Question

can someone help me with this?

[(16/m-3)-(4/m-4)]
/
[(16/m^2)-(m-4/m-3)}

Answer 1

\[[\frac{16}{m-3}-\frac{4}{m-4}] \div [\frac{16}{m^2}-\frac{m-4}{m-3}] ?\]

Answer 2

yes its a complex fraction

Answer 3

do you know how to combine the first set of fractions
that do you know how to write
\[\frac{16}{m-3}-\frac{4}{m-4} \text{ as one fraction }?\]

Answer 4

[16(m-4)-4(m-3)/(m-3)(m-4)] ? i havent figured out how to use the equation writer yet sorry! so it becomes 16m-64-4m+12,

12m-52/(m-3)(m-4) ?????

Answer 5

\[\frac{16(m-4)-4(m-3)}{(m-3)(m-4)}=\frac{16m-64-4m+12}{(m-3)(m-4)} \\ =\frac{12m-52}{(m-3)(m-4)}\]
that is awesome now we have to work with the second pair of fractions that follow the division sign

Answer 6

\[\frac{16}{m^2}-\frac{m-4}{m-3}=\frac{16(m-3)-m^2(m-4)}{m^2(m-3)} \\ =\frac{16m-48-m^3+4m^2}{m^2(m-3)}=\frac{-m^3+4m^2+16m-48}{m^2(m-3)}\]


so you have:
\[\frac{12m-52}{(m-3)(m-4)} \div \frac{-m^3+4m^2+16m-48}{m^2(m-3)} \\ \text{ which we can change \to multiplication } \\ \text{ so we have} \\ \frac{12m-52}{(m-3)(m-4)} \times \frac{m^2(m-3)}{-m^3+4m^2+16m-48}\]
notice I just flipped the second fraction

Answer 7

you should see something already that cancels

Answer 8

m-3

Answer 9

hello?

Answer 10

yes the (m-3) cancels

Answer 11

now whhat

Answer 12

@freckles

Answer 13

multiply top and bottom

Answer 14

we know to this because there is a operation of multiplication between the fractions

Answer 15

\[\frac{(12m-52)(m^2)}{(m-4)(-m^3+4m^2+16m-48)}\]