I just need some confirmation to whether or not these are right,
can I simplify $\displaystyle -\frac{3x^2}{2x}$ into $\displaystyle \frac{3x}{2} $?
can $\displaystyle \frac{x+3}{x+6} $ be simplified to $\displaystyle \frac1{2} $?
I just need some confirmation to whether or not these are right,
can I simplify $\displaystyle -\frac{3x^2}{2x}$ into $\displaystyle \frac{3x}{2} $?
can $\displaystyle \frac{x+3}{x+6} $ be simplified to $\displaystyle \frac1{2} $?
on the first you're correct. and let me tell you why.
you have: $\displaystyle \frac{ - 3x^2}{2x}$
Pull out the x on top and bottom: $\displaystyle \frac{ x(-3x)}{x(2)}$
this is the same as: $\displaystyle \frac{x}{x}\cdot \frac{-3x}{2}$
$\displaystyle \frac{x}{x}$ is one. anything divided by itself is one (except zero).
So you're left with: $\displaystyle 1\cdot\frac{-3x}{2} = \frac{-3x}{2} $
But there's another way to think about this, a more easier way.
you can just cancel the x's: $\displaystyle \frac{ \not x(-3x)}{\not x(2)}$
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On the second problem, you're wrong.
There's nothing you can factor out on the top and bottom.
but say you had: $\displaystyle \frac{4x + 8}{2x + 6}$
now you can pull out factors: $\displaystyle \frac{2(2x + 4)}{2(x + 3)} = \frac{ \not 2 (2x + 4)}{ \not 2 (x + 3)}$
$\displaystyle = \frac{2x + 4}{x + 3}$