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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

x,y are negative integers

given :

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

find all the possible solutions of y

given :

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

find all the possible solutions of y

- Thread starter Albert
- Start date

- Thread starter
- #1

- Jan 25, 2013

- 1,225

x,y are negative integers

given :

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

find all the possible solutions of y

given :

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

find all the possible solutions of y

- Mar 31, 2013

- 1,346

y(100-x) = 10 xx,y are negative integers

given :

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

find all the possible solutions of y

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

Last edited:

- Jan 26, 2012

- 183

- Feb 13, 2012

- 1,704

Because is...x,y are negative integers

given :

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

find all the possible solutions of y

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

... 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

$\chi$ $\sigma$