How do you solve #\frac { 2} { P } = \frac { 1} { 4p - 3} + \frac { 1} { 4p ^ { 2} - 3p }#?

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Answer 1 · 2017-12-03T04:42:49

#P=(2p(4p- 3))/((p+ 1) #

Solve for P

#2/P=1/(4p−3) +1/(4p^2−3p)#

1) Factor the second denominator to find the LCD

#2/P=1/(4p−3) +1/(p(4p−3))#

2) Change both fractions on the right to equivalents
with an LCD of #p(4p + 3)#

#2/P=(p+ 1)/(p(4p−3) #

3) Get #P# out of the denominator by multiplying both sides by #P#

#2=(P(p+ 1))/(p(4p−3) #

4) Clear the denominator by multiplying both sides by #p(4p-3)# and letting the denominator cancel

#2p(4p-3) = P(p+ 1) #

5) Divide both sides by #(p + 1)# to isolate #P#

#P=(2p(4p- 3))/((p+ 1) #