Fractions - PSAT Math

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:

Another way to represent this question is:

In the one's column, and add to produce a number with a two in the one's place. In the ten's column, we can see that a one must carry in order to get a digit in the hundred's place. Together, we can combine these deductions to see that the sum of and must be twelve (a one in the ten's place and a two in the one's place).

In the one's column:

The one carries to the ten's column.

In the ten's column:

The three goes into the answer and the one carries to the hundred's place. The final answer is 132. From this, we can see that because .

Using this information, we can solve for .

You can check your answer by returning to the original addition and plugging in the values of and .

_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.

First we want to simplify the expression: .

One way to simplify this complex fraction is to find the least common multiple of all the denominators, i.e. the least common denominator (LCD). If we find this, then we can multiply every fraction by the LCD and thereby be left with only whole numbers. This will make more sense in a little bit.

The denominators we are dealing with are 2, 3, 4, 5, and 6. We want to find the smallest multiple that these numbers have in common. First, it will help us to notice that 6 is a multiple of both 2 and 3. Thus, if we find the least common multiple of 4, 5, and 6, it will automatically be a multiple of both 2 and 3. Let's list out the first several multiples of 4, 5, and 6.

4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60

5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60

6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60

The smallest multiple that 4, 5, and 6 have in common is 60. Thus, the least common multiple of 4, 5, and 6 is 60. This also means that the least common multiple of 2, 3, 4, 5, and 6 is 60. Therefore, the LCD of all the fractions is 60.

Let's think of the expression we want to simplify as one big fraction. The numerator contains the fractions 1/4, 1/3, and –1/5. The denominator of the fraction is 1/2, –1/6 and 1. Remember that if we have a fraction, we can multiply the numerator and denominator by the same number without changing the value of the fraction. In other words, x/y = (xz)/(yz). This will help us because we can multiply the numerator (which consists of 1/4, 1/3, and –1/5) by 60, and then mutiply the denominator (which consists of 1/2, –1/6, and 1) by 60, thereby ridding us of fractions in the numerator and denominator. This process is shown below:

= =

=

This means that a/b = 23/80. We are told that a and b are both positive and that their greatest common factor is 1. In other words, a/b must be the simplified form of 23/80. When a fraction is in simplest form, the greatest common factor of the numerator and denominator equals one. Since 23/80 is simplified, a = 23, and b = 80. The sum of a and b is thus 23 + 80 = 103.

The answer is 103.

(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.

A negative exponent in the numerator of a fraction can be rewritten with a positive exponent in the denominator. The same is true for a negative exponent in the denominator. Thus, .

When is multiplied by , the numerators and denominators cancel out, and you are left with 1.

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:

To simply a fraction with variables, subtract exponents of like bases:

This leaves us with the expression:

Next, we know that to change negative exponents in the numerator into positive exponents, we place them in the denominator. Thus, our expression simplifies to:

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:

The and variables with negative exponents can be rewritten with positive exponents by moving them from the denominator to the numerator, and vice versa. Therefore, the expression can be rewritten as

.

The exponents on the denominator can then be subtracted from the exponent in the numerator to give

The expression can be rewritten as

Now the expression can be combined by adding and subtracting exponents

We can combine the expression into

The combined expression can then be simplified to

The expression can be rewritten as

We can now move the variables to the numerator and combine alike variables

This becomes