1. simplifying the expression

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

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

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

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

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.

3. Hello, jvignacio!

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

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

4. Originally Posted by o_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.

wow thats for the help.. u really explained it well..

so the answer to this would be

$\displaystyle \frac{-x+y}{yx}$

?? the top one like that because u solve -(x-y) which is -x+y

5. Yep, just as Soroban has it.

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