How to find a solution set - SAT Math
Card 0 of 232
Let 
be an integer that can be represented by 
. If 
, 
, and 
are integers such that 
, and 
, what is a possible value for 
?
Let
In order to find n, we need to figure out possible values for a, b, and c.
We are told that a, b, and c are integers. We are also told that 0 < a < b < c, which means that a, b, and c are all positive. It also means that a, b, and c are different numbers (they cannot be equal). It also means that a is the smallest, and c is the largest.
We are now told that a + b + c = 7.
Let's start with the smallest possible value that a could be. The smallest integer greater than 0 is 1.
Let us assume that a = 1.
If a = 1, then b > 1. The smallest integer that b can be is 2. So let's assume b = 2.
If a =1 and b = 2, then 1 + 2 + c = 7. This means that c = 4.
If a =1, b = 2, and c = 4, then all of the conditions for a, b, and c are met. Thus, we can use these values to find n.
What if we had assumed that a was 2? If a is 2, then the smallest value that b could be is 3. If a + b + c = 7, then c would have to be 2. But we are told that c is the largest number, so it can't be equal to 2.
Thus, the only value of a that works is 1.
Let's assume that if a = 1, then b = 3. If a + b + c = 7, then c = 3. However, we are told that c > b. Thus, b cannot be 3.
Thus, b must equal 2, and c must equal 4, and a must be 1. This is the only possibility that satisfies all of the criteria given in the question.
Now we can use the values of a, b, and c to find n.
n =2a3b5c
n = 213254 = 2(9)(625) = 11250.
The answer is 11250.
In order to find n, we need to figure out possible values for a, b, and c.
We are told that a, b, and c are integers. We are also told that 0 < a < b < c, which means that a, b, and c are all positive. It also means that a, b, and c are different numbers (they cannot be equal). It also means that a is the smallest, and c is the largest.
We are now told that a + b + c = 7.
Let's start with the smallest possible value that a could be. The smallest integer greater than 0 is 1.
Let us assume that a = 1.
If a = 1, then b > 1. The smallest integer that b can be is 2. So let's assume b = 2.
If a =1 and b = 2, then 1 + 2 + c = 7. This means that c = 4.
If a =1, b = 2, and c = 4, then all of the conditions for a, b, and c are met. Thus, we can use these values to find n.
What if we had assumed that a was 2? If a is 2, then the smallest value that b could be is 3. If a + b + c = 7, then c would have to be 2. But we are told that c is the largest number, so it can't be equal to 2.
Thus, the only value of a that works is 1.
Let's assume that if a = 1, then b = 3. If a + b + c = 7, then c = 3. However, we are told that c > b. Thus, b cannot be 3.
Thus, b must equal 2, and c must equal 4, and a must be 1. This is the only possibility that satisfies all of the criteria given in the question.
Now we can use the values of a, b, and c to find n.
n =2a3b5c
n = 213254 = 2(9)(625) = 11250.
The answer is 11250.
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Find the sum of all of the integer values of x that satisfy the following inequality:
|4 – 2x| < 5
Find the sum of all of the integer values of x that satisfy the following inequality:
|4 – 2x| < 5
In general, when we take the absolute value of a quantity, we can represent it as either itself, or as its additive inverse.
In other words, |x| = x (if x > 0) or |x| = –x (if x < 0).
Therefore, we can represent |4 – 2x| as either 4 – 2x or as –(4 – 2x). We must consider both of these cases and solve for x. Let’s us first consider the case that |4 – 2x| = 4 – 2x.
4 – 2x < 5
We can add 2x to both sides
4 < 2x + 5
Then subtract 5 from both sides.
–1 < 2x
Divide by 2.
–1/2 < x
x > –0.5
Now, we consider the case that |4 – 2x | = –(4 – 2x).
–(4 – 2x) < 5
Multiply both sides by negative one. Remember, whenever we multiply or divide an inequality by a negative number, we must flip the inequality sign.
4 – 2x > –5
Add 2x to both sides.
4 > 2x – 5
Add 5 to both sides.
9 > 2x
Divide by 2,
9/2 > x
4.5 > x
This means that the values of x that satisfy the original quality must be greater than -0.5 AND less than 4.5.
The question asks us to find the sum of the integer values of x that satisfy the inequality. The only integers between -0.5 and 4.5 are the following: 0, 1, 2, 3, and 4.
The sum of 0, 1, 2, 3, and 4 is 10.
The answer is 10.
In general, when we take the absolute value of a quantity, we can represent it as either itself, or as its additive inverse.
In other words, |x| = x (if x > 0) or |x| = –x (if x < 0).
Therefore, we can represent |4 – 2x| as either 4 – 2x or as –(4 – 2x). We must consider both of these cases and solve for x. Let’s us first consider the case that |4 – 2x| = 4 – 2x.
4 – 2x < 5
We can add 2x to both sides
4 < 2x + 5
Then subtract 5 from both sides.
–1 < 2x
Divide by 2.
–1/2 < x
x > –0.5
Now, we consider the case that |4 – 2x | = –(4 – 2x).
–(4 – 2x) < 5
Multiply both sides by negative one. Remember, whenever we multiply or divide an inequality by a negative number, we must flip the inequality sign.
4 – 2x > –5
Add 2x to both sides.
4 > 2x – 5
Add 5 to both sides.
9 > 2x
Divide by 2,
9/2 > x
4.5 > x
This means that the values of x that satisfy the original quality must be greater than -0.5 AND less than 4.5.
The question asks us to find the sum of the integer values of x that satisfy the inequality. The only integers between -0.5 and 4.5 are the following: 0, 1, 2, 3, and 4.
The sum of 0, 1, 2, 3, and 4 is 10.
The answer is 10.
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How many distinct solutions does the following polynomial have?
x(x_2 – 14_x + 49) = 0
How many distinct solutions does the following polynomial have?
x(x_2 – 14_x + 49) = 0
There are 3 solutions: 0, 7, 7.
The correct answer is 2 distinct solutions, however, because 7 occurs twice. So the two distinct solutions are 0 and 7.
There are 3 solutions: 0, 7, 7.
The correct answer is 2 distinct solutions, however, because 7 occurs twice. So the two distinct solutions are 0 and 7.
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Find the product of the values of 
that satisfy the following equation:
Find the product of the values of
Because we are dealing with an absolute value on the left side, we are going to have to consider two possible cases. Remember that, in general, |a| = a if a > 0, and |a| = –a if a < 0. In other words, we need to consider two cases for the left side: x_2 – 2_x – 8 and –(x_2 – 2_x – 8).
Case 1: |x_2 – 2_x – 8| = x_2 – 2_x – 8 = 2_x_ + 4
x_2 – 2_x – 8 = 2_x_ + 4
Subtract 2_x_ from both sides.
x_2 – 4_x – 8 = 4
Subtract 4 from both sides.
x_2 – 4_x – 12 = 0
We must think of two numbers that multiply to give us –12 and add to give –4. These two numbers are –6 and 2. This means we can factor the left side as follows:
x_2 – 4_x – 12 = (x – 6)(x + 2) = 0
Set each of the factors equal to 0.
x – 6 = 0
Add 6 to both sides.
x = 6
Next, x + 2 = 0
Subtract 2 from both sides.
x = –2
The values of x that solve the equation are –2 and 6.
However, before moving on to the next case, it's always important to check our work. Let's put x = –2 and x = 6 into our original equation and make sure that both sides are the same.
|x_2 – 2_x – 8| = 2_x_ + 4
Let x = –2:
|x_2 – 2_x – 8| = |(–2)2 – 2(–2) – 8| = |4 + 4 – 8| = 0
2_x_ + 4 = 2(–2) + 4 = 0
x = –2 is a solution.
Let x = 6:
|62 – 2(6) – 8| = |36 – 12 – 8| = 16
2_x_ + 4 = 2(6) + 4 = 16
x = 6 is also a solution.
Case 2: |x_2 – 2_x – 8| = –(x_2 – 2_x – 8) = 2_x_ + 4.
–(x_2 – 2_x – 8) = 2_x_ + 4
Distribute the –1.
–x_2 + 2_x + 8 = 2_x_ + 4
Subtract 2_x_ from both sides.
_–x_2 + 8 = 4
Subtract 8 from both sides.
_–x_2 = –4
Divide both sides by negative 1.
_x_2 = 4
Take the square root.
x = 2 or –2
We have already established that x = –2 is a solution. Let's check to see if 2 is also a solution by going back to the original equation |x_2 – 2_x – 8| = 2_x_ + 4.
|x_2 – 2_x – 8| = |22 – 2(2) – 8| = |–8| = 8
2_x_ + 4 = 2(2) + 4 = 8
This means x = 2 is also a solution.
To summarize, the solutions to x are –2, 2, and 6.
We are asked to find their product, which is –2(2)(6) = –24.
The answer is therefore –24.
Because we are dealing with an absolute value on the left side, we are going to have to consider two possible cases. Remember that, in general, |a| = a if a > 0, and |a| = –a if a < 0. In other words, we need to consider two cases for the left side: x_2 – 2_x – 8 and –(x_2 – 2_x – 8).
Case 1: |x_2 – 2_x – 8| = x_2 – 2_x – 8 = 2_x_ + 4
x_2 – 2_x – 8 = 2_x_ + 4
Subtract 2_x_ from both sides.
x_2 – 4_x – 8 = 4
Subtract 4 from both sides.
x_2 – 4_x – 12 = 0
We must think of two numbers that multiply to give us –12 and add to give –4. These two numbers are –6 and 2. This means we can factor the left side as follows:
x_2 – 4_x – 12 = (x – 6)(x + 2) = 0
Set each of the factors equal to 0.
x – 6 = 0
Add 6 to both sides.
x = 6
Next, x + 2 = 0
Subtract 2 from both sides.
x = –2
The values of x that solve the equation are –2 and 6.
However, before moving on to the next case, it's always important to check our work. Let's put x = –2 and x = 6 into our original equation and make sure that both sides are the same.
|x_2 – 2_x – 8| = 2_x_ + 4
Let x = –2:
|x_2 – 2_x – 8| = |(–2)2 – 2(–2) – 8| = |4 + 4 – 8| = 0
2_x_ + 4 = 2(–2) + 4 = 0
x = –2 is a solution.
Let x = 6:
|62 – 2(6) – 8| = |36 – 12 – 8| = 16
2_x_ + 4 = 2(6) + 4 = 16
x = 6 is also a solution.
Case 2: |x_2 – 2_x – 8| = –(x_2 – 2_x – 8) = 2_x_ + 4.
–(x_2 – 2_x – 8) = 2_x_ + 4
Distribute the –1.
–x_2 + 2_x + 8 = 2_x_ + 4
Subtract 2_x_ from both sides.
_–x_2 + 8 = 4
Subtract 8 from both sides.
_–x_2 = –4
Divide both sides by negative 1.
_x_2 = 4
Take the square root.
x = 2 or –2
We have already established that x = –2 is a solution. Let's check to see if 2 is also a solution by going back to the original equation |x_2 – 2_x – 8| = 2_x_ + 4.
|x_2 – 2_x – 8| = |22 – 2(2) – 8| = |–8| = 8
2_x_ + 4 = 2(2) + 4 = 8
This means x = 2 is also a solution.
To summarize, the solutions to x are –2, 2, and 6.
We are asked to find their product, which is –2(2)(6) = –24.
The answer is therefore –24.
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Solve _x_2 – 48 = 0.
Solve _x_2 – 48 = 0.
No common terms cancel out, and this isn't a difference of squares.
Let's move the 48 to the other side: _x_2 = 48
Now take the square root of both sides: x = √48 or x = –√48. Don't forget the second (negative) solution!
Now √48 = √(3*16) = √(3*42) = 4√3, so the answer is x = 4√3 or x = –4√3.
No common terms cancel out, and this isn't a difference of squares.
Let's move the 48 to the other side: _x_2 = 48
Now take the square root of both sides: x = √48 or x = –√48. Don't forget the second (negative) solution!
Now √48 = √(3*16) = √(3*42) = 4√3, so the answer is x = 4√3 or x = –4√3.
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What is the sum of all solutions to the equation
?
What is the sum of all solutions to the equation
If 
, then either
or 
.
These two equations yield 
and 
as answers.
If
These two equations yield
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Using the ordered pairs listed below, which of the following equations is true?
(0, –4)
(2, 0)
(4, 12)
(8, 60)
Using the ordered pairs listed below, which of the following equations is true?
(0, –4)
(2, 0)
(4, 12)
(8, 60)
You can solve this problem using the guess and check method by substituting the first number in the ordered pair for "x" and the second number for "y". Therfore the correct answer is 
–4 = 0 – 4
0 = 4 – 4
12 = 16 – 4
60 = 64 – 4
You can solve this problem using the guess and check method by substituting the first number in the ordered pair for "x" and the second number for "y". Therfore the correct answer is
–4 = 0 – 4
0 = 4 – 4
12 = 16 – 4
60 = 64 – 4
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Solve for x: 
Solve for x:
First, take the square root of both sides:
Therefore, 
or 
Add 8 to both sides of the equation; therefore, 
or 
First, take the square root of both sides:
Therefore,
Add 8 to both sides of the equation; therefore,
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What is a possible solution to this equation: 
?
What is a possible solution to this equation:
This equation can be solved using the guess and check method.
Therefore, the ordered pair (3, 16) is the correct answer.
This equation can be solved using the guess and check method.
Therefore, the ordered pair (3, 16) is the correct answer.
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First multiply each decimal number in each term by 100 to remove the decimals (to get a whole number you have to multiply 0.03 by 100 to get 3). You need to do this for terms on both sides of the equal sign.
The second method would be to look for the number of digits to the right of the decimal point (e.g., 0.03 has two digits). So in this method shift the decimal point to the right two places.
Now the equation looks as follows:
Now solve for 
and 
will be equal to 
.
First multiply each decimal number in each term by 100 to remove the decimals (to get a whole number you have to multiply 0.03 by 100 to get 3). You need to do this for terms on both sides of the equal sign.
The second method would be to look for the number of digits to the right of the decimal point (e.g., 0.03 has two digits). So in this method shift the decimal point to the right two places.
Now the equation looks as follows:
Now solve for
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Solve 
Solve
Absolute value is the distance from the point to the origin, so the value itself is always poisitive. The only solution that makes sense for this problem is when the absolute value is equal to zero, or 
and that happens when 
.
Absolute value is the distance from the point to the origin, so the value itself is always poisitive. The only solution that makes sense for this problem is when the absolute value is equal to zero, or
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In this question we describe the solution set in the form of a line diagram. Remember a solution to an inequality can be described in the form of (1) Inequality notation, (2) A line diagram, (3) and or an interval notation.
Given a solution set for a linear inequality as shown below:
what would be the correct representation of the above set in the form of a line diagram
In this question we describe the solution set in the form of a line diagram. Remember a solution to an inequality can be described in the form of (1) Inequality notation, (2) A line diagram, (3) and or an interval notation.
Given a solution set for a linear inequality as shown below:
what would be the correct representation of the above set in the form of a line diagram
The solution lies in the set of real numbers less than 
or the set of real numbers greater than 
.
The solution lies in the set of real numbers less than
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Solve and describe your answer in both inequality notation and interval notation:
Solve and describe your answer in both inequality notation and interval notation:
This is a question with double inequality.
First solve the left side which will be 
which will give you 
and then solve the right side which is 
and solution is 
which is really equal to 
This is a question with double inequality.
First solve the left side which will be
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This question deals with absolute value inequalities and as a result you get a set of solutions.
The first solution involves solving 
which gives you 
.
Next solve for 
which will give you 
.
Since 
and 
overlap, the correct solution is \[-1,4\]
This question deals with absolute value inequalities and as a result you get a set of solutions.
The first solution involves solving
Next solve for
Since
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: What is the 22nd digit after the decimal?
The repeating pattern after the decimal is 63475, so the 22nd number would be 3.
The repeating pattern after the decimal is 63475, so the 22nd number would be 3.
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This problem deals with absolute value inequalities.
An absolute value expression can never be less than 0. So the correct answer is "No Solution"
This problem deals with absolute value inequalities.
An absolute value expression can never be less than 0. So the correct answer is "No Solution"
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In the two equations below, 
and 
. What is the value of 
?
In the two equations below,
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The set 
contains all multiples of 
. Which of the following sets are contained within 
?
I. The set of all multiples of 
.
II. The set of all multiples of 
.
III. The set of all multiples of 
.
The set
I. The set of all multiples of
II. The set of all multiples of
III. The set of all multiples of
Think of the multiples of 10: 10, 20, 30, 40, 50, 60, 70, . . .
I. Multiples of 2: 2, 4, 6, 8, 10, 12, 14, . . .
Some of these already are not contained in S.
II. Multiples of 5: 5, 10, 15, 20, 25, . . .
Some of these already are not contained in S.
III. Multiples of 20: 20, 40, 60, 80, 100, . . .
All of these are also multiples of 10. Thus, our answer must be III only.
Think of the multiples of 10: 10, 20, 30, 40, 50, 60, 70, . . .
I. Multiples of 2: 2, 4, 6, 8, 10, 12, 14, . . .
Some of these already are not contained in S.
II. Multiples of 5: 5, 10, 15, 20, 25, . . .
Some of these already are not contained in S.
III. Multiples of 20: 20, 40, 60, 80, 100, . . .
All of these are also multiples of 10. Thus, our answer must be III only.
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Different colored marbles are placed in a bag. There are 
red marbles, 
black marbles, and 
green marbles in the bag. What is the probability that a green marble will be chosen?
Different colored marbles are placed in a bag. There are
When doing probability problems, we are looking for the number of successes over number of possible outcomes. There are 4 chances to successfully choose a green marble. The number of possible outcomes are 11, one for each of the 11 marbles in the bag. When we write the fraction, we get our answer.
In mathematical words we get the following:
When doing probability problems, we are looking for the number of successes over number of possible outcomes. There are 4 chances to successfully choose a green marble. The number of possible outcomes are 11, one for each of the 11 marbles in the bag. When we write the fraction, we get our answer.
In mathematical words we get the following:
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Which of the following is a true statement?
Which of the following is a true statement?
Looking at the second statement, isolate x on one side with all other constants and variables on the other side.
Looking at the third statement, isolate z on one side with all other constants and variables on the other side.
Looking at the first statement, isolate y on one side with all other constants and variables on the other side.
From here, use these equivalencies so solve for y.
Substituting twice:
Looking at the second statement, isolate x on one side with all other constants and variables on the other side.
Looking at the third statement, isolate z on one side with all other constants and variables on the other side.
Looking at the first statement, isolate y on one side with all other constants and variables on the other side.
From here, use these equivalencies so solve for y.
Substituting twice:
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