Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

Login

Username

Password

Not registered yet?

Post a reply

Go back

Write your message and submit
:) :| :( :D :o ;) :/ :P :lol: :mad: :rolleyes: :cool: | :dizzy :eek :kiss :roflol :rolleyes :shame :down :up :touched :sleep :wave :swear :tongue :what :faint :dunno
Options

Go back

Topic review (newest first)

noelevans
2012-08-23 12:51:07

Hi SlowlyFading!

Clicking on your LaTex I think I can see what the problems look like:

1)  (21x^3+14x)/7  =  21x^3/7  +  14x/7  =  ?

2)  (3(2x-3)-x(2x-3))/(2x-3)  =  3(2x-3)/(2x-3)  -  x(2x-3)/(2x-3)   =  ?

3)  ((x^2(5x+6)-3(5x+6))/(5x+6)  =  x^2(5x+6)/(5x+6)  -  3(5x+6)/(5x+6)  =  ?

In each of these we are dividing the denominator into each of the two terms.  Then in
each of these cancel to obtain the final answer.

Or you can take the original numerators and factor them and then cancel:

1)  7x(3x^2+2)/7   =  ?

2)  (2x-3)(3-x)/(2x-3)  =  ?

3)  (5x+6)(x^2-3)/(5x+6)  =  ?

The first approach is probably easier.

Can you take them from here?  smile

bobbym
2012-08-23 12:37:34

Hi;

The latexing did not turn out well so we are going to need to put the problems in an understandable form.

I am guessing for the first one:



For the second one,



The third

SlowlyFading
2012-08-23 06:42:01

These are some more problems of the same type that I got wrong in my newest lesson! I'll need the answers and your work for it.





Board footer

Powered by FluxBB