Linear Inequalities - College Algebra

To evaluate, we will need to multiply both sides by one-fifth.

The sign does not need to be switched unless we multiply or divide by a negative value.

The answer is:

Upon solving for x, we find that x is larger than -2. The left-hand term of the interval is -2 since it is the lower bound for our set, and it has a parenthesis around it because it is not included in our set (-2 is not greater than -2). The right-hand term of the interval is positive infinity because any number larger than -2 is in the set. There is always a parenthesis around infinity or negative infinity.

Upon solving for x, we find that x is less than or equal to 5. The left-hand term of the interval is negative infinity since any number less than 5 is in our set, and infinity always has a parenthesis around it. The right-hand term of the interval is 5 since it is the upper bound of our set. There is a bracket around it because 5 is included in our set (5 is less than or equal to 5).

Upon expanding and solving for x, we find that x is greater than -1. The left-hand term of the interval is -1 since it is the lower bound of our set, and has a parenthesis around it because it is not in our set (-1 is not greater than -1). The right-hand term of the interval is infinity because any number larger than -1 is in our set, and there is always a parenthesis around infinity.

Upon solving for x, we find that 8 is less than or equal to x which is less than or equal to 10, or x is greater than or equal to 8 AND x is less than or equal to 10. The left-hand term of our interval is 8 since it is the lower bound for our set, and there is a bracket around it since it is in our set (8 is greater than or equal to 8). The right-hand term of our interval is 10 since it is the upper bound for our set, and there is also a bracket around it since it is in our set (10 is less than or equal to 10).

Upon solving for x, we find that 1 is less than x which is less than or equal to 5, or x is greater than 1 AND x is less than or equal to 5. The left-hand term of our interval is 1 since it is the lower bound for our set, and there is a parenthesis around it since it is not in our set (1 is not greater than 1). The right-hand term of our interval is 5 since it is the upper bound for our set, and there is a bracket around it since it is in our set (5 is less than or equal to 5). Remember when dividing or multiplying an inequality by a negative number to switch the direction of the inequalities.

Upon solving for x, we find that -1 is less than or equal to x which is less than 2, or x is greater than or equal to -1 AND x is less than 2. The left-hand term of our interval is -1 since it is the lower bound for our set, and there is a bracket around it since it is in our set (-1 is greater than or equal to -1). The right-hand term of our interval is 2 since it is the upper bound for our set, and there is a parenthesis around it since it is not in our set (2 is not less than 2). Remember when dividing or multiplying an inequality by a negative number to switch the direction of the inequalities.

Upon solving for x, we find that x is less than or equal to -2 OR x is greater than 1. Since x may be in either of those sets, we must take the union of the sets. The first interval begins with negative infinity, since x can be anything less than -2. The first interval ends at -2 and includes it in the set, so there is a bracket around it. We join this interval with the second interval beginning with 1, and there is a parenthesis around 1 since it is not in the set. The second interval ends with infinity since any number larger than 1 is in our set. There are always parenthesis around infinities.

Upon solving for x, we find that x is less than or equal to -7/2 OR x is greater than or equal to 1/2. Since x may be in either of those sets, we must take the union of the sets. The first interval begins with negative infinity, since x can be anything less than -7/2. The first interval ends at -7/2 and includes it in the set, so there is a bracket around it. We join this interval with the second interval beginning with 1/2, and there is a bracket around 1/2 since it is in the set. The second interval ends with infinity since any number larger than 1/2 is in our set. There are always parenthesis around infinities.

Upon solving for x, we find that 6 is less than x which is less than 7, or that x is greater than 6 AND x is less than 7. The left-hand term of our interval is 6 since it is the lower bound for our set, and there is a parenthesis around it since it is not in our set (6 is not greater than 6). The right-hand term of our interval is 7 since it is the upper bound for our set, and there is a parenthesis around it since it is not in our set (7 is not less than 7). Remember to flip the direction of the inequalities when dividing or multiplying by a negative number.

Upon solving for x, we find that x is greater than or equal to 10/3. The left-hand term of the interval is 10/3 since it is the lower bound of our set, and there is a bracket around it because it is in our set (10/3 is greater than or equal to 10/3). The right-hand term of the interval is infinity since any number larger than 10/3 is in our set, and there is always parenthesis around infinity.

Our variable x represents the number of hours to move. Marvin's rate can be expressed as 30x+100. Elliot's rate can be expressed as 55x. We must set the expression for Marvin's rate less than Elliot's rate: . Upon solving for x, we find that x is greater than 4, so Marvin's is cheaper for jobs greater than 4 hours.

To solve this inequality, we treat is a a regular equation. First, we add 8 to both sides.

.

We now divide both sides by 2 to get

.

To evaluate, we will need to multiply both sides by one-fifth.

The sign does not need to be switched unless we multiply or divide by a negative value.

The answer is:

Upon solving for x, we find that x is larger than -2. The left-hand term of the interval is -2 since it is the lower bound for our set, and it has a parenthesis around it because it is not included in our set (-2 is not greater than -2). The right-hand term of the interval is positive infinity because any number larger than -2 is in the set. There is always a parenthesis around infinity or negative infinity.

Upon solving for x, we find that x is less than or equal to 5. The left-hand term of the interval is negative infinity since any number less than 5 is in our set, and infinity always has a parenthesis around it. The right-hand term of the interval is 5 since it is the upper bound of our set. There is a bracket around it because 5 is included in our set (5 is less than or equal to 5).

Upon expanding and solving for x, we find that x is greater than -1. The left-hand term of the interval is -1 since it is the lower bound of our set, and has a parenthesis around it because it is not in our set (-1 is not greater than -1). The right-hand term of the interval is infinity because any number larger than -1 is in our set, and there is always a parenthesis around infinity.

Upon solving for x, we find that 8 is less than or equal to x which is less than or equal to 10, or x is greater than or equal to 8 AND x is less than or equal to 10. The left-hand term of our interval is 8 since it is the lower bound for our set, and there is a bracket around it since it is in our set (8 is greater than or equal to 8). The right-hand term of our interval is 10 since it is the upper bound for our set, and there is also a bracket around it since it is in our set (10 is less than or equal to 10).

Upon solving for x, we find that 1 is less than x which is less than or equal to 5, or x is greater than 1 AND x is less than or equal to 5. The left-hand term of our interval is 1 since it is the lower bound for our set, and there is a parenthesis around it since it is not in our set (1 is not greater than 1). The right-hand term of our interval is 5 since it is the upper bound for our set, and there is a bracket around it since it is in our set (5 is less than or equal to 5). Remember when dividing or multiplying an inequality by a negative number to switch the direction of the inequalities.