Algebra for College || Graph the set on the number line. Then, write the set using interval notation
Algebra for College || Graph the set on the number line. Then, write the set using interval notation

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Showing Inequalities on Number Line: In mathematics, an inequality occurs when two mathematical expressions or two numbers are compared in a way that is not equal. Inequalities can be either numerical or algebraic. Two numbers are compared according to their values on the number line, resulting in a numerical inequality. This occurs when one expression is greater or smaller than another. Unevenness can be represented in a variety of ways.

When two values are compared using the inequality symbols, a linear inequity is created.
Between any two quantities (say \(x\) and \(y\)) in the universe, we can compare them using the symbols \( < ,\, > ,\, = ,\, \ge ,\, \le ,\, \ne \), etc., i.e. \(x < y,\,x > y\) or \(x \ne y\). The symbols \( > ,\, < ,\, \le ,\, \ge \ne \) are called the signs of inequality or inequation.

An inequation is linear if each variable’s exponent is only of the first degree and there is no term involving the product of the variables. If two real numbers or algebraic expressions are related by symbols such as “less than,” “greater than”, and “equal to,” then we have inequality.

Inequalities with one variable are plotted on a number line, with the output representing the solution to the inequality. A number line is used to graph a linear inequality with only one variable. Two-dimensional graphs plot linear inequalities with two variables because the output gives the solution to both variables.

Each of the following is called a linear inequation in one variable if \(a,\,b\) and \(c\) are real numbers.

In an inequation, the signs \( > ,\, < ,\, \ge ,\, \le \) and \( \ne \) are known as the signs of inequality.
\(ax + b > c,\,ax + b < c\) are considered as strict iequalities. \(ax + b \ge c,\,ax + b \le c\) are not strict inequalities. Let us consider some examples for inequalities for integers.

A linear inequation with one variable can be represented as \(ax + b < 0\) or \(ax + b > 0\) or \(ax + b \ne 0\) or \(ax + b \ge 0\) or \(ax + b \le 0\), where \(a,\,b\) are real numbers and \(a \ne 0\).

\(ax \le 0,\,ax + by + c > 0,\,ax \ge 5,\,by \le 10\) etc are few examples of linear inequations.
When we represent linear inequation on a number line, we use two types of dots: a solid dot and a hollow dot.

Let us see some numericals and understand the representation of linear inequalities on the number line.

Example 1: Represent \(4x > – 3x + 21\) on the number line.

The solution, in this case, is simple.
Transposing \(-3x\) from \({\rm{RHS}}\) to \({\rm{LHS}}\) we have,
\(4x + 3x > 21\)
\( \Rightarrow 7x > 21\)
Now, dividing both sides by \(7\) we get,
\( \Rightarrow x > 3\)

This inequality holds for any point on the green part of the number line. We have drawn a hollow red dot at point \(3\) in this case.
Note: \(3\) is not included in the solution set (because the given inequality has a strict inequality). This inequity is satisfied by the green part of the number line.

Example 2: Represent \(2 – 3x > 7 – 5x\) on the number line.
Transposing \(2\) in \({\rm{LHS}}\) to \({\rm{RHS}}\) we have,
\( \Rightarrow – 3x > – 2 + 7 – 5x\)

Transposing \(-x\,{\rm{RHS}}\) to \({\rm{LHS}}\) we have,
\( \Rightarrow – 3x + 5x > – 2 + 7\)
\( \Rightarrow 2x > 5\)

Now, dividing both sides by \(5\) we get,
\( \Rightarrow x > \frac{5}{2}\)

\( \Rightarrow x > 2.5\)

This inequality holds for any point on the red part of the number line. We have drawn a hollow dot at point \(2.5\) in this case.
Note: \(2.5\) is not included in the solution set (because the given inequality has a strict inequality). This inequity is satisfied by the red part of the number line.

Example 3: Represent \(x + 3 \le 2x + 9\) on the number line.

Transposing \(2x\,{\rm{RHS}}\) to \({\rm{LHS}}\) we have,
\( \Rightarrow x – 2x + 3 \le 9\)
Transposing \({\rm{3}}\,{\rm{LHS}}\) to \({\rm{RHS}}\) we have,
\( \Rightarrow – x \le 6\)

Now, multiplying both sides by \(-1\) we get,
\( \Rightarrow x \ge – 6\)

This inequality holds for any point on the bold black part of the number line. We have drawn a solid dot at point \(-6\) here.
Note: \(-6\) is included in the solution set (this is because the given inequality has no strict inequality). This inequity is satisfied by the bolded black part of the number line.

Example 4: Represent \( – 1 < 3 – 2x \le 7\) on the number line.

We can split the given inequality into two parts.

We have \( – 1 < 3 – 2x\) and \(3 – 2x \le 7\)
Let us first solve \( – 1 < 3 – 2x\).
Transposing \(3\) from \({\rm{RHS}}\) to \({\rm{LHS}}\) we have,
\( \Rightarrow – 1 – 3 < – 2x\)
\( \Rightarrow – 4 < – 2x\)
Now, dividing both sides by \(-2\) we have,
\( \Rightarrow 2 > x\)

(The inequality sign has been changed as we divided both sides by a negative number)
Let us solve the next part.
\(3 – 2x \le 7\)
Transposing \(3\) from \({\rm{LHS}}\) to \({\rm{RHS}}\) we have,

\( \Rightarrow – 2x \le 7 – 3\)
\( \Rightarrow – 2x \le 4\)
Now, dividing both sides by \(-2\) we have,
\( \Rightarrow x \ge – 2\)

Hence, the solution set lies between \({ – 2 \le x < 2,\,x \in R} \)
Therefore, the values for \(x\) are \( – 2,\, – 1,\,0,\,1\).

This inequality holds for any point on the bold black part of the number line. We have drawn a solid dot at point \(-2\) and a hollow dot for \(2\).
Note: \(-2\) is included in the solution set (this is because \(3 – 2x \le 7\) has not a strict inequality), and \(2\) is not included in the solution set (this is because \( – 1 < 3 – 2x\) has a strict inequality).

Q.1. Find the solution of the linear equation for \( – 2x – 39 \ge – 15\)
Ans: Given,
\( – 2x – 39 \ge – 15\)
Transposing \(-39\) from \({\rm{LHS}}\) to \({\rm{RHS}}\) we have,
\( \Rightarrow – 2x \ge 24\)
Now, dividing both sides by \(-2\) we have,
\( \Rightarrow x \le – 12\)

Q.2. Determine the solution set of \(50 – 3(2x – 5) < 25\), given that \(x \in W\). Also, represent the solution set on a number line.
Ans: Given: \(50 – 3(2x – 5) < 25\)
\( \Rightarrow 50 – 6x + 15 < 25\)
\( \Rightarrow 65 – 6x < 25\)
\( \Rightarrow – 6x < 25 – 65\)
\( \Rightarrow – 6x < – 40\)
\( \Rightarrow \frac{{ – 6x}}{{ – 6}} > \frac{{ – 40}}{{ – 6}}\) (Division by a negative number reverse the sign of inequality).
\( \Rightarrow x > 6\frac{2}{3}\)
Required solution set \( = { 7,\,8,\,9,\, \ldots \ldots \ldots } \)
And, the required number line is

Q.3. Find the solution of the linear equation for \( – 4 \le 3x – 1 < 8\) and plot it on a number line.
Ans: Given,
\( – 4 \le 3x – 1 < 8\)
That is, \( – 4 < 3x – 1\) and \(3x – 1 < 8\)
\( – 4 \le 3x – 1\)
Transposing \(-1\) from \({\rm{RHS}}\) to \({\rm{LHS}}\) we have,
\( \Rightarrow – 3 \le 3x\)
Now, dividing both sides by \(1\) we have,
\( \Rightarrow – 1 \le x\)
Let us solve the next part.
\(3x – 1 < 8\)
Transposing \(-1\) from \({\rm{LHS}}\) to \({\rm{RHS}}\) we have,
\( \Rightarrow 3x < 8 + 1\)
\( \Rightarrow 3x < 9\)
Now, dividing both sides by \(3\) we have,
\( \Rightarrow x < 3\)
Hence, the solution set lies between \({ – 1 \le x < 3,\,x \in R} ]\)
Therefore, the values for \(x\) are \( – 1,\,0,\,1,\,2\).

Q.4. Find the solution of the linear equation for \( – 5 \le 2x – 3 < x + 2\) and plot it on a number line.
Ans: Given,
\( – 5 \le 2x – 3 < x + 2\)
\( – 5 < 2x – 3\) and \(2x – 3 \le x + 2\)
\( – 5 < 2x – 3\)
Transposing \(-3\) from \({\rm{RHS}}\) to \({\rm{LHS}}\) we have,
\( \Rightarrow – 5 + 3 < 2x \Rightarrow – 2 < 2x\)
Now, dividing both sides by \(2\) we have,
\( \Rightarrow – 1 < x\)
Let us solve the next part.
\(2x – 3 \le x + 2\)
Transposing \(-3\) from \({\rm{LHS}}\) to \({\rm{RHS}}\) we have,
\( \Rightarrow 2x \le x + 2 + 3 \Rightarrow 2x \le x + 5\)
Now, transposing \(x\) from \({\rm{RHS}}\) to \({\rm{LHS}}\)
\( \Rightarrow x \le 5\)
Hence, the solution set lies between \({ – 1 < x \le 5,\,x \in R} \)
Therefore, the values for \(x\) are \( – 1,\,0,\,1,\,2,\,3,\,4,\,5\).

Q.5. Represent \( – 4 < x < 4\) on a number line.
Ans: This inequality holds for any point on the bold black part of the number line. We have drawn hollow dots at points \(-4\) and \(4\).
Note: \(-4\) is not included in the solution set (this is because it has a strict inequality), and \(4\) is not included in the solution set (this is because it has a strict inequality).

Linear inequalities are defined here, along with an example and rules for solving them. We also learnt how to solve a single-variable linear inequalities problem and represent the solutions to a single-variable linear inequalities problem on the number line, among other topics covered. We have solved linear inequalities examples and frequently asked questions about linear inequalities.

Learn Inequalities in a Triangle

Q.1. Define linear inequalities with an example.
Ans: An inequation is linear if each variable’s exponent is only of the first degree and there is no term involving the product of the variables.
If \(a,\,b\) and \(c\) are real numbers, then each of the following is called a linear inequation in one variable.
Example: \(x + 1 \le 2,\,2x + 5 > 7x + 2,\,6x \ge 5,\,y \le 1\).
Q.2. What are the symbols to represent the linear inequalities?
Ans: In inequality, the signs \( > ,\, < ,\, \ge \) and \( \le \) are known as the signs of inequality.

Q.3. How could we easily solve an inequality problem?
Ans: Transpose the variables to one side and the constants to the other side of the inequation and deduce the variable’s value.

Q.4. How to represent the solution set of linear inequation on the number line?
Ans: When we represent linear inequation on a number line, we use two types of dots: a solid dot and a hollow dot. To express the inequalities that have no strict inequalities, we use “solid dot” on the number line.
To express the strict inequalities, we use “hollow dot” on the number line.

Q.5. What are the real-life uses of linear inequalities?
Ans: Inequalities are more commonly used than equalities in many real-life problems to determine the best solution. It is used to determine the maximum or minimum values of a situation with multiple constraints. This solution can be as simple as determining how many units of a product should be produced to maximise profit or minimise loss.

We hope this detailed article on showing inequalities on number line helped you in your studies. If you have any doubts, queries or suggestions regarding this article, feel free to ask us in the comment section and we will be more than happy to assist you. Happy learning!

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