Finding negative integer solutions to y = 10x/(100−x)

[Albert1]

x,y are negative integers

given :

$y=\dfrac {10x}{100-x}$

find all the possible solutions of y

 

[kaliprasad]

x,y are negative integers

given :

$y=\dfrac {10x}{100-x}$

find all the possible solutions of y

y(100-x) = 10 x
or xy + 10x - 100y = 0
or x(y+10) - 100(y+10) = - 1000
or (x-100)( y+ 10) = - 1000 ..1

y < 0 so y + 10 < 10
x < 0 so x - 100 < -100

- 1000 = 1 * (- 1000) => y = - 9 , x = -900
similarly you can find other solutions
= 2 * (-500) => y = - 8, x = - 400
= 5 * (-200) => y = -5
= 4 * (-250) => y = - 6
= 8 * (-125) => y = -2

y + 10 has to be positive and < 10 so these are only solutions

 

[Mathwebster]

Another way to look at it is from the graph. Here you will see that if $x,y < 0$ then $ -10 < y < 0$ and so it's a matter of setting $y$ equal to a negative integer between $-9$ and $-1$ and solving for $x$ to see which one is a negative integer. It's not the most elegant but it works.

 

[chisigma]

x,y are negative integers

given :

$y=\dfrac {10x}{100-x}$

find all the possible solutions of y

Because is...

$$\lim_{x \rightarrow - \infty} \frac{10 x}{100 - x} = -10\ (1)$$

... the possible solutions will be with $-10 < y < 0$. Proceeding from -9 we can verify...

$$10 x = -9\ (100 - x) \implies x = - 900\ ,\ y = -9\ \text{is solution}$$

$$10 x = -8\ (100 - x) \implies x = - \frac{800}{2}= -400\ ,\ y = -8\ \text{is solution}$$

$$10 x = -6\ (100 - x) \implies x = - \frac{600}{4}= - 150\ ,\ y = -6\ \text{is solution}$$

$$10 x = -5\ (100 - x) \implies x = - \frac{500}{5}= -100\ ,\ y = -5\ \text{is solution}$$

$$10 x = -2\ (100 - x) \implies x = - \frac{200}{8}= -25\ ,\ y = -2\ \text{is solution}$$

Kind regards

$\chi$ $\sigma$