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- Mar 7th 2006, 03:46 PMautopimpHelp Help!
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- Mar 7th 2006, 04:05 PMtopsquarkQuote:

Originally Posted by**autopimp**

First, anything to the zeroth power is 1, so:

Now, and , so:

Cancelling the x's in the first fraction:

Now, , so:

Now we add the fractions by getting a common denominator, which is :

I am leaving the answer in this form because nothing will cancel upon multiplying the .

-Dan - Mar 7th 2006, 04:26 PMtopsquarkQuote:

Originally Posted by**autopimp**

We need to remove the from the denominator. So I will multiply the numerator and denominator by :

.

3.

We need to get a common denominator, which in this case is 15:

Now we multiply both sides of the equation by 15:

.

4.

Let's simplify some of these radicals:

etc.

Once you get those, plug them into your original expression:

Then just collect like terms. My final answer was .

-Dan - Mar 7th 2006, 04:34 PMtopsquarkQuote:

Originally Posted by**autopimp**

First, factor the denominators:

A good common denominator for this would be: . So:

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

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

7. divided by

When you divide by a fraction, you are multiplying by the inverse of the fraction:

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

-Dan - Mar 7th 2006, 04:46 PMtopsquarkQuote:

Originally Posted by**autopimp**

8.

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:

Now we put this in as the value for x in the first equation:

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