All SAT Math Resources
Example Questions
Example Question #884 : Arithmetic
Solve:
To solve this problem, we want to set what's inside the absolute value signs equal to the positive and negative value on the right side of the equation. That's because the value inside the absolute value symbols could be equivalent to or , and the equation would still hold true.
So let's set equal to and separately and solve for our unknown.
First:
Second:
Therefore, our answers are and .
Example Question #21 : Absolute Value
Evaluate the expression if and .
To solve, we replace each variable with the given value.
Simplify. Remember that terms inside of the absolute value are always positive.
Example Question #891 : Arithmetic
Simplify the following:
Simplify the following:
Begin with basic subtraction:
Next, remember what we do with absolute value signs; we change negative values to positive values, and positive values remain positive.
So our answer is positive 51
Example Question #14 : How To Find Absolute Value
Solve for all possible values of x.
When solving for x in the presence of absolute value, there are always two answers.
To eliminate the absolute value, the equation must be re-written two ways:
and
and
and
and
and
Example Question #3 : How To Find Absolute Value
Define an operation as follows:
For all real numbers ,
Evaluate
Both and
Example Question #892 : Arithmetic
If , what is the value of ?
Substitute – 4 in for x. Remember that when a negative number is raised to the third power, it is negative. - = – 64. – 64 – 36 = – 100. Since you are asked to take the absolute value of – 100 the final value of f(-4) = 100. The absolute value of any number is positive.
Example Question #1 : Algebra
If 6x = 42 and xk = 2, what is the value of k?
1/7
2/7
5
1/6
7
2/7
Solve the first equation for x by dividing both sides of the equation by 6; the result is 7. Solve the second equation for k by dividing both sides of the equation by x, which we now know is 7. The result is 2/7.
Example Question #1 : Algebra
If 4x + 5 = 13x + 4 – x – 9, then x = ?
8
0
5/8
5/4
–5/4
5/4
Start by combining like terms.
4x + 5 = 13x + 4 – x – 9
4x + 5 = 12x – 5
–8x = –10
x = 5/4
Example Question #2 : Algebra
If 3 – 3x < 20, which of the following could not be a value of x?
–2
–3
–6
–4
–5
–6
First we solve for x.
Subtracting 3 from both sides gives us –3x < 17.
Dividing by –3 gives us x > –17/3.
–6 is less than –17/3.
Example Question #1 : Algebra
Let x be a number. Increasing x by twenty percent yields that same result as decreasing the product of four and x by five. What is x?
The problem tells us that increasing x by twenty percent gives us the same thing that we would get if we decreased the product of four and x by five. We need to find expressions for these two situations, and then we can set them equal and solve for x.
Let's find an expression for increasing x by twenty percent. We could represent this as x + 20%x = x + 0.2x = 1.2x = 6x/5.
Let's find an expression for decreasing the product of four and x by five. First, we must find the product of four and x, which can be written as 4x. Then we must decrease this by five, so we must subtract five from 4x, which could be written as 4x - 5.
Now we must set the two expressions equal to one another.
6x/5 = 4x - 5
Subtract 6x/5 from both sides. We can rewrite 4x as 20x/5 so that it has a common denominator with 6x/5.
0 = 20x/5 - 6x/5 - 5 = 14x/5 - 5
0 = 14x/5 - 5
Now we can add five to both sides.
5 = 14x/5
Now we can multiply both sides by 5/14, which is the reciprocal of 14/5.
5(5/14) = (14x/5)(5/14) = x
25/14 = x
The answer is 25/14.