Question

How do you solve `(200/r)-1= 200/(r-10)`?

Answer 1

`r = 5 +- sqrt(-7900)/2`

Explanation:

`200/r - 1 = 200/(r - 10)`

Rearrange the equation

`200/r - 200/(r - 10) = 1`

Take `200` common

`200[1/r - 1/(r - 10)] = 1`

Bring `200` to other side

`1/r - 1/(r - 10) = 1/200`

Make denominators equal

`(1/r × (r - 10)/(r - 10)) - (1/(r - 10) xx r/r) = 1/200`

`(r - 10)/(r(r - 10)) - r/(r(r - 10)) = 1/200`

Now, denominators are same. Numerators can be added

`(r - 10 - r) / (r(r - 10)) = 1/200`

`(-10)/(r(r - 10)) = 1/200`

`r(r - 10) = -2000`

`r^2 - 10r = -2000`

`r^2 - 10r + 2000 = 0 color(white)(.)……[1]`

Use quadratic equation formula to find value(s) of `r`

`r = (-b +- sqrt(b^2 - 4ac))/(2a)`

In equation `[1]`

• `a = 1`

• `b = -10`

• `c = 2000`

`r = (-(-10) +- sqrt((-10)^2 - (4 × 1 × 2000)))/(2 × 1)`

`r = (10 +- sqrt(100 - 8000))/2`

`r = (10+- sqrt(-7900))/2`

`r = 5 +-sqrt(-7900)/2`