The Central Idea
Consider
\[ 3x+5=20. \]We want the input that makes the two sides equal.
Subtract \(5\):
\[ 3x=15. \]Divide by \(3\):
\[ \boxed{x=5}. \]Now compare the inequality
\[ 3x+5<20. \]Subtract \(5\):
\[ 3x<15. \]Divide by \(3\):
\[ \boxed{x<5}. \]The equation identifies one boundary value. The inequality identifies an entire region of values on one side of that boundary.
What Is a Linear Equation?
A linear equation in one variable can be written in a form such as
\[ \boxed{ax+b=c}, \]or more generally
\[ \boxed{ax+b=cx+d}, \]where the variable appears only to the first power.
Solving the equation means finding every input that makes the equality true.
What Is a Linear Inequality?
A linear inequality compares affine expressions using symbols such as
\[ <,\qquad \le,\qquad >,\qquad \ge. \]Examples include:
\[ 2x+1<9, \] \[ 5-3x\ge14, \]and
\[ 4x+7\le2x+15. \]The solution is usually a set of many real numbers rather than one value.
Equations as Balance Conditions
An equation states that two expressions represent the same quantity:
\[ L=R. \]Equivalent operations preserve that equality.
If we add the same quantity \(k\) to both sides,
\[ L+k=R+k. \]If we multiply both sides by the same nonzero quantity \(k\),
\[ kL=kR. \]This balance idea underlies equation solving.
Inverse Operations
To isolate a variable, we often undo operations in reverse order.
For
\[ 4x-7=21, \]first undo subtraction:
\[ 4x=28. \]Then undo multiplication:
\[ \boxed{x=7}. \]Check the Solution
Substitute
\[ x=7 \]into
\[ 4x-7=21. \]Then
\[ 4(7)-7=28-7=21. \]The equality is true, so the solution is verified.
Variables on Both Sides
Consider
\[ 5x+2=2x+17. \]Subtract \(2x\):
\[ 3x+2=17. \]Subtract \(2\):
\[ 3x=15. \]Divide by \(3\):
\[ \boxed{x=5}. \]Why Moving a Term Works
People often say “move \(2x\) to the other side and change the sign.”
The underlying operation is actually subtraction of \(2x\) from both sides:
\[ 5x+2=2x+17 \] \[ 5x-2x+2=2x-2x+17. \]Then
\[ 3x+2=17. \]The shortcut is useful, but the equality property explains why it works.
Distribution in Linear Equations
Consider
\[ 3(x+4)=24. \]One method is to divide by \(3\) first:
\[ x+4=8, \]so
\[ \boxed{x=4}. \]Another method is to distribute:
\[ 3x+12=24, \] \[ 3x=12, \] \[ x=4. \]Equivalent methods may differ in efficiency.
Collecting Like Terms
Consider
\[ 2x+5+3x-1=19. \]Combine like terms:
\[ 5x+4=19. \]Then
\[ 5x=15, \] \[ \boxed{x=3}. \]Fractions in Linear Equations
Consider
\[ \frac{x}{3}+2=7. \]Subtract \(2\):
\[ \frac{x}{3}=5. \]Multiply by \(3\):
\[ \boxed{x=15}. \]Clearing Fractions
Consider
\[ \frac{x}{2}+\frac{x}{3}=10. \]The least common denominator is \(6\).
Multiply every term by \(6\):
\[ 3x+2x=60. \]So
\[ 5x=60, \] \[ \boxed{x=12}. \]Decimals in Linear Equations
Consider
\[ 0.4x+1.2=5.6. \]Subtract \(1.2\):
\[ 0.4x=4.4. \]Divide by \(0.4\):
\[ \boxed{x=11}. \]We could also multiply the original equation by \(10\) first to remove decimals.
One Solution, No Solution, or Infinitely Many Solutions
A linear equation can have:
- one solution;
- no solution;
- infinitely many solutions.
The algebra reveals which case occurs.
Identity: Infinitely Many Solutions
Consider
\[ 2(x+3)=2x+6. \]Distribute:
\[ 2x+6=2x+6. \]Subtract \(2x\):
\[ 6=6. \]This is always true.
Therefore, every real number satisfies the equation:
\[ \boxed{x\in\mathbb{R}}. \]Contradiction: No Solution
Consider
\[ 2(x+3)=2x+9. \]Distribute:
\[ 2x+6=2x+9. \]Subtract \(2x\):
\[ 6=9. \]This is impossible.
Therefore, the solution set is
\[ \boxed{\varnothing}. \]Graph Interpretation of a Linear Equation
Consider
\[ 2x+1=5. \]This asks when the affine function
\[ y=2x+1 \]has output
\[ y=5. \]Graphically, we are finding the intersection of
\[ y=2x+1 \]and the horizontal line
\[ y=5. \]The intersection occurs at
\[ \boxed{(2,5)}. \]Equation as an Intersection of Two Lines
Consider
\[ 3x+2=x+10. \]This asks when
\[ y=3x+2 \]and
\[ y=x+10 \]have the same output.
Solve:
\[ 3x+2=x+10, \] \[ 2x=8, \] \[ x=4. \]The common output is
\[ 14. \]So the lines intersect at
\[ \boxed{(4,14)}. \]Parallel Lines and No Solution
Consider
\[ 2x+1=2x+5. \]Subtract \(2x\):
\[ 1=5. \]No solution exists.
Graphically,
\[ y=2x+1 \]and
\[ y=2x+5 \]are distinct parallel lines.
Identical Lines and Infinitely Many Solutions
Consider
\[ 2(x+1)=2x+2. \]Both sides describe the same affine expression.
Graphically, the corresponding lines coincide.
Every input produces the same output on both sides.
Linear Equations as Zero Problems
Any equation
\[ L(x)=R(x) \]can be rewritten as
\[ L(x)-R(x)=0. \]For example,
\[ 3x+2=x+10 \]becomes
\[ 2x-8=0. \]So solving the original equation is equivalent to finding the zero of the difference function.
Linear Inequalities as Comparison Problems
Consider
\[ 3x+2>x+10. \]This asks when the first affine expression lies above the second.
Solve:
\[ 3x+2>x+10, \] \[ 2x>8, \] \[ \boxed{x>4}. \]The value \(x=4\) is the boundary where the expressions are equal.
Inequality Symbols
| Symbol | Meaning |
|---|---|
| \(<\) | less than |
| \(\le\) | less than or equal to |
| \(>\) | greater than |
| \(\ge\) | greater than or equal to |
Strict Versus Inclusive Inequalities
The inequality
\[ x<5 \]excludes \(5\).
The inequality
\[ x\le5 \]includes \(5\).
This difference matters when graphing and writing interval notation.
Solving a Basic Linear Inequality
Consider
\[ 4x-3\le17. \]Add \(3\):
\[ 4x\le20. \]Divide by \(4\):
\[ \boxed{x\le5}. \]Adding or Subtracting Preserves Inequality Direction
If
\[ aThis is why ordinary addition and subtraction steps work much like they do in equations.
Multiplying by a Positive Number Preserves Direction
If
\[ aand \[ c>0, \]then
\[ acMultiplying by a Negative Number Reverses Direction
If
\[ aThe order reverses.
Therefore, multiplying or dividing an inequality by a negative number requires reversing the inequality symbol.
Why the Inequality Reverses
On the number line,
\[ 2<5. \]Multiply both numbers by \(-1\):
\[ -2>-5. \]Reflection across zero reverses left-right order.
This geometric fact explains the sign-reversal rule.
Example: Divide by a Negative Number
Solve
\[ -3x>12. \]Divide by \(-3\).
Because the divisor is negative, reverse the inequality:
\[ \boxed{x<-4}. \]Example with Several Steps
Solve
\[ 7-2x\ge15. \]Subtract \(7\):
\[ -2x\ge8. \]Divide by \(-2\) and reverse the inequality:
\[ \boxed{x\le-4}. \]Check an Inequality Solution
For
\[ 7-2x\ge15, \]we found
\[ x\le-4. \]Test a value inside the solution set, such as
\[ x=-5. \]Then
\[ 7-2(-5)=17\ge15. \]Test a value outside, such as
\[ x=0. \]Then
\[ 7\not\ge15. \]The solution direction is consistent.
Number-Line Graphs of Inequalities
For
\[ x<5, \]use an open endpoint at \(5\) and shade to the left.
For
\[ x\le5, \]use a closed endpoint at \(5\) and shade to the left.
The endpoint style records whether the boundary is included.
Interval Notation
The solution
\[ x<5 \]is written
\[ \boxed{(-\infty,5)}. \]The solution
\[ x\le5 \]is written
\[ \boxed{(-\infty,5]}. \]Infinity always uses a parenthesis because it is not a real endpoint.
Rightward Intervals
The inequality
\[ x>2 \]corresponds to
\[ \boxed{(2,\infty)}. \]The inequality
\[ x\ge2 \]corresponds to
\[ \boxed{[2,\infty)}. \]Compound Inequalities
A compound inequality combines conditions.
For example,
\[ 2and
\[ x<8. \]The solution is the interval
\[ \boxed{(2,8)}. \]Inclusive Compound Inequality
The inequality
\[ 2\le x\le8 \]includes both endpoints.
Its interval notation is
\[ \boxed{[2,8]}. \]Mixed Endpoint Conditions
The inequality
\[ 2The endpoint \(2\) is excluded, while \(8\) is included.
Solving a Three-Part Inequality
Solve
\[ 3<2x+1\le9. \]Subtract \(1\) from all three parts:
\[ 2<2x\le8. \]Divide all three parts by \(2\):
\[ \boxed{1Reversing a Three-Part Inequality
Solve
\[ -6<-2x\le4. \]Divide all parts by \(-2\).
Because the divisor is negative, both inequality directions reverse:
\[ 3>x\ge-2. \]Rewrite in increasing order:
\[ \boxed{-2\le x<3}. \]Interval notation:
\[ \boxed{[-2,3)}. \]“And” Inequalities
An and condition requires both inequalities to be true.
For example:
\[ x>1 \]and
\[ x\le6. \]The overlap is
\[ \boxed{1“Or” Inequalities
An or condition accepts values satisfying either inequality.
For example:
\[ x<-2 \]or
\[ x\ge5. \]Interval notation is
\[ \boxed{ (-\infty,-2)\cup[5,\infty) }. \]Union and Intersection Language
Compound inequalities connect naturally with set language.
- And corresponds to intersection.
- Or corresponds to union.
For example,
\[ x>1 \]and
\[ x<4 \]means
\[ (1,\infty)\cap(-\infty,4) = \boxed{(1,4)}. \]Inequality Identities and Contradictions
Linear inequalities can also simplify to statements with no variable.
The result may be always true or always false.
Always-True Inequality
Consider
\[ 2x+3<2x+8. \]Subtract \(2x\):
\[ 3<8. \]This is always true.
Therefore, every real number is a solution:
\[ \boxed{\mathbb{R}}. \]Always-False Inequality
Consider
\[ 2x+9<2x+4. \]Subtract \(2x\):
\[ 9<4. \]This is false.
Therefore,
\[ \boxed{\varnothing}. \]Comparing Two Affine Functions
Suppose
\[ A(x)=4x+20 \]and
\[ B(x)=6x+8. \]When is
\[ A(x)>B(x)? \]Solve:
\[ 4x+20>6x+8. \]Subtract \(4x\):
\[ 20>2x+8. \]Subtract \(8\):
\[ 12>2x. \]Divide by \(2\):
\[ \boxed{x<6}. \]Interpret the Comparison Graphically
The functions
\[ A(x)=4x+20 \]and
\[ B(x)=6x+8 \]are equal at
\[ x=6. \]For
\[ x<6, \]the graph of \(A\) lies above the graph of \(B\).
For
\[ x>6, \]the graph of \(B\) lies above \(A\).
Break-Even Equations
Suppose revenue and cost are
\[ R(q)=25q \]and
\[ C(q)=15q+400. \]Break-even occurs when
\[ R(q)=C(q). \]So
\[ 25q=15q+400. \]Then
\[ 10q=400, \] \[ \boxed{q=40}. \]Profit Inequality
Using the same models, profit is positive when
\[ R(q)>C(q). \]So
\[ 25q>15q+400, \] \[ 10q>400, \] \[ \boxed{q>40}. \]The equation gives the threshold; the inequality gives the profitable region.
Threshold Problems
Linear inequalities often describe thresholds.
Examples include:
- minimum required production;
- maximum allowed cost;
- temperature safety ranges;
- budget limits;
- minimum scores;
- capacity constraints.
The boundary value often comes from the corresponding equality.
Example: Budget Limit
A service charges \$30 initially plus \$12 per hour.
How many hours can be used if the budget is at most \$90?
Model:
\[ 12h+30\le90. \]Subtract \(30\):
\[ 12h\le60. \]Divide by \(12\):
\[ \boxed{h\le5}. \]If time cannot be negative, the practical solution is
\[ \boxed{0\le h\le5}. \]Example: Minimum Requirement
A student has \(40\) points and earns \(8\) points per completed task.
How many tasks are needed to reach at least \(80\) points?
Let \(t\) be tasks:
\[ 40+8t\ge80. \]Then
\[ 8t\ge40, \] \[ t\ge5. \]Because tasks are counted in whole numbers,
\[ \boxed{t\ge5,\quad t\in\mathbb{Z}}. \]Discrete Domains Matter
An inequality may produce a real-number solution such as
\[ x\ge4.3. \]If \(x\) counts people, boxes, or completed tasks, then only integer values may be meaningful.
The smallest practical solution may be
\[ \boxed{x=5}. \]Example: Capacity Constraint
An elevator can carry at most \(1200\) kilograms.
If the equipment already weighs \(300\) kilograms and each crate weighs \(75\) kilograms, then
\[ 300+75c\le1200. \]Subtract \(300\):
\[ 75c\le900. \]So
\[ c\le12. \]For nonnegative whole crates,
\[ \boxed{0\le c\le12,\quad c\in\mathbb{Z}}. \]Inequalities and Domain Restrictions
Algebraic solving may produce a broad interval, but the model may require additional restrictions.
For example, if time \(t\) satisfies
\[ t\le8 \]but negative time is not meaningful, the practical solution is
\[ \boxed{0\le t\le8}. \]Solving Before or After Simplifying
Consider
\[ 4(x-2)+3\le2x+9. \]Distribute:
\[ 4x-8+3\le2x+9. \]Combine:
\[ 4x-5\le2x+9. \]Subtract \(2x\):
\[ 2x-5\le9. \]Add \(5\):
\[ 2x\le14. \]So
\[ \boxed{x\le7}. \]Fractions in Inequalities
Consider
\[ \frac{x}{3}-2>4. \]Add \(2\):
\[ \frac{x}{3}>6. \]Multiply by positive \(3\):
\[ \boxed{x>18}. \]Clearing Fractions with a Positive Common Denominator
Consider
\[ \frac{x}{2}+\frac{x}{4}\le9. \]Multiply every term by positive \(4\):
\[ 2x+x\le36. \]So
\[ 3x\le36, \] \[ \boxed{x\le12}. \]Be Careful When Multiplying by an Expression
Multiplying an inequality by a known positive number is safe.
Multiplying by an expression whose sign is unknown requires more care because the inequality direction depends on whether that expression is positive or negative.
This becomes important later in rational inequalities and more advanced algebra.
Absolute-Value Connection Preview
A bounded interval such as
\[ -3\le x\le3 \]can also be written
\[ |x|\le3. \]Likewise,
\[ |x|<5 \]means
\[ -5Equation Solving as Rewriting to an Equivalent Statement
At each correct step, the new equation should have the same solution set as the previous one.
For example:
\[ 3x+6=18 \] \[ 3x=12 \] \[ x=4. \]Each statement is equivalent to the previous one.
Inequality Solving Also Preserves Solution Sets
Correct inequality transformations preserve the same set of solutions.
The sign-reversal rule is necessary precisely because without it, the solution set would change incorrectly.
Example: Detect a Sign-Reversal Error
Suppose someone solves
\[ -2x<8 \]and writes
\[ x< -4. \]Test \(x=0\).
The original inequality gives
\[ 0<8, \]which is true.
But \(0<-4\) is false.
The proposed solution cannot be correct.
The correct solution is
\[ \boxed{x>-4}. \]Boundary Testing
For an inequality such as
\[ x<5, \]the boundary is
\[ x=5. \]Substitute into the original strict inequality:
\[ 5<5, \]which is false.
So the endpoint is excluded.
Inclusive Boundary Testing
For
\[ x\le5, \]the boundary value gives
\[ 5\le5, \]which is true.
So the endpoint is included.
Solution Sets as Mathematical Objects
An equation or inequality does not merely produce an answer number.
It defines a solution set.
Examples:
\[ \{5\}, \] \[ (-\infty,5), \] \[ [2,8], \] \[ \mathbb{R}, \]or
\[ \varnothing. \]Thinking in terms of sets becomes increasingly important in later mathematics.
A Reliable Linear-Equation Workflow
- Simplify each side.
- Distribute and combine like terms when needed.
- Move variable terms toward one side.
- Move constants toward the other side.
- Undo multiplication or division.
- Recognize identity or contradiction cases.
- Substitute to verify the solution.
A Reliable Linear-Inequality Workflow
- Simplify both sides.
- Collect variable terms and constants.
- Use addition or subtraction normally.
- If multiplying or dividing by a negative number, reverse the inequality.
- Write the solution clearly.
- Convert to interval notation or a number-line graph when useful.
- Check a sample point and the boundary.
A Reliable Comparison Workflow
- Write the two expressions or functions being compared.
- Use equality to find the boundary or intersection.
- Use the inequality to determine which side of the boundary satisfies the comparison.
- Interpret the result graphically or in context.
A Reliable Applied-Inequality Workflow
- Define the variable.
- Translate words such as at most, at least, more than, and less than carefully.
- Construct the inequality.
- Solve algebraically.
- Apply domain restrictions such as nonnegativity or integrality.
- Interpret the final solution in words and units.
Common Translation Phrases
| Phrase | Symbol |
|---|---|
| less than | \(<\) |
| at most | \(\le\) |
| no more than | \(\le\) |
| greater than | \(>\) |
| at least | \(\ge\) |
| no less than | \(\ge\) |
Common Errors
Changing Signs Without Performing an Operation
Terms do not literally jump across an equation. The shorthand reflects addition or subtraction on both sides.
Forgetting to Reverse an Inequality
Multiplying or dividing by a negative number reverses the inequality direction.
Reversing an Inequality When Dividing by a Positive Number
Positive scaling preserves order.
Confusing One Solution with All Real Numbers
An identity such as \(6=6\) means every input works, not \(x=6\).
Confusing a Contradiction with \(x=0\)
A false statement such as \(4=9\) means no solution.
Ignoring Domain Restrictions
A real-number inequality may need to be restricted to nonnegative or integer values in context.
Using a Closed Endpoint for a Strict Inequality
\(<\) and \(>\) exclude the boundary.
Using Brackets with Infinity
Infinity is not an included endpoint, so interval notation always uses parentheses at \(\pm\infty\).
Practice
1. Solve a Linear Equation
Solve
\[ 5x-7=18. \]Check Answer
\[ 5x=25, \] \[ \boxed{x=5}. \]2. Variables on Both Sides
Solve
\[ 7x+4=3x+20. \]Check Answer
\[ 4x+4=20, \] \[ 4x=16, \] \[ \boxed{x=4}. \]3. Identity or Contradiction?
Solve
\[ 3(x+2)=3x+6. \]Check Answer
Distribute:
\[ 3x+6=3x+6. \]The equation is always true.
\[ \boxed{x\in\mathbb{R}}. \]4. No Solution
Solve
\[ 4x+1=4x-3. \]Check Answer
Subtract \(4x\):
\[ 1=-3. \]This is impossible.
\[ \boxed{\varnothing}. \]5. Basic Inequality
Solve
\[ 3x+2<17. \]Check Answer
\[ 3x<15, \] \[ \boxed{x<5}. \]Interval notation:
\[ \boxed{(-\infty,5)}. \]6. Negative Division
Solve
\[ -4x\le20. \]Check Answer
Divide by \(-4\) and reverse the inequality:
\[ \boxed{x\ge-5}. \]Interval notation:
\[ \boxed{[-5,\infty)}. \]7. Compound Inequality
Solve
\[ 1<2x+3\le11. \]Check Answer
Subtract \(3\):
\[ -2<2x\le8. \]Divide by \(2\):
\[ \boxed{-18. Compare Two Functions
For what values of \(x\) is
\[ 4x+10>6x-2? \]Check Answer
\[ 12>2x, \] \[ \boxed{x<6}. \]9. Budget Problem
A rental costs \$25 plus \$15 per hour. If the budget is at most \$100, how many hours can be rented?
Check Answer
\[ 25+15h\le100. \] \[ 15h\le75. \] \[ h\le5. \]With nonnegative time:
\[ \boxed{0\le h\le5}. \]10. Minimum Whole-Number Requirement
A person has \$30 and saves \$12 per week. How many whole weeks are needed to have at least \$100?
Check Answer
\[ 30+12w\ge100. \] \[ 12w\ge70. \] \[ w\ge\frac{70}{12}\approx5.83. \]Whole weeks are required, so
\[ \boxed{w\ge6}. \]11. Always True or Always False?
Solve
\[ 5x-2\le5x+7. \]Check Answer
Subtract \(5x\):
\[ -2\le7. \]This is always true.
\[ \boxed{\mathbb{R}}. \]12. Solve and Write Interval Notation
Solve
\[ 9-3x>0. \]Check Answer
\[ -3x>-9. \]Divide by \(-3\) and reverse the inequality:
\[ \boxed{x<3}. \]Interval notation:
\[ \boxed{(-\infty,3)}. \]Key Ideas
- A linear equation asks when two affine expressions are equal.
- A linear inequality asks when one affine expression is greater than or less than another.
- Equation solving preserves equality through equivalent operations.
- Linear equations may have one solution, no solution, or infinitely many solutions.
- Graphically, solving \(L(x)=R(x)\) finds intersections of two lines.
- Any equation \(L(x)=R(x)\) can be rewritten as a zero problem \(L(x)-R(x)=0\).
- Adding or subtracting the same quantity preserves inequality direction.
- Multiplying or dividing by a positive number preserves inequality direction.
- Multiplying or dividing by a negative number reverses inequality direction.
- Strict inequalities exclude boundary values; inclusive inequalities include them.
- Interval notation and number-line graphs represent inequality solution sets.
- Compound inequalities combine conditions through intersection or union.
- Equations often identify thresholds, while inequalities identify the region on one side of a threshold.
- Applied solutions may require nonnegative, integer, or otherwise restricted domains.
- Verification through substitution, boundary tests, and sample points helps catch sign and direction errors.
Where This Leads
Linear equations and inequalities turn affine relationships into decision tools. Equations locate exact thresholds and intersections; inequalities describe intervals where conditions hold. The same algebra can therefore answer questions about equality, comparison, limits, feasibility, and change.
From here, these ideas lead naturally into interpolation, extrapolation, and trend analysis, where lines are used to estimate unknown values and interpret patterns across data. They also prepare for systems of equations and inequalities, where several linear conditions must be satisfied together.