Question

suppose x y - 5 x + 2 Y =30 x and y are positive integers find the number of possible values of x

Answer 1

Main thing to keep in mind is x and y are positive and integers.

Now lest solve ..

[math]xy-5x+2y=30[/math]

Rearrange

[math]xy+2y=30+5x[/math]

Separate x and y

[math]y(x+2)=30+5x[/math]

[math]y=(30+5x)/(x+2)[/math]

Now lets use trial and error

First observtion is that numerator is divisible by 5, therefore if the denominator is 5 we can get an integer. x+2=5 i.e. x=3

lets see,

y= (30+5*3)/(3+2)=45/5=9

One solution is (x=3,y=9)

Lets try another , what if x=1

y=(30+5*1)/(1+2)=35/3 Not an integer therefore it cannot be a solution

Lets try another, put x=2

y=(30+5*2)/(2+2)=40/4=10

Another solution (x=2,y=10)

We can do trial and error and can find many solutions!

Answer 2

xy−5x+2y=30

x(y−5)+2y−10=20

x(y−5)+2(y−5)=20

(x+2)(y−5)=20=4⋅5=5⋅4=10⋅2=20⋅1

{(2,10),(3,9),(8,7),(18,6)}

4 Natural solutions.

Answer putted by me.