How do you solve #8/(2m+1) - 1/(m-2) = 5/(2m+1)# and find any extraneous solutions?

1 Answer
May 5, 2018

Answer:

#m=7#

Explanation:

#8/(2m+1) - 1/(m-2) = 5/(2m+1)#

The restrictions for #m# are:
#m!=-1/2 and m!=2#

When you have an equation with fractions, you can get rid of the denominators immediately by multiplying each term by the LCM of the denominators. In this case it would be #color(blue)((2m+1)(m-2))#

#(color(blue)(cancel((2m+1))(m-2)xx)8)/cancel((2m+1)) - (color(blue)((2m+1)cancel((m-2)))xx1)/cancel((m-2)) = (color(blue)(cancel((2m+1))(m-2))xx5)/cancel((2m+1))#

This leaves us with an equation with no fractions.

#8(m-2) -(2m+1) = 5(m-2)#

#" "8m-16 -2m-1 =5m-10#

#" "8m-2m-5m = 16+1-10" "larr#re-arrange

#color(white)(xxxxxxxxxxxx)m=7#

This is not one of the resrcticed vales for #m# so there are no extraneous solutions.