COMPOUND INEQUALITIES

Examples

(i) $x > 2$ and $x \le 6$

This means that $x$ must be greater than 2 at the same time that it is less than or equal to 6.

(ii) $-3 \le x < 5$

This is a short way of writing an "and" inequality. It means $x$ sits completely between $-3$ and $5$.

(iii) $x < -1$ or $x \ge 4$

This means that $x$ can be any number less than $-1$, or any number greater than or equal to $4$.

Examples

(i) For the compound inequality $-1 < x \le 4$:

The connector meaning is "And" because $x$ is trapped in the middle of both constraints.

The interval notation is written as (-1, 4].

The number line graph will have a hollow circle at $-1$ and a solid circle at $4$.

(ii) For the compound inequality $x \le 2$ or $x > 5$:

The connector meaning is "Or" because the ranges point in opposite directions independently.

The interval notation is written as $(-\infty, 2] \cup (5, \infty)$.

The number line graph will have a solid circle at $2$ pointing left, and a hollow circle at $5$ pointing right.

Example

Solve the compound inequality: $-5 \le 2x + 1 < 7$

Step 1: Identify the starting compound inequality statement.

$-5 \le 2x + 1 < 7$

This is our three-part expression with the variable term in the middle.

Step 2: Subtract 1 from all three parts of the inequality expression.

$-5 - 1 \le 2x + 1 - 1 < 7 - 1$

This eliminates the positive constant 1 from the middle section while keeping the entire chain balanced.

Step 3: Simplify the values across all parts.

$-6 \le 2x < 6$

Subtracting 1 from -5 gives -6, the middle constants cancel out, and 7 minus 1 leaves 6.

Step 4: Divide all three parts by the positive center coefficient 2.

$$\frac{-6}{2} \le \frac{2x}{2} < \frac{6}{2}$$

This removes the multiplier from the variable without changing the direction of any inequality signs.

Step 5: Simplify the fraction operations to find the final range.

$-3 \le x < 3$

Dividing -6 by 2 yields -3, and dividing 6 by 2 yields 3.

Final Answer :

$-3 \le x < 3$

Example

Solve the compound inequality: $3x - 2 < -11$ or $-2x + 5 \le -1$

Step 1: Isolate the first inequality expression on the left side.

$3x - 2 < -11$

We focus purely on solving this first independent segment.

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

$3x - 2 + 2 < -11 + 2$

This eliminates the negative constant away from our first variable term.

Step 3: Simplify both sides of the first inequality.

$3x < -9$

The left constants cancel out, and adding 2 to -11 leaves -9.

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

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

This isolates our variable x for the first solution set portion.

Step 5: Simplify the calculation to find the first solution range.

$x < -3$

Dividing -9 by 3 gives -3.

Step 6: Isolate the second inequality expression on the right side.

$-2x + 5 \le -1$

Now we switch focus entirely to solving this second independent segment.

Step 7: Subtract 5 from both sides of the second inequality.

$-2x + 5 - 5 \le -1 - 5$

This moves our constant terms over to the right side of the sign.

Step 8: Simplify both sides of the second inequality.

$-2x \le -6$

Subtracting 5 from 5 leaves 0, and combining -1 and -5 results in -6.

Step 9: Divide both sides of the second inequality by the negative coefficient -2 and reverse the sign.

$$\frac{-2x}{-2} \ge \frac{-6}{-2}$$

Because we are dividing by a negative number, the inequality sign must completely flip from less than or equal to (≤) to greater than or equal to (≥).

Step 10: Simplify the expressions to find the second solution range.

$x \ge 3$

Dividing -2x by -2 leaves x, and dividing -6 by -2 leaves a positive 3.

Step 11: Combine both independent ranges using the word connector "or".

$x < -3$ or $x \ge 3$

This structural set forms our total combined range profile.

Final Answer :

$x < -3$ or $x \ge 3$

Example 1: Overlapping "AND" Range Layout

Represent the compound solution $-2 \le x < 3$ on a number line and list all valid integral values.

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

Since the left operator is less than or equal to (≤), we place a solid circle (shaded circle) at position -2.

Step 2: Identify the circle style for the upper boundary value.

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

Step 3: Determine the indicator path tracking method.

This is an "AND" condition, so we draw a single solid line bar connecting our two boundary values together.

Step 4: Identify and list all integers bounded within the drawn track range.

The solid circle includes -2, while the hollow circle excludes 3. The integers inside are -2, -1, 0, 1, and 2.

Final Answer :

The valid integral values are -2, -1, 0, 1, 2.

Example 2: Diverging "OR" Range Layout

Represent the compound solution $x \le -1$ or $x > 2$ on a number line and list the 4 integral values closest to zero.

Step 1: Identify the circle style and direction for the left statement.

Since the operation is less than or equal to (≤), we place a solid circle at -1 and draw our track arrow pointing left.

Step 2: Identify the circle style and direction for the right statement.

Since the operation is strictly greater than (>), we place a hollow circle at 2 and draw our track arrow pointing right.

Step 3: Identify whole numbers falling into either branch closest to zero.

On the left branch, our valid numbers are -1, -2, -3, etc. On the right branch, our valid numbers are 3, 4, 5, etc. Note that 0, 1, and 2 are completely excluded empty spaces.

Step 4: Select the 4 integer values nearest to the zero point layout.

Comparing distances to zero from both active paths, our closest valid integers are -1, -2, 3, and 4.

Final Answer :

The 4 integral values closest to zero are -1, -2, 3, 4.

Example 1: An "AND" Word Problem (Range Limits)

A manufacturing plant produces steel rods. To pass quality control inspection, a rod's length must be at least 15 centimetres and at most 18 centimetres. Find the range of acceptable lengths.

Step 1: Define the unknown variable item.

Let $l$ represent the length of a steel rod in centimetres.

This gives a clear variable letter designation to the target item we are tracking.

Step 2: Translate the lower boundary condition phrasing.

$l \ge 15$

The phrase "at least 15" means the length must be greater than or exactly equal to 15.

Step 3: Translate the upper boundary condition phrasing.

$l \le 18$

The phrase "at most 18" means the length must be less than or exactly equal to 18.

Step 4: Join both operational constraints into a single combined "AND" statement chain.

$15 \le l \le 18$

Since both conditions must be true at the same time, the variable sits trapped between the values.

Final Answer :

The range of acceptable steel rod lengths is $15 \le l \le 18$.

Example 2: An "OR" Word Problem (Safety Trigger Limits)

An industrial holding tank heating system contains an automated safety valve alert mechanism. The system triggers an emergency warning alarm if the core temperature drops below 10°C or exceeds above 90°C. Find the range of temperature values that trigger the warning alarm.

Step 1: Define the unknown variable item.

Let $t$ represent the core tank holding temperature in degrees Celsius.

This links a tracking letter container directly to our physical measurement variable.

Step 2: Translate the lower threshold warning condition.

$t < 10$

The phrase "below 10" describes values strictly less than 10, requiring an open exclusion border.

Step 3: Translate the upper threshold warning condition.

$t > 90$

The phrase "exceeds above 90" describes values strictly greater than 90, requiring an open exclusion border.

Step 4: Combine both distinct condition tracks using the mandatory connector logic word.

$t < 10$ or $t > 90$

Because the temperature cannot be under 10 and over 90 simultaneously, we use the "or" logical framework.

Final Answer :

The range of alarm triggering temperatures is $t < 10$ or $t > 90$.

Example 1: Solving a Combined "AND" Chain with Fractional Multiplication

Solve the compound inequality $-3 \le \frac{2x - 1}{3} < 3$, write the solution in interval notation, and represent it on a number line.

Step 1: Identify the starting three-part compound inequality statement.

$-3 \le \frac{2x - 1}{3} < 3$

This is a fractional "AND" chain inequality with the variable trapped in the center fraction numerator.

Step 2: Multiply all three parts of the inequality expression by the positive denominator 3.

$-3 \times 3 \le \left(\frac{2x - 1}{3}\right) \times 3 < 3 \times 3$

This eliminates the fractional divider across all sections without reversing any inequality signs.

Step 3: Simplify the results of the multiplication steps.

$-9 \le 2x - 1 < 9$

Multiplying $-3$ by 3 gives $-9$, the center denominator cancels out, and 3 times 3 leaves 9.

Step 4: Add 1 to all three sections of the chain expression.

$-9 + 1 \le 2x - 1 + 1 < 9 + 1$

This removes the negative constant from the center section to isolate the variable term.

Step 5: Simplify the terms across all sides.

$-8 \le 2x < 10$

Adding 1 to $-9$ leaves $-8$, the center constants cancel out, and 9 plus 1 results in 10.

Step 6: Divide all three parts by the positive center variable coefficient 2.

$$\frac{-8}{2} \le \frac{2x}{2} < \frac{10}{2}$$

This completely isolates the single variable x in the center region.

Step 7: Simplify the fraction operations to find the final algebraic range.

$-4 \le x < 5$

Dividing $-8$ by 2 gives $-4$, and dividing 10 by 2 gives 5.

Step 8: Translate the solved algebraic range into interval notation format.

[-4, 5)

We use a solid square bracket at $-4$ because it is included (≤) and a hollow parenthesis at 5 because it is excluded (<).

Step 9: Plot the solution range onto a number line.

Final Answer :

Algebraic range: $-4 \le x < 5$; Interval: [-4, 5).

Example 2: Independent "OR" Solution with Sign Reversal

Solve the compound inequality $2x - 3 \le -9$ or $-3x + 1 < -11$, list the 3 integral values closest to zero, and display the solution on a number line.

Step 1: Set up the first independent inequality statement from the left branch.

$2x - 3 \le -9$

We solve this left section completely on its own first.

Step 2: Add 3 to both sides of the first inequality expression.

$2x - 3 + 3 \le -9 + 3$

$2x \le -6$

The constants cancel on the left side, and adding 3 to $-9$ yields $-6$.

Step 3: Divide both sides by the positive number 2 to isolate the variable.

$$\frac{2x}{2} \le \frac{-6}{2}$$

$x \le -3$

Dividing $-6$ by 2 gives our first isolated solution section boundary range.

Step 4: Set up the second independent inequality statement from the right branch.

$-3x + 1 < -11$

Now we switch to solving this independent right section entirely.

Step 5: Subtract 1 from both sides of the second inequality expression.

$-3x + 1 - 1 < -11 - 1$

$-3x < -12$

The constants cancel on the left side, and subtracting 1 from $-11$ results in $-12$.

Step 6: Divide both sides by the negative number -3 and reverse the inequality sign.

$$\frac{-3x}{-3} > \frac{-12}{-3}$$

$x > 4$

Because we divided by a negative number, the sign flips direction from less than (<) to greater than (>).

Step 7: Combine both branch solution blocks using the connector word "or".

$x \le -3$ or $x > 4$

This forms our complete algebraic compound solution layout profile.

Step 8: List all valid whole numbers falling inside either active branch closest to zero.

From the left branch ($x \le -3$), valid integers include -3, -4, -5. From the right branch ($x > 4$), valid integers include $5, 6, 7$. The three numbers closest to zero are $-3, -4$, and $5$.

Step 9: Plot the solution range onto a number line.

Final Answer :

Solution range: $x \le -3$ or $x > 4$ ; 3 closest integral values to zero: -3, -4, 5.

Example 3: Solving Multi-Step "AND" Systems Separately

Solve the compound inequality system given as two separate conditions: $3x + 1 \ge -5$ and $x - 4 < 0$. Write the answer in interval notation.

Step 1: Set up the first linear inequality expression from the left side.

$3x + 1 \ge -5$

We isolate the variable for the first linear expression condition branch.

Step 2: Subtract 1 from both sides of the first inequality expression.

$3x + 1 - 1 \ge -5 - 1$

This eliminates the positive constant 1 from the first inequality.

Step 3: Simplify the constants on the right side of the expression.

$3x \ge -6$

Combining $-5$ and $-1$ leaves $-6$ on the right side boundary.

Step 4: Divide both sides of the first inequality by the positive number 3.

$$\frac{3x}{3} \ge \frac{-6}{3}$$

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

Step 5: Simplify the division calculation to find the first solution set.

$x \ge -2$

Dividing $-6$ by 3 gives $-2$, completing our first bound constraint segment.

Step 6: Set up the second linear inequality expression from the right side.

$x - 4 < 0$

Now we calculate the individual boundary rule for the second branch.

Step 7: Add 4 to both sides of the second inequality expression.

$x - 4 + 4 < 0 + 4$

This balances the expression and eliminates the negative 4 from the variable side.

Step 8: Simplify both sides to find the second solution set.

$x < 4$

Adding 0 and 4 finishes isolating the variable x for the second constraint segment.

Step 9: Merge both solved inequality statements back together using the logical "and" layout.

$x \ge -2$ and $x < 4$

This describes a variable that is simultaneously greater than or equal to $-2$ and strictly under 4.

Step 10: Condense the statement into a unified three-part chain inequality form.

$-2 \le x < 4$

Putting the lower bound on the left and the upper bound on the right wraps x securely in the center overlap zone.

Step 11: Express the overlapping solution chain using interval notation bracket containers.

[-2, 4)

A closed bracket is attached to $-2$ because it is included (≤), and an open parenthesis is placed at 4 because it is excluded (<).

Final Answer :

$-2 \le x < 4$

The compound solution in interval notation format is [-2, 4).

Example 4: Word Problem on "AND" Range Boundaries

A chemist is monitoring a sensitive chemical compound solution. For the solution to remain completely stable, its temperature must be kept at least 40°F. Safety protocols also dictate that the fluid temperature must remain under 70°F. Formulate the valid stability range context, write it in interval notation, and find the total number of integer temperature degrees available for storage.

Step 1: Define the unknown tracking value variable letter.

Let $T$ represent the stable compound fluid temperature in degrees Fahrenheit.

This sets an algebraic character representation to map out our problem requirements.

Step 2: Translate the lower thermal baseline storage parameter condition phrasing.

$T \ge 40$

The condition "at least 40" signifies that values must be greater than or exactly equal to 40.

Step 3: Translate the upper thermal cutoff protocol boundary condition phrasing.

$T < 70$

The condition "under 70" describes a strict cutoff constraint that is strictly less than 70.

Step 4: Construct the complete mutual system compound inequality with the logical "AND" connector chain.

$40 \le T < 70$

Because the fluid must meet both specific restrictions simultaneously, it is locked into a single overlapping chain block format.

Step 5: Convert the target inequality expression into an explicit interval notation string layout.

$[40, 70)$

We use a solid bracket at 40 for inclusion and an open parenthesis symbol at 70 for strict exclusion.

Step 6: Identify the sequence of valid integers inside the stability boundaries.

The valid integers starting from 40 up to the excluded boundary 70 are 40, 41, 42, all the way to 69.

Step 7: Calculate the total count of numbers inside this sequence.

Total integers = $70 - 40$

Since the lower bound 40 is included but the top value 70 is completely excluded, subtracting the lower bound from the upper bound gives the exact count.

Step 8: Simplify the subtraction calculation to get the total integer count.

Total integers = 30

Subtracting 40 from 70 yields exactly 30 total integral storage choices.

Final Answer :

The stability parameter range is $40 \le T < 70$, interval format is $[40, 70)$, and there are exactly 30 integer temperature degree options available.

Example 5: Word Problem on "OR" Disjoint Limits

An electronic component manufacturing sensor triggers a flashing dashboard status warning if the structural voltage in a power circuit drops at most 12 volts or matches at least 18 volts. Formulate the warning trigger solution expression range and list the 3 lowest positive integer voltage measurements that keep the machine completely safe (no warning triggered).

Step 1: Define the unknown variable letter container asset.

Let $v$ represent the structural power circuit line voltage measured in volts.

This assigns our mathematical character focus to the physical voltage measurement.

Step 2: Map out the lower warning alarm threshold criteria rule context.

$v \le 12$

The parameter phrase "at most 12 volts" establishes an alarm trigger constraint that is less than or exactly equal to 12.

Step 3: Map out the upper warning alarm threshold criteria rule context.

$v \ge 18$

The parameter phrase "at least 18 volts" establishes an alarm trigger constraint that is greater than or exactly equal to 18.

Step 4: Link both independent alarm trigger profiles using the disjoint connector concept word.

$v \le 12$ or $v \ge 18$

This combined expression outlines the exact zones where a status warning alarm will flash.

Step 5: Determine the safe operating range inverse condition profile where no warning is triggered.

$12 < v < 18$

To avoid triggering any alarms, the circuit voltage must stay safely inside the un-alerted empty gap space completely between 12 and 18.

Step 6: List the positive integer voltage sequence values that sit entirely inside that safe operating zone.

The valid integers strictly between 12 and 18 are 13, 14, 15, 16, and 17.

Step 7: Select the first 3 lowest values from the safe operational integer list.

Looking at the safe sequence, the first three smallest integer values are 13, 14, and 15.

Final Answer :

The warning trigger parameters are $v \le 12$ or $v \ge 18$, and the 3 lowest safe positive integer operations are 13, 14, 15.