# Help Help!

• March 7th 2006, 03:46 PM
autopimp
Help Help!
• March 7th 2006, 04:05 PM
topsquark
Quote:

Originally Posted by autopimp

$\frac{y^0x^{-8}}{x^{-7}z} \left ( \frac{(y^5)^{-2}}{(z^0)^9} - \frac{x^{11}}{y^4} \right )$

First, anything to the zeroth power is 1, so:
$\frac{1*x^{-8}}{x^{-7}z} \left ( \frac{(y^5)^{-2}}{(1)^9} - \frac{x^{11}}{y^4} \right )$

Now, $x^{-a} \equiv \frac{1}{x^a}$ and $\frac{1}{x^{-a}} \equiv x^a$, so:
$\frac{x^7}{x^8z} \left ( \frac{1}{(y^5)^2} - \frac{x^{11}}{y^4} \right )$

Cancelling the x's in the first fraction:
$\frac{1}{xz} \left ( \frac{1}{(y^5)^2} - \frac{x^{11}}{y^4} \right )$

Now, $(x^a)^b = x^{ab}$, so:
$\frac{1}{xz} \left ( \frac{1}{y^{10}} - \frac{x^{11}}{y^4} \right )$

Now we add the fractions by getting a common denominator, which is $y^{10}$:
$\frac{1}{xz} \left ( \frac{1}{y^{10}} - \frac{x^{11}}{y^4} * \frac{y^6}{y^6} \right )$
$\frac{1}{xz} \left ( \frac{1}{y^{10}} - \frac{x^{11}y^6}{y^4y^6} \right )$
$\frac{1}{xz} \left ( \frac{1}{y^{10}} - \frac{x^{11}y^6}{y^{10}} \right )$
$\frac{1}{xz} \left ( \frac{1-x^{11}y^6}{y^{10}} \right )$

I am leaving the answer in this form because nothing will cancel upon multiplying the $\frac{1}{xz}$.

-Dan
• March 7th 2006, 04:26 PM
topsquark
Quote:

Originally Posted by autopimp

2. $\frac{5-\sqrt2}{\sqrt2}$

We need to remove the $\sqrt2$ from the denominator. So I will multiply the numerator and denominator by $\sqrt2$:
$\frac{5-\sqrt2}{\sqrt2}*\frac{\sqrt2}{\sqrt2}$
$\frac{5 \sqrt2-\sqrt2 \sqrt2}{\sqrt2 \sqrt2}$
$\frac{5 \sqrt2-2}{2}$.

3. $\frac{x-2}{5}-\frac{x-7}{3}=-1$
We need to get a common denominator, which in this case is 15:
$\frac{x-2}{5}*\frac{3}{3}-\frac{x-7}{3}*\frac{5}{5}=-1$

$\frac{3x-6}{15}-\frac{5x-35}{15}=-1$

$\frac{3x-6-(5x-35)}{15}=-1$

$\frac{3x-6-5x+35}{15}=-1$

$\frac{-2x+29}{15}=-1$

Now we multiply both sides of the equation by 15:
$-2x+29=-15$

$-2x+29-29=-15-29$

$-2x=-44$

$\frac{-2x}{-2}=\frac{-44}{-2}$

$x=22$.

4. $3\sqrt{32}-5\sqrt{50}-8\sqrt{45}+7\sqrt{20}$

Let's simplify some of these radicals:
$\sqrt{32}=\sqrt{2*16}=4\sqrt2$
$\sqrt{50}=\sqrt{2*25}=5\sqrt2$
etc.

Once you get those, plug them into your original expression:
$3(4\sqrt2)-5(5\sqrt2)+...$
$12\sqrt2-25\sqrt2 + ...$

Then just collect like terms. My final answer was $-13\sqrt2-10\sqrt5$.

-Dan
• March 7th 2006, 04:34 PM
topsquark
Quote:

Originally Posted by autopimp

5. $\frac{5x-2}{x^2-5x-14}-\frac{3}{x^2+2x}$

First, factor the denominators:
$\frac{5x-2}{(x-7)(x+2)}-\frac{3}{x(x+2)}$

A good common denominator for this would be: $x(x-7)(x+2)$. So:
$\frac{5x-2}{(x-7)(x+2)}*\frac{x}{x}-\frac{3}{x(x+2)}*\frac{(x-7)}{(x-7)}$

$\frac{x(5x-2)}{x(x-7)(x+2)}-\frac{3(x-7)}{x(x-7)(x+2)}$

$\frac{x(5x-2)-3(x-7)}{x(x-7)(x+2)}$

By the way, you rarely want to multiply out the denominator. Just leave it like it is.

$\frac{(5x^2-2x)-(3x-21)}{x(x-7)(x+2)}$

$\frac{5x^2-2x-3x+21}{x(x-7)(x+2)}$

$\frac{5x^2-5x+21}{x(x-7)(x+2)}$

At this stage ALWAYS look to see if the numerator factors. In this case, it doesn't so we are done.

7. $\frac{x}{x^2-16}$ divided by $\frac{x^3}{x^2-14x+40}$

When you divide by a fraction, you are multiplying by the inverse of the fraction:
$\frac{x}{x^2-16}*\frac{x^2-14x+40}{x^3}$

At this point, factor what you can, then terms will cancel.

-Dan
• March 7th 2006, 04:46 PM
topsquark
Quote:

Originally Posted by autopimp

6. This is similar to what I did in 1. Group the powers of 10 together and apply the exponent rules I listed in 1. I get 250000.

8.
$3x+5y=5$
$x-2y=9$

For the substitution method we pick one equation to solve for one variable and put it into the remaining equation. I usually pick the easiest one to solve, so I'll solve the second equation for x:
$x=2y+9$

Now we put this in as the value for x in the first equation:
$3(2y+9)+5y=5$
This is now one equation in one unknown. Solve this for y and use your x equation to find x. I get (x,y)=(5,-2).

-Dan