Sciencing Video Vault Switch the inequality if you multiply or divide by a negative number. Then solve the linear inequality that arises. So your solution ranges from negative infinity up to but not including 2 and would be written as -inf, 2.
The inequality symbol suggests that the solution are all values of x between -3 and 7, and also including the endpoints -3 and 7.
It makes sense that it must always be greater than any negative number. The answer in interval notation makes more sense if you see how it looks on the number line. In case 2, the arrows will always be in opposite directions.
You might also be interested in: We can also write the answer in interval notation using a parenthesis to denote that -8 and -4 are not part of the solutions.
This statement must be false, therefore, there is no solution. No other number will substitute for x and make that a true statement.
The solution set describing all numbers between -2 and 3 is expressed as: This is case 4. Do this by subtracting 6 from both sides. What do you get? Solve the absolute value inequality. The answer to this case is always no solution.
Because this is subtraction, the inequality sign does not change. Set your right side equal to zero, as you would when solving a polynomial equation. Now factor the left side: The solution to this would be written as: And finally, we will use closed or shaded circles to show that -3 and 7 are included.
The answer to this case is always all real numbers. In case 2, the arrows will always point to opposite directions. Use interval notation to express the range of numbers making your inequality a true statement.
Pick some test values to verify: The absolute value of any number is either zero 0 or positive. As you can see, we are solving two separate linear inequalities. Clear out the absolute value symbol using the rule and solve the linear inequality. Then divide both sides by -3, leaving x on the left side of the inequality, and 5 on the right.
Use brackets instead of parentheses to indicate that either or both of the numbers serving as boundaries for the range of your solution set are included in the solution set.
Multiply or divide both sides by the same positive number just as you would in an equation. If the solution set were all numbers between -2 and 3, including -2 and 3, the solution set would be written as: The shaded or closed circles signifies that -2 and 3 are part of the solution.
To describe this infinite set of solutions, you would use interval notation, and provide the boundaries of the range of numbers constituting a solution to this inequality. Therefore, the answer is all real numbers. Or, write the answer on a number line where we use open circles to exclude -8 and -4 from the solution.In this lesson, we are going to learn how to solve absolute value inequalities using the standard approach usually taught in an algebra class.
That is, learn the rules and apply it correctly. We can also write the answer in interval notation using a parenthesis to denote that -8 and.
Use interval notation to express the range of numbers making your inequality a true statement. The solution set describing all numbers between -2 and 3 is expressed as: (-2,3). For the inequality x + 2. Whereas the inequality $$\left | x \right |>2$$ Represents the distance between x and 0 that is greater than 2.
You can write an absolute value inequality as a compound inequality. $$\left | x \right |0$$ $$=-cc,\: where\: c>0$$ $$=ax+bc$$ You can replace >. Since 5 is at 5 units distance from the origin 0, the absolute value of 5 is 5, |5|=5 Since -5 is also at a distance of 5 units from the origin, the absolute value of -5 is.
This calculator will solve the linear, quadratic, polynomial, rational and absolute value inequalities. It can handle compound inequalities and system Inequality Calculator - eMathHelp. Absolute value inequalities will produce two solution sets due to the nature of absolute value.
We solve by writing two equations: one equal to a positive value and one equal to a negative value. Absolute value inequality solutions can be verified by graphing.Download