Equations / Inequalities - SAT Math
Card 0 of 2800
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In the equation below,
,
, and
are non-zero numbers. What is the value of
in terms of
and
?

In the equation below, ,
, and
are non-zero numbers. What is the value of
in terms of
and
?
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Solve for x:

Solve for x:
The first step is to cancel out the denominator by multiplying both sides by 7:


Subtract 3 from both sides to get
by itself:


The first step is to cancel out the denominator by multiplying both sides by 7:
Subtract 3 from both sides to get by itself:
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Solve for
and
using elimination:


Solve for and
using elimination:
When using elimination, you need two factors to cancel out when the two equations are added together. We can get the
in the first equation to cancel out with the
in the second equation by multiplying everything in the second equation by
:


Now our two equations look like this:


The
can cancel with the
, giving us:


These equations, when summed, give us:


Once we know the value for
, we can just plug it into one of our original equations to solve for the value of
:





When using elimination, you need two factors to cancel out when the two equations are added together. We can get the in the first equation to cancel out with the
in the second equation by multiplying everything in the second equation by
:
Now our two equations look like this:
The can cancel with the
, giving us:
These equations, when summed, give us:
Once we know the value for , we can just plug it into one of our original equations to solve for the value of
:
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Give the solution set of the rational equation 
Give the solution set of the rational equation
Multiply both sides of the equation by the denominator
:


Rewrite both expression using the binomial square pattern:


This can be rewritten as a linear equation by subtracting
from both sides:


Solve as a linear equation:




Multiply both sides of the equation by the denominator :
Rewrite both expression using the binomial square pattern:
This can be rewritten as a linear equation by subtracting from both sides:
Solve as a linear equation:
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Solve:

Solve:

Multiply by
on each side

Subtract
on each side

Multiply by
on each side

Multiply by on each side
Subtract on each side
Multiply by on each side
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Let
be an integer that can be represented by
. If
,
, and
are integers such that
, and
, what is a possible value for
?
Let be an integer that can be represented by
. If
,
, and
are integers such that
, and
, what is a possible value for
?
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 that satisfy the following equation:
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 , then either
or
.
These two equations yield and
as answers.
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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, or
Add 8 to both sides of the equation; therefore, or
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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 and
will be equal to
.
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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 and that happens when
.
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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 or the set of real numbers greater 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 which will give you
and then solve the right side which is
and solution is
which is really equal to
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