All SAT Math Resources
Example Questions
Example Question #143 : Expressions
for .
Which of the following is equal to ?
Square both expressions
Add 6 to both sides:
Take the square root of both sides:
Example Question #144 : Expressions
. is a positive number.
Which of the following is equal to ?
None of the other choices gives the correct response.
, so, taking the square root of both sides:
is positive, so is as well; therefore,
Add 4 to both sides:
Square both sides, and apply the binomial square pattern to the right expression:
Example Question #145 : Expressions
. .
Which of the following is equal to ?
Take the reciprocal of both expressions:
Add 4 to both sides:
Rewrite the expression at left and simplify it:
Take the reciprocal of both expressions:
Example Question #5 : How To Simplify Expressions
Simplify the following expression: x3 - 4(x2 + 3) + 15
To simplify this expression, you must combine like terms. You should first use the distributive property and multiply -4 by x2 and -4 by 3.
x3 - 4x2 -12 + 15
You can then add -12 and 15, which equals 3.
You now have x3 - 4x2 + 3 and are finished. Just a reminder that x3 and 4x2 are not like terms as the x’s have different exponents.
Example Question #2611 : Sat Mathematics
The expression
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.
Example Question #833 : Algebra
Two two-digit numbers, and , sum to produce a three-digit number in which the second digit is equal to . The addition is represented below. (Note that the variables are used to represent individual digits; no multiplication is taking place).
What is ?
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 .
Example Question #2612 : Sat Mathematics
Let = , where and are both positive integers whose greatest common factor is one. What is the value of ?
103
73
44
34
115
103
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.
Example Question #2 : Algebraic Fractions
Change to a mixed number
To convert from a fraction to a mixed number we must find out how many times the denominator goes into the numerator using division and the remainder becomes the new fraction.
Example Question #2613 : Sat Mathematics
What is the average of and ?
To average, we have to add the values and divide by two. To do this we need to find a common denomenator of 6. We then add and divide by 2, yielding 4.5/6. This reduces to 3/4.
Example Question #4 : Algebraic Fractions
Which of the following is equivalent to ?
None of the answers are correct
This problem is solved the same way ½ + 1/3 is solved. For example, ½ + 1/3 = 3/6 + 2/6 = 5/6. Find a common denominator then convert each fraction into an equivalent fraction using that common denominator. The final step is to add the two new fractions and simplify.
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