Combining Inequalities in One Variable: Definition, Properties, Examples

Combining Inequalities in One Variable: A combining inequality is a sentence with two inequality statements connected by the words “or” or “and.” The word “and” denotes that both compound sentences’ statements are true at the same time. It’s the intersection or overlap of the solution sets for each statement. The word “or” signifies that the entire compound sentence is true if either statement is true.

A conjunction is a compound inequality that contains the word “and.” Although the words “and” and “or” are both conjunctions in speech, the mathematical conjunction has a distinct meaning than the grammatical one. To demonstrate the idea, the conjunction (part of speech) “or” generates a disjunction when used in a compound inequality. Remember that “con” implies “with another “and” “dis” indicates “one OR the other.”

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Linear Inequalities

Linear inequalities are defined as expressions that compare two values using inequality symbols.

We know the larger (greater) or smaller (lesser) numbers for each other or any third amount. Only three conditions hold between any two quantities (say \(x\) and \(y\)) in the universe: \(‘ < ‘,’ > ‘\) and \(‘ = ‘\) i.e.,\(x < y,x > \) and \(x = y.\) The symbols \(‘ > ‘,\,’ < ‘,\,’ \le ‘,\,’ \ge ‘\) are called the sign of inequality or inequation.

Properties of Inequalities: Addition, Subtraction, Multiplication, Division

Addition Property of Inequality Definition

The addition property of linear inequality states that adding the same number to each side of the inequality produces an equivalent inequality, in which the inequality symbol remains the same.
Example: \(x < 5 \Rightarrow x + 2 < 5 + 2\) and \(x > y \Rightarrow x + a > y + a.\)

Subtraction Property of Inequality Definition

The linear inequality subtraction property states that removing the same number from both sides results in an identical inequality, i.e., the inequality symbol remains the same. Example: \(x < 3 \Rightarrow x – 2 < 3 – 2\) and \(x > y \Rightarrow x – b > y – b.\)

Multiplication Property of Inequality Definition

Multiplying both sides of an inequality by a positive number always results in an equivalent inequality, i.e., the inequality symbol remains the same, according to the linear inequality multiplication property.
Example: \(y < 4 \Rightarrow y \times 2 < 4 \times 2\)
However, the inequality symbol is reversed if multiplication is done with a negative value on both sides of the inequality.
Example: \(y < 3\)
Multiply the given inequation by \( – 2.\)
Thus, \( – 2y > – 6\)

Division Property of Inequality Definition

The division property of linear inequality states that dividing both sides of an inequality by a positive number produces an analogous inequality, i.e., the inequality symbol remains the same.
Example: \(3y < 6\)
Divide the given inequation by \(3.\)
Thus, \(\frac{{3y}}{3} < \frac{6}{3}.\)
However, the inequality sign is reversed if we divide both sides of an inequality with a negative number.
Example: \(- 2x > 6\)
Divide the given inequation by \( – 2.\)
Thus, \(\frac{{ – 2x}}{{ – 2}} < \frac{6}{{ – 2}}.\)

Learn Properties of Inequalities

Equivalent Inequality

If one side of the inequality is multiplied by a negative number, the inequality’s “direction” must be changed to get an equivalent inequality (for example,\( \le \) must be changed to \( \ge \)).
That is, if \(p\) is negative.
1. \(x < y \Rightarrow px > py\) and \(\left({\frac{x}{p}}\right) > \left({\frac{y}{p}} \right)\)
2. \(x \ge y \Rightarrow px \le py\) and \(\left( {\frac{x}{p}} \right) \le \left( {\frac{y}{p}} \right)\)

Compound Inequalities

When the word “and” is used to unite two inequalities, the compound inequality is solved when both inequalities are true simultaneously. It is the intersection, or overlap, of the solutions to each inequality. When the two inequalities are connected by the word or, the compound inequality is solved when one of the inequalities is true.
A compound inequality comprises at least two inequalities connected by “and” or “or.” The combination, or union, of the two distinct solutions, is the solution.
Example: Consider the inequality \(x > 2\) and the inequality \(x < 6.\) What would the interval look like if we were to interpret what numbers \(x\) could be?
In other words, \(x\) must be less than \(6\) while being greater than \(2.\) Let’s look at a graph to see what figures we can get with these restrictions.

The intersection of the two inequalities \(x\) refers to the numbers shared by both lines on the graph \(x < 6\) and \(x > 2.\) This is a bounded inequality, and it’s written like this: \(2 < x < 6.\) Take a moment to consider that \(x\) must be less than \(6\) and greater than \(2;\) the values for \(x\) will be in the middle of the two integers. This looks like this in interval notation \(\left({2,6} \right).\)

For example, the inequality \(x > 6\) or \(x <2\) is connected by the term “or”.
We may describe each of these inequalities separately using interval notation: \(x > 6\) is equivalent to \(\left({6,\infty } \right)\) while \(x <2\) is equivalent to \(\left({\infty ,2} \right).\) What effect do they have when describing solutions to inequalities? Solutions are either real numbers less than two or real numbers higher than \(6,\) we say.

Applications of Linear Inequalities

The applications of linear inequalities are:

1. Inequalities, rather than equalities, are probably utilized more frequently in “real life.” Businesses use inequalities to manage inventory, arrange production lines, develop pricing models, and ship and store goods and materials. Look up the terms “linear programming” and “Simplex technique.”
2. The maximum and minimum values of a scenario with many restrictions are frequently determined using a system of linear inequalities. For example, you might be trying to figure out how many units of a product should be produced to maximize profit.

Learn Concepts on Linear Inequalities

Solved Examples – Combining Inequalities in One Variable

Q.1. Solve for \(x:2x \ge 3 + 7\) OR \(2x + 9 > 11.\)
Ans: By solving the first inequality for \(x,\) we get
\(2x \ge 3 + 7\)
\( \Rightarrow 2x + 3 – 3 \ge 7 – 3\)
\( \Rightarrow 2x \ge 4\)
\( \Rightarrow x \ge 2\)
By solving the second inequality for \(x,\) we get
\(2x + 9 > 11\)
\( \Rightarrow 2x + 9 – 9 > 11 – 9\)
\( \Rightarrow 2x > 2\)
\( \Rightarrow x > 1\)
In terms of graphs, we have:

As a result, we may represent our compound inequality as a simple inequality: \(x > 1.\)

Q.2. Solve for \(x:4x – 39 > – 43\) AND \(8x + 31 < 23.\)
Ans: By solving the first inequality for \(x,\) we get
\(4x – 39 > – 43\)
\( \Rightarrow 4x – 39 + 39 > – 43 + 39\)
\( \Rightarrow 4x > – 4\)
\( \Rightarrow x > – 1\)
By solving the second inequality for \(x,\) we get
\(8x + 31 < 23\)
\( \Rightarrow 8x + 31 – 31 < 23 – 31\)
\( \Rightarrow 8x < – 8\)
\( \Rightarrow x < – 1\)
In terms of graphs, we have:

From the above graph, we interpret that there are no solutions to the compound inequality because there is no value of \(x\) that is both greater than the negative one and less than the negative one.

Q.3. Solve for \(x:2x + 7 < – 11\) OR \( – 3x – 2 < 13.\)
Ans: By solving the first inequality for \(x,\) we get
\(2x + 7 < – 11\)
\( \Rightarrow 2x + 7 – 7 < – 11 – 7\)
\( \Rightarrow 2x < – 18\)
\( \Rightarrow x < – 9\)
By solving the second inequality for \(x,\) we get
\( – 3x – 2 < 13\)
\( \Rightarrow – 3x – 2 + 2 < 13 + 2\)
\( \Rightarrow – 3x < 15\)
\( \Rightarrow x > – 5\)
All the numbers to the left of \( – 9\) are represented by \(x < – 9,\) while those to the right of \( – 5\) are represented by \(x > – 5.\) The solution set is written as follows:
{\(x|x < – 9\) or \(x > – 5\)}

Q.4. Solve for \(x:2 + x < 5\) AND \( – 1 < 2 + x.\)
Ans: By solving each inequality separately, we get
\(2 + x < 5\)
To isolate the variable from the first equation, divide both sides by two, obtaining;\(x < 3.\)
Now, by subtracting \(2\) on both sides of inequation, we get
\( – 1 – 2 < 2 + x – 2\)
\(\Rightarrow – 3 < x\)
Hence, the compound inequality is \( – 3 < x < 3.\)

Q.5. Solve: \(5 + x > 7\) OR \(x – 3 < 5.\)
Ans: By solving each inequality separately, we get
Consider the inequation \(5 + x > 7,\) and subtract both the sides of inequation by \(5.\)
So, \(5 + x – 5 > 7 – 5\)
\( \Rightarrow x > 2\)
Consider the inequation \(x – 3 < 5,\) and add both the sides of inequation by \(3.\)
So, \(x – 3 + 3 < 5 + 3\)
\( \Rightarrow x < 8\)
Hence, the solution to this combined inequation is \(x > 2\) or \(x < 8.\)

Summary

In this article, we learnt about linear inequalities, properties of inequalities (addition, subtraction, multiplication, and division property of inequalities), equivalent inequality, compound inequalities, applications of linear inequalities, solved examples on combining inequalities in one variable, FAQs on combining inequalities in one variable.

This article’s learning outcome is used to find the maximum, and minimum values of a situation with many restrictions are frequently determined using a system of linear inequalities.

Inequalities in One Variable – Frequently Asked Questions (FAQs)

Q.1. How do you combine two inequalities?
Ans: When the word “and” is used to unite two inequalities, the compound inequality is solved when both inequalities are true simultaneously. It is the intersection, or overlap, of the solutions to each inequality. When the two inequalities are connected by the word or, the compound inequality is solved when one of the inequalities is true.
Example: Solve: \(3\left( {2x + 5} \right) \ge 18\) and \(2\left({x – 7} \right) < – 6\)
Solution: By solving each inequality separately, we get
\(3\left( {2x + 5} \right) \ge 18 \Rightarrow 6x + 15 \ge 18\)
\( \Rightarrow 6x \ge 3 \Rightarrow x \ge \frac{1}{2}\) and
\(2\left({x – 7}\right) < – 6 \Rightarrow 2x – 14 < – 6\)
\( \Rightarrow 2x < 8 \Rightarrow x < 4\)
Therefore, the solution is \(x \ge \frac{1}{2}\) and \(x < 4.\)
Hence, the combined inequation is \(\frac{1}{2} \le x < 4.\)

Q.2. How do you solve an inequality with one variable?
Ans: There are the following steps used to solve the linear inequation in one variable.
1. Remove the fractions (or decimals) by multiplying both sides by an appropriate factor.
2. Put all variable terms on one side and all constants on the other side.
3. Make the coefficient of the variable \(1.\)
4. Choose the solution set from the replacement set.

Q.3. What is inequality with one variable?
Ans: If and only if the exponent of each variable in the inequation is one, it is said to be linear (or each variable appears in first degree only). No word involves the variables’ product.
Example: \(2y + 3 \le 2,\,x + 2y + 5 > 5x + 2,\,2x \ge 3,\,x \le 1.\)

Q.4. What are the two types of compound inequalities?    
Ans: In other circumstances, inequality can be used to express more than one restricting value. In these cases, a compound inequality is used.
As a result, a compound inequality can be defined as an expression having two inequality statements connected by the terms “AND” or “OR.”
The conjunction “and” denotes the truth of two statements at the same time.
On the other hand, the term “or” denotes that the entire compound statement is true if one of the statements is true.
A combination of the solution sets for the individual statements is denoted by the term “or.”

Q.5. What is the difference between properties of equality and properties of inequality?
Ans: The solution set will not change if you use the properties of equality on any equation. The inequality symbol must be reversed when multiplying or dividing an inequality by a negative number; otherwise, the solution set will be incorrect.

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