solve for by simplifying both sides of the equation, then isolation the variable.
r = 0
:)))
The writing is a bit ambiguous, so I'm assuming that the equation is
[tex] \frac{1}{4}r - \frac{1}{16}r + \frac{1}{2}r = \frac{1}{2}r [/tex]
Since [tex]\frac{1}{2}r [/tex] appears as an additive term on both sides, you can simplify it, remaining with
[tex] \frac{1}{4}r - \frac{1}{16}r = 0 [/tex]
Factor the common r:
[tex] (\frac{1}{4} - \frac{1}{16})r = 0 [/tex]
If a multiplication equals zero, one of the factors equal zero. Since the numeric fraction is not zero, the only option is [tex] r=0 [/tex].
I don't believe this is the case, but in case the equation actually was
[tex] \frac{1}{4r} - \frac{1}{16r} + \frac{1}{2r} = \frac{1}{2r} [/tex]
proceed as before to get rid of terms appearing on both sides and factor the r (except now it is at the denominator):
[tex] \frac{1}{4r} - \frac{1}{16r} = 0 [/tex]
[tex] \frac{1}{r}(\frac{1}{4} - \frac{1}{16}) = 0 [/tex]
Again, one of the factor has to be zero, but none of the factor can equal zero, so the equation has no solutions.
