How do you solve (200/r)-1= 200/(r-10)(200r)−1=200r−10?
1 Answer
Explanation:
200/r - 1 = 200/(r - 10)200r−1=200r−10
Rearrange the equation
200/r - 200/(r - 10) = 1200r−200r−10=1
Take
200[1/r - 1/(r - 10)] = 1200[1r−1r−10]=1
Bring
1/r - 1/(r - 10) = 1/2001r−1r−10=1200
Make denominators equal
(1/r × (r - 10)/(r - 10)) - (1/(r - 10) xx r/r) = 1/200(1r×r−10r−10)−(1r−10×rr)=1200
(r - 10)/(r(r - 10)) - r/(r(r - 10)) = 1/200r−10r(r−10)−rr(r−10)=1200
Now, denominators are same. Numerators can be added
(r - 10 - r) / (r(r - 10)) = 1/200r−10−rr(r−10)=1200
(-10)/(r(r - 10)) = 1/200−10r(r−10)=1200
r(r - 10) = -2000r(r−10)=−2000
r^2 - 10r = -2000r2−10r=−2000
r^2 - 10r + 2000 = 0 color(white)(.)……[1]r2−10r+2000=0.……[1]
Use quadratic equation formula to find value(s) of
r = (-b +- sqrt(b^2 - 4ac))/(2a)r=−b±√b2−4ac2a
In equation
a = 1a=1
b = -10b=−10
c = 2000c=2000
r = (-(-10) +- sqrt((-10)^2 - (4 × 1 × 2000)))/(2 × 1)r=−(−10)±√(−10)2−(4×1×2000)2×1
r = (10 +- sqrt(100 - 8000))/2r=10±√100−80002
r = (10+- sqrt(-7900))/2r=10±√−79002
r = 5 +-sqrt(-7900)/2r=5±√−79002
