Fractions - GRE Quantitative Reasoning

With this problem the first thing to do is cancel out variables. The x2 can all be divided by each other because they are present in each system. The equation will now look like this:

Now we can see that the equation can all be divided by y, leaving the answer to be:

_x_2 – _y_2 can be also expressed as (x + y)(x – y).

Therefore, the fraction now can be re-written as (x + y)(x – y)/(x + y).

This simplifies to (x – y).

Following this equation, you divide 4 by 8 to get 1/2. When a variable is raised to an exponent, and you are dividing, you subtract the exponents, so 6 – 3 = 3.

125 goes into 2000 evenly 16 times. 1/16 is the fraction in its simplest form.

120 goes into 6000 evenly 50 times, so we get 1/50 as our simplified fraction.

(2_x_ + 4)/(x + 2)

To simplify you must first factor the top polynomial to 2(x + 2). You may then eliminate the identical (x + 2) from the top and bottom leaving 2.

Set up the conversions as fractions and solve:

To simplify exponents which are being divided, subtract the exponents on the bottom from exponents on the top. Remember that only exponents with the same bases can be simplified

Factor both the numerator and the denominator:

After reducing the fraction, all that remains is:

Notice that the term appears frequently. Let's try to factor that out:

Now multiply both the numerator and denominator by the conjugate of the denominator:

The numerator and denominator are both divisible by 12. Thus, we divide both by 12 to get our final answer.

If we instead divide by another common factor, we may need to complete the process again to make sure that we have completely reduced the fraction.

In mathematical words we get the following:

This can be solved using 2 methods.

The most time-efficient solution would recognize that is the largest value and nearly equals the sum the other fraction by itself.

The more time consuming method would be to convert each fraction to decimal form and calculate the sum of each quantity.

Quantity A:

Quantity B:

When we factor the numerator and denominator, we get:

.

After cancelling , we are left with .

With common denominators, the range is from

or

.

The only fraction that falls in either of these ranges is .

Begin by simplifying all terms inside the parentheses. Begin with the innermost set. Find a common denominator for the two terms. In this case, the common denominator will be twenty:

Simplify to and convert to not a mixed fraction:

Multiply the two fractions in the parentheses by multiplying straight across (A quick shortcut would be to factor out the 10 on top and bottom).

Now convert to a non-mixed fraction. It will become .

In order to subtract the two fractions, find a common denominator. In this case, it will be 70.

Now subtract, and find the answer!

is the answer

To simplify a complex fraction, simply invert the denomenator and multiply by the numerator:

Multiplying the numerator by the reciprocal of the denominator for each term we get:

Since we have a common denominator we can now add these two terms.

Although you could look for the common denominator of the fraction as it has been written, it is probably easiest to rewrite the fraction in slightly simpler terms. Thus, recall that you can rewrite your fraction as:

Using the rule for dividing fractions, you can rewrite your expression as:

Then, you can multiply each set of fractions, getting:

This makes things very easy, for then your value is:

For this problem, begin by rewriting the complex fraction, using the rule for dividing fractions:

This is much easier to work on. Cancel out the s and the and the , this gives you:

, which is merely . Thus, your problem is:

The common denominator is , so you can rewrite this as:

Whenever there are decimals in fractions, we remove them by shifting the decimal place over however many it takes to make number an integer.

In this case we have to move the decimal in the numerator to the right one place.

Then, we add just one zero to the denominator.

Final answer becomes:

.

With the numerator having more decimal spots than the denominator, we need to move the decimal point in the numerator two places to the right.

Then in the denominator, we move the decimal point also two to the right. Since there's only one decimal place we just add one more zero.

Then we can reduce by dividing top and bottom by .