How do you solve #p/(p-2) - 1/2 = 3/(3p-6)#?

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Answer 1 · 2017-04-14T19:18:54

#p=0#

Find a common denominator.

I can see that #3p-6# is actually #3(p-2)# There's also a #2# in #1/2#. So a common denominator is #6(p-2)#

Take this common denominator and multiply everything by that:

#6p-3(p-2)=6#

Distribute the #3#

#6p-3p+6=6#

Combine the #p#s:

#3p+6=6#

Subtract #6# on both sides:

#3p=0#

Divide #3# on both sides to solve for #p#:

#p=0#

Plug #p=0# back into the equation to make sure it works:

#(0/(0-2))-(1/2)=3/(3(0)-6)#

#-1/2=3/-6#

Simplifying #3/-6# would get #-1/2# so the answer works!

Answer 2 · 2017-04-14T19:20:26

#p = 0#

Multiply both sides by #3 p - 6#:
#1/2 (6 - 3 p) + (p (3 p - 6))/(p - 2) = 3#

Rewrite the left hand side by combining fractions. #1/2 (6 - 3 p) + (p (3 p - 6))/(p - 2) = (3 (p + 2))/2#:

#(3 (p + 2))/2 = 3#

Multiply both sides by #2/3#:
#p + 2 = 2#

Subtract 2 from both sides:
Answer:
#p = 0#