+m)×n = c

And I also know that

(1) 16875×29 + -15750×31 = 1125

(2) (16875 - 31)×29 + (-15750 + 29)×31 = 1125

(3) 16844×29 + -15721×31 = 1125

I just have to find out how many 31s to subtract for 16875 and how many 29s to add to -15750 so that the x and y would fulfill the equation while being positive.

Thank you so much if you answer this for me!

Frooglie Apr 25, 2020

#1**+2 **

"*A pair of positive integers (x, y) satisfies the equation 31x + 29y = 1125. What is x + y?*"

Write 31x as 30x + x

Write 29y as 30y - y

Then you can write 31x + 29y = 1125 as 30x + x + 30y - y = 1125 or 30(x+y) + (x-y) = 1125

Subtract x-y from both sides: 30(x+y) = 1125 - (x-y)

Divide both sides by 30: x+y = 1125/30 - (x-y)/30

Now 1125/30 = 37 + 15/30 so we have x+y = 37 +15/30 -(x-y)/30

If x and y are positive integers, the fractional terms on the right hand side must cancel each other out.

For this to happen we set (x-y) = 15, then we are left with

x+y = 37

Alan Apr 25, 2020

#3**+2 **

Look at the right hand side of x+y = 37 +15/30 -(x-y)/30

There are two fractions on that side, namely 15/30 and -(x-y)/30

We could rewrite them as [15 - (x-y)]/30

We want the fractional parts to vanish because we must be left with an integer on the right hand side in order to be compatible with the integers on the left hand side. For this to happen we simply need to make 15 - (x-y) vanish, which happens if we set x-y equal to 15.

Alan
Apr 25, 2020