## anonymous one year ago can someone help me with this? [(16/m-3)-(4/m-4)] / [(16/m^2)-(m-4/m-3)}

1. freckles

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

2. anonymous

yes its a complex fraction

3. freckles

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

4. anonymous

[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) ?????

5. freckles

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

6. freckles

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

7. freckles

you should see something already that cancels

8. anonymous

m-3

9. anonymous

hello?

10. freckles

yes the (m-3) cancels

11. anonymous

now whhat

12. anonymous

@freckles

13. freckles

multiply top and bottom

14. freckles

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

15. freckles

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