1. Plug in wherever there is an in the above equation.

2. Perform the above operations.

The common denominator of the left side is x(x–1). Multiplying the top and bottom of 1/x by (x–1) yields

Since this statement is true, there are infinitely many solutions.

Begin by simplifying the numerator and the denominator.

Numerator

has a common denominator of . Therefore, we have:

Denominator

has a common denominator of . Therefore, we have:

Now, reconstructing our fraction, we have:

To make this division work, you multiply the numerator by the reciprocal of the denominator:

Solve for the variables, the plug into formula.

x = 12/3 = 4

y = 10 * 4 = 40

z= 9/4 = 2 1/4

10xyz = 3600

Xy = 160

3600/160 = 22 1/2

To begin a problem like this, you must first convert everything to and alone. This way, you can begin to cancel and combine to its most simplified form.

Since and , we insert those identities into the equation as follows.

From here we combine the numerator and denominators of each fraction together to easily see what we can combine and cancel.

Since there is a in the numerator and the denominator, we can cancel them as they divide to equal 1. All we have left is , the answer.

In order to solve , we must first find the values of , , and using the initial equations provided. Starting with :

Then:

Finally:

With the values of , , and in hand, we can solve the final equation:

Recall that dividing is equivalent multiplying by the reciprocal. Therefore, ((x - 4) / (1 / 2)) / (1 / (x + 4)) = ((x - 4) * 2) * (x + 4) / 1.

Let's simplify this further:

(2x – 8) * (x + 4) = 2x2 – 8x + 8x – 32 = 2x2 – 32

To raise to the exponent , square and then take the cube root.

Begin by simplifying the numerator.

has a common denominator of . Therefore, we can rewrite it as:

Now, in our original problem this is really is:

When you divide by a fraction, you really multiply by the reciprocal:

If 10 pizzas cost x dollars, then each pizza costs x/10. Similarly, each soda costs y/6. We can add the pizzas and sodas together by finding a common denominator: