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

1 Answer
Apr 5, 2018

Answer:

#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#