LINEAR INEQUALITIES

Examples

(i) $$x < 7$$ → This means that $$x$$ can be any number strictly less than $$7$$.

(ii) $$x > 3$$ → This means that $$x$$ can be any number strictly greater than $$3$$.

(iii) $$x \le 5$$ → This means that $$x$$ can be any number less than or exactly equal to $$5$$.

(iv) $$x \ge -2$$ → This means that $$x$$ can be any number greater than or exactly equal to $$-2$$.

Examples

(i) $$2x + 5 < 15$$ — Linear (The variable $$x$$ has a power of 1)

(ii) $$x^2 > 9$$ — Not linear (The variable has a power of 2)

(iii) $$x^3 \le 8$$ — Not linear (The variable has a power of 3)

Example

Solve the inequality: $$x - 5 \le 9$$

Step 1: Identify the inequality statement.

$$x - 5 \le 9$$

This is our starting problem.

Step 2: Add 5 to both sides of the inequality.

$$x - 5 + 5 \le 9 + 5$$

This balances the expression and eliminates the negative 5 from the variable side.

Step 3: Simplify the left side of the inequality.

$$x \le 9 + 5$$

The numbers -5 and +5 cancel each other out to zero.

Step 4: Simplify the right side to find the final value range.

$$x \le 14$$

Adding 9 and 5 gives 14.

Final Answer :

$$x \le 14$$

Example

Solve the inequality: $$2x + 5 < 13$$

Step 1: Identify the inequality statement.

$$2x + 5 < 13$$

This is our starting problem.

Step 2: Subtract 5 from both sides.

$$2x + 5 - 5 < 13 - 5$$

This isolates the term with the variable on the left side.

Step 3: Simplify both sides of the inequality.

$$2x < 8$$

Subtracting 5 from 5 leaves 0, and subtracting 5 from 13 leaves 8.

Step 4: Divide both sides by the positive number 2.

$$\frac{2x}{2} < \frac{8}{2}$$

This removes the multiplier from the variable.

Step 5: Simplify the fractions to find the final answer.

$$x < 4$$

Dividing 2x by 2 gives x, and dividing 8 by 2 gives 4.

Final Answer :

$$x < 4$$

Example

Solve the inequality: $$-3x + 6 \le 15$$

Step 1: Identify the inequality statement.

$$-3x + 6 \le 15$$

This is our starting problem.

Step 2: Subtract 6 from both sides.

$$-3x + 6 - 6 \le 15 - 6$$

This moves the constant values away from our variable term.

Step 3: Simplify both sides of the inequality.

$$-3x \le 9$$

Subtracting 6 from 6 leaves 0, and subtracting 6 from 15 leaves 9.

Step 4: Divide both sides by the negative number -3 and flip the sign.

$$\frac{-3x}{-3} \ge \frac{9}{-3}$$

Because we divide by a negative number, the sign flips from less than or equal to ($$\le$$) to greater than or equal to ($$\ge$$).

Step 5: Simplify the expressions to find the final answer.

$$x \ge -3$$

Dividing -3x by -3 leaves x, and dividing 9 by -3 gives -3.

Final Answer :

$$x \ge -3$$

Example

Verify if the solution range for $$x + 2 > 6$$ is correct, which is solved as $$x > 4$$.

Step 1: Identify the solved solution range.

$$x > 4$$

This tells us that any number strictly greater than 4 should make the original statement true.

Step 2: Choose a sample test number within that range.

Let $$x = 5$$

We choose 5 because it is a simple whole number that is strictly greater than 4.

Step 3: Write down the original inequality statement.

$$x + 2 > 6$$

This is the initial expression we need to test.

Step 4: Substitute the chosen test number into the expression.

$$5 + 2 > 6$$

We replace the variable x with our chosen value of 5.

Step 5: Simplify the left side of the statement.

$$7 > 6$$

Adding 5 and 2 gives a total of 7.

Step 6: Evaluate the final statement to check if it is true.

$$7 > 6$$ is True.

Since 7 is greater than 6, our chosen test number satisfies the condition.

Final Answer :

The solution set $$x > 4$$ is verified and correct.

Example 1: Strictly Less Than (<)

Represent $x < 3$ on a number line and list the 3 largest positive integral values.

Step 1: Identify the circle style for the boundary value.

Since the operator is strictly less than (<), we place a hollow circle at position 3.

Step 2: Determine the arrow path direction.

Numbers less than 3 are located to the left, so the indicator line draws to the left.

Step 3: Identify the largest valid integers below the boundary number 3.

Integers strictly less than 3 are 2, 1, 0, -1, and so on.

Final Answer :

The 3 largest positive integral values are 2, 1, 0.

Example 2: Greater Than or Equal To (≥)

Represent $x \ge -1$ on a number line and list the 3 smallest integral values.

Step 1: Identify the circle style for the boundary value.

Since the operator is greater than or equal to (≥), we place a filled, solid circle (shaded circle) at position -1.

Step 2: Determine the arrow path direction.

Numbers greater than -1 are located to the right, so the indicator line draws to the right.

Step 3: Identify the smallest valid integers within the range, including the boundary.

Because the circle is solid, -1 is included as the absolute smallest value. The numbers increase progressively up from there.

Final Answer :

The 3 smallest integral values are -1, 0, 1.

Example

James has a budget of $50 to spend at a bookstore. He buys a notebook for $8 and wants to spend the remaining amount on pens that cost $6 each. Find the maximum number of pens he can buy.

Step 1: Define the unknown variable.

Let $$x$$ be the number of pens James can buy.

This gives a letter designation to the value we are trying to find.

Step 2: Express the cost of the pens mathematically.

$$6x$$

Since each pen costs $6, multiplying 6 by the number of pens gives the total price for the pens.

Step 3: Set up the total spending expression.

$$6x + 8$$

We add the $8 notebook cost to the total cost of the pens to find his total expenditure.

Step 4: Form the linear inequality based on the budget limit.

$$6x + 8 \le 50$$

Because he has a budget of $50 , his total spending must be less than or exactly equal to $50.

Step 5: Subtract 8 from both sides of the inequality.

$$6x + 8 - 8 \le 50 - 8$$

This removes the fixed notebook cost from his budget total.

Step 6: Simplify the expression.

$$6x \le 42$$

Subtracting 8 from 50 leaves him with $42 to spend purely on pens.

Step 7: Divide both sides by the positive number 6.

$$\frac{6x}{6} \le \frac{42}{6}$$

This helps find the upper limit for the single variable $$x$$.

Step 8: Simplify the final division calculation.

$$x \le 7$$

Dividing 42 by 6 gives 7.

Final Answer :

The maximum number of pens James can buy is 7.

Example 1

Solve the linear inequality: $$x + 3 > 7$$

Step 1: Identify the inequality statement.

$$x + 3 > 7$$

This is our starting problem.

Step 2: Subtract 3 from both sides of the sign.

$$x + 3 - 3 > 7 - 3$$

This helps move the constant number away from our variable.

Step 3: Simplify the left side of the expression.

$$x > 7 - 3$$

The positive 3 and negative 3 cancel out to 0.

Step 4: Simplify the right side to get the final answer.

$$x > 4$$

Subtracting 3 from 7 gives 4.

Final Answer :

$$x > 4$$

Example 2

Solve the linear inequality $$3x - 4 \ge 11$$ and list the 3 smallest integral values that satisfy the solution range.

Step 1: Identify the starting inequality statement.

$$3x - 4 \ge 11$$

This is our starting problem.

Step 2: Add 4 to both sides of the expression.

$$3x - 4 + 4 \ge 11 + 4$$

This eliminates the negative constant from the variable side.

Step 3: Simplify the terms on both sides of the sign.

$$3x \ge 15$$

The left side constants cancel out to 0, and adding 11 and 4 on the right side gives 15.

Step 4: Divide both sides by the positive coefficient 3.

$$\frac{3x}{3} \ge \frac{15}{3}$$

This isolates our single variable x without changing the direction of the inequality sign.

Step 5: Calculate the final simplified values to find the range.

$$x \ge 5$$

Dividing 3x by 3 gives x, and dividing 15 by 3 gives us 5.

Step 6: List the sequence of whole numbers within this range.

Since the inequality symbol is greater than or equal to ($$\ge$$), our boundary integer 5 is included. The valid whole numbers are 5, 6, 7, 8, 9, 10, and so on.

Step 7: Select the first 3 smallest values from your list.

Looking at our sequence, the three lowest integer values that fit are 5, 6, and 7.

Final Answer :

The solution range is $$x \ge 5$$ and the 3 smallest integral values are 5, 6, 7.

Example 3

Solve the linear inequality $$-2x > 10$$ and illustrate the solution range clearly on a number line.

Step 1: Identify the starting inequality statement.

$$-2x > 10$$

This is our starting problem.

Step 2: Divide both sides by the negative number -2 and reverse the sign direction.

$$\frac{-2x}{-2} < \frac{10}{-2}$$

Because we are dividing by a negative number, the sign must completely flip from greater than ($$>$$) to less than ($$<$$).

Step 3: Simplify the expression lines to find the solution set.

$$x < -5$$

Dividing -2x by -2 leaves x, and dividing 10 by -2 gives us -5.

Step 4: Identify the type of boundary circle needed for the graph.

Since the solved solution uses a strictly less than symbol ($$<$$), we will use an open, hollow circle at position -5 to show that -5 itself is excluded.

Step 5: Determine the path direction of the indicator line.

Values less than -5 go towards smaller, more negative values, so the indicator line must point to the left side of the number line.

Step 6: Plot the solution values visually onto the responsive axis layout.

Final Answer :

The solution range is $$x < -5$$ as shown on the chart track above.

Example 4

Solve the linear inequality: $$\frac{x}{4} \le 6$$

Step 1: Identify the inequality statement.

$$\frac{x}{4} \le 6$$

This is our starting problem.

Step 2: Multiply both sides by the positive denominator 4.

$$\left(\frac{x}{4}\right) \times 4 \le 6 \times 4$$

This removes the division fraction format from our variable side.

Step 3: Simplify the left side computation.

$$x \le 6 \times 4$$

The multiplying 4 cancels out the dividing 4.

Step 4: Multiply the constants on the right side.

$$x \le 24$$

Multiplying 6 by 4 gives us 24.

Final Answer :

$$x \le 24$$

Example 5 (Word Problem)

A farmer wants to fence a rectangular garden plot. The length of the plot must be 12 metres. If the total perimeter of the fence can be at most 50 metres, find the maximum possible width of the garden plot.

Step 1: Define the unknown variable for the width.

Let $$w$$ be the width of the rectangular plot.

This sets a letter container for our missing value.

Step 2: Identify the formula for the perimeter of a rectangle.

Perimeter = $$2(\text{Length} + \text{Width})$$

This is the standard geometric rule to find the distance around a rectangle.

Step 3: Substitute the known length value into the perimeter expression.

$$2(12 + w)$$

We replace the length item with our known value of 12 metres.

Step 4: Set up the inequality using the boundary limit word.

$$2(12 + w) \le 50$$

The term "at most" means the perimeter must be less than or exactly equal to 50.

Step 5: Expand the bracket grouping by multiplying through by 2.

$$24 + 2w \le 50$$

Multiplying 2 by 12 gives 24, and multiplying 2 by w gives 2w.

Step 6: Subtract 24 from both sides of the expression.

$$24 - 24 + 2w \le 50 - 24$$

This isolates our variable term on the left side.

Step 7: Simplify both sides after subtraction.

$$2w \le 26$$

Subtracting 24 from 50 leaves a value of 26.

Step 8: Divide both sides by the positive value 2.

$$\frac{2w}{2} \le \frac{26}{2}$$

This isolates our single width variable.

Step 9: Simplify the division terms to find the final value.

$$w \le 13$$

Dividing 26 by 2 yields 13.

Final Answer :

The maximum possible width of the garden plot is 13 metres.