hey guys, need help simplifying this expression. any help i would appreciate it! thank uu

$\displaystyle

\frac{x^2-y^2}{4xy-y^2} \cdot \frac{y-4x}{x^2+xy}

$

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- Aug 19th 2008, 07:48 AMjvignaciosimplifying the expression
hey guys, need help simplifying this expression. any help i would appreciate it! thank uu

$\displaystyle

\frac{x^2-y^2}{4xy-y^2} \cdot \frac{y-4x}{x^2+xy}

$ - Aug 19th 2008, 08:00 AMo_O
$\displaystyle \frac{x^2-y^2}{4xy-y^2} \cdot \frac{y-4x}{x^2+xy}$

$\displaystyle = \frac{(x-y)(x+y)}{y(4x-y)} \cdot \frac{y-4x}{x(x+y)}$

I'll break this into pieces so you'll see what is happening:

- $\displaystyle x^2 - y^2 = (x-y)(x+y)$. This is what the difference of squares is (ring a bell?).
- $\displaystyle 4xy - y^2 = y(4x - y)$. The most you can take out from the whole expression (i.e. BOTH terms) is a single $\displaystyle y$. You can't factor out an $\displaystyle x$ because the second term does not have one and you can't factor out another $\displaystyle y$ because the first term only has one $\displaystyle y$.
- $\displaystyle y-4x$. Not much we can do here but notice how this looks like $\displaystyle 4x - y$ in the second bullet above? They look similar but are not equal so that you can
*not*directly cancel them. However, (remember this trick), if you factor out a -1 from this expression, we get: $\displaystyle -(4x - y)$ which we can now cancel with the other (4x - y). - $\displaystyle x^2 + xy = x(x+y)$. The most you can factor out from
*both*terms is an $\displaystyle x$. The reasoning is the same as the second bullet.

Now, see what you can cancel and hopefully you'll be able to do your other problems similarly as well. - $\displaystyle x^2 - y^2 = (x-y)(x+y)$. This is what the difference of squares is (ring a bell?).
- Aug 19th 2008, 08:03 AMSoroban
Hello, jvignacio!

I must assume that you know that we must factor and reduce . . .

Quote:

Simplify: .$\displaystyle \frac{x^2-y^2}{4xy-y^2} \cdot \frac{y-4x}{x^2+xy}$

Factor: . $\displaystyle \frac{(x-y)(x+y)}{y\,(4x-y)} \cdot \frac{-(4x-y)}{x\,(x+y)} $

Reduce: . $\displaystyle \frac{(x-y){\color{red}\rlap{//////}}(x+y)}{y\,{\color{blue}\rlap{///////}}(4x-y)} \cdot \frac{-{\color{blue}\rlap{///////}}(4x-y)}{x\,{\color{red}\rlap{//////}}(x+y)} $ .$\displaystyle = \;\frac{-(x-y)}{xy} \;=\;\frac{y-x}{xy} $

- Aug 19th 2008, 08:11 AMjvignacio
- Aug 19th 2008, 10:20 AMo_O
Yep, just as Soroban has it.

$\displaystyle -x + y = y - x$ (order doesn't matter)