Question

I need help understanding how to solve a linear equation with fractions
1/6 (2/c -1/6) = 1/3c -1/6

Answer 1

is the problem\[\frac16(\frac2c-\frac16)=\frac1{3c}-\frac16\]?

Answer 2

yes

Answer 3

you need to find the GCF of the denominators and multiply the whole equation by that to get rid of the fractions

do you know how to find the GCF of a set of expressions?

Answer 4

that would be 6?

Answer 5

actually it would be 6c
if we don't include the c, then we leave out c as a factor and we will still have fractions, which we don't want

Answer 6

the denominators are
6=2*3
c
3

so their GCF is 2*3*c=6c

Answer 7

now multiply everything by 6c

Answer 8

okay could you show me how to set the problem up to be solved

Answer 9

\[6c[\frac16(\frac2c-\frac16)]=6c[\frac1{3c}-\frac16]\]now simplify and solve for c

Answer 10

so i can cancel out the 6 with 1 and cancel out the 6c leaves 1c, then i multiplu 1c x 2 =2/c

Answer 11

couls you show what it look like after i simplify plz

Answer 12

what is\[6c(\frac16)\]?

Answer 13

cancel out the bottom 6 by 6 and get 1, cancel out the 6c with 6 and get 1c

Answer 14

1/6 (2c-1) =1/3c-1/6 this is how it looks. Srry i made a mistake with the equation