Question

help please! ive been stuck on this for a while ;_;
http://www.clipular.com/c?6515160=7_CJs6NMh5fHH4Re57wDDT07kJE&f=.png

Answer 1

it might help to realize that \[A\frac{b}{c}=A+\frac{b}{c}\]
\[-~A\frac{b}{c}=-A-\frac{b}{c}\]

Answer 2

i dont really understand that o_o

Answer 3

for 6*(1/9) you multiply the 6 and 9 and add 1, all while keeping it over the denominator of 9.

Answer 4

for 3*(1/3) you do the same thing. multiply 3*3 and add 1, keeping it over 3.

Answer 5

\[6\frac{1}{9}-3\frac{1}{3}\]Convert both to improper fractions:
\[\frac{6*9+1}{9} - \frac{3*3+1}{3} = \frac{73}{9}-\frac{10}{3}\]Multiply the second fraction by \(\frac{3}{3}\) to change the denominator to 9:\[\frac{73}{9}-\frac{10*3}{3*3} = \frac{73}{9}-\frac{30}{9} =\]you do the rest

Answer 6

\[6\frac{1}{9}-3\frac{1}{3}=n\]
\[6+\frac{1}{9}-3-\frac{1}{3}=n\]
\[54+1-27-3=9n\]
\[55-30=9n\]
\[25=9n\]solve for n

Answer 7

Uh, duh, Bill, lets do our simple multiplication correctly :-)
\[\frac{6*9+1}{9} - \frac{3*3+1}{3} = \frac{55}{9}-\frac{10}{3}\]
\[\frac{55}{9}-\frac{10*3}{3*3} = \frac{55}{9}-\frac{30}{9} =\]

Answer 8

I'm getting old, that 6 looked like an 8!

Answer 9

I have to say I like @amistre64's approach here!