I need help: Simplify 1/4(8m - 4n) +1/3 (6m + 3n).
factor both denominators
all it asks is to simplify, i don't know how to do it
Multiply 4(8m-4n) and 3(6m+3n)
i just wanted to make sure it look like the way i wrote it earlier .... and NOT: \[\frac 14 (8m-4n) + \frac 13(6m+3n)\]
@jewjewbri ?
@jewjewbri could you clarify, does your expression look like the first one or the second one?
the second one
then it is much easier: what is the common factor of (8m-4n)
2?
a/c+b/d=ad+cb/cd
\[\large 8m+4n= 2(4m+2n) = \dots\] keep going ... there is still some thing in common left
If the question look like the first one.
8m+4n=8m+2n
we are simplifying:\[\Large \frac 14 (8m-4n) + \frac 13(6m+3n)\] all right @jewjewbri let's try this another way... Are you familiar with the distributive property?
no
Hey for the fist expression you can factor ou as (8m-4n)= 4(2m-n). It will be more simple
\[\large a(b+c) = ab+ ac\]
And the second, using the distribuitive property got (6m+3n)=3(2m+n)
You got (6m+3n)=3(2m+n)
Make some algebric now jewjewbri.
6m+3n=6m+3n
try it another way: \[\frac 14 \left( 8m-4n \right)=\frac14\cdot8m-\frac14\cdot 4n=...\] instead of doing one-fourth of the whole thing do 1 fourth of the parts...
what is : \[\frac 14 \cdot 8m = {8m \over 4}=???\]
Great tip.
Just make the division of 8 by 4. Jewjewbri.
@jewjewbri ?
what is 8 divided by 4? @jewjewbri
2
so, 1-fourth of 8m is \(2m\)...
same way 1-fourth of 4n is \(1n\), or just \(n\)
so, \[\large \frac 14 \left( 8m-4n \right)=\color{green}{2m-n}\]
you can do the second part the same way yourself: \[\large \frac 13 \left( 6n+3n \right)=\]
2m+n
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