Question

please solve for m
(m+8/m^2-3m-4) - (8/m^2-8m+16) = (m-8)/m^2-3m-4

Answer 1

first thing factorize all denominators

Answer 2

how can i factorize the denominator because for example m^2-3m-4.. nothing will go into all three of these terms

Answer 3

rewrite -3m as -4m+m and factorize by grouping

Answer 4

ok,i'll do this one for you:
m^2-3m-4:
you look for 2 numbers whose product is -4 and whose sum is -3:-4,1
now rewrite it as m^2-4m+m-4
now factor each two terms alone,that's grouping
m(m-4)+(m-4)=(m-4)(m+1)

Answer 5

like a quadratic equation?

Answer 6

yes,there are many methods to factorize these,i don't know which one you use

Answer 7

i'm not really sure what to do with this problem at that point. should i divide the (m-4)(m+1) by the other side... can i even do this since it is already a fraction??

Answer 8

just factorize the other denominators for now

Answer 9

can you do it?

Answer 10

yes i'm working on factorizing right now

Answer 11

ok,take your time:)

Answer 12

m(m-4)+(m-4)=(m-4)(m+1)

m(m-4)-4(m-4)=(m-4)(m-4)

m(m-4)+(m+4)=(m-4)(m+1)

Answer 13

the last one should be m(m-4)+(m-4)=(m-4)(m+1)

Answer 14

that's the one i did!
factorize m^2-8m+16 and m^2-3m-4

Answer 15

sorry i can't believe i did the same one three times. all of this algebra is getting to my head. okay i'm doing it now

Answer 16

relax..

Answer 17

so for the m^2-8m+16 -4*-4 give me 16... and -4+-4 = 8....

m^2-4m-4m+16..

m(m-4)-4(m-4)=(m-4)(m-4)

Answer 18

m^2-3m-4 -4*1 = -4 and added gives me -3 so..

m^2-4m+1m-4
m(m-4)+(m-4)=(m-4)(m+1)

Answer 19

good