How do you combine #4/(3p)-5/(2p^2)#?

Question

Abigail Allen · Accepted Answer

#(8p-15)/(6p^2)#

Take L CM for the denominator:

Factors of 3p are #3, color(red)p# Factors of #2p^2# are #2, p, color(red)p#

L C M of the Denominator is #3* 2* p* color(red)p= 6p^2# #color(red)p# is used only once as it is appearing in both.

Combining the two terms, #(4/(3p))-(5/(2p^2)) = ((4*2*p)-(5*3))/(6p^2)# #=(8p-15)/(6p^2)#

Hazel Anglin · Answer

#(8p-15)/(6p^2)#

First get common denominators.

#3p# and #2p^2# will have the least common multiple of #6p^2#

We can find this by looking at one term, determining what factors it is missing from the other term, and then multiplying those factors in.

#3p = 3 * p# #2p^2 = 2 * p * p#

They both have a #p#, so let's take the first one and give it the ones the second one has except for a #p#. #3*p color(blue)( * 2 * p) = 6p^2#

Now we multiply each fraction by a unit factor to get the common denominator.

#4/(3p) - 5/(2p^2)#

#color(blue)((2p)/(2p))*4/(3p) - color(blue)(3/3)*5/(2p^2)#

#(8p)/(6p^2) - 15/(6p^2)#

Now we can finally combine the fractions

#(8p-15)/(6p^2)#

Genesis Allen · Answer

undefined To combine 4/(3p) and -5/(2p^2), we need a common denominator. The least common denominator is 6p^2. Multiplying the first fraction by 2p/2p and the second fraction by 3/3, we get (8p)/(6p^2) - (15)/(6p^2). Combining the fractions, we have (8p - 15)/(6p^2).