Solve and visualize linear inequalities with ease.
Linear Inequality Solver
Enter the coefficient for the variable 'x'.
Enter the constant term.
< (Less Than)
<= (Less Than or Equal To)
> (Greater Than)
>= (Greater Than or Equal To)
Select the inequality symbol.
Enter the value on the right side of the inequality.
Calculation Results
What is a Linear Inequality?
A linear inequality is a mathematical statement that compares two linear expressions using an inequality symbol (, ≤, ≥). Unlike linear equations, which represent a single point or line, linear inequalities represent a region of solutions. They are fundamental in various fields, including optimization problems, economics, and engineering, where we often deal with constraints and boundaries rather than exact values.
Who should use it? Students learning algebra, mathematicians, engineers, economists, data scientists, and anyone needing to model situations with constraints or ranges of possible values. Understanding linear inequalities is crucial for grasping concepts like feasible regions in linear programming.
Common misconceptions: A frequent misunderstanding is that a linear inequality has a single solution. In reality, it typically defines an infinite set of solutions, often represented graphically as a shaded region. Another misconception is that the direction of the inequality sign never changes; while it usually stays the same, it flips when you multiply or divide both sides by a negative number.
Linear Inequality Formula and Mathematical Explanation
The standard form of a simple linear inequality in one variable is typically expressed as:
ax + b c
Where:
'a' is the coefficient of the variable 'x'.
'b' is a constant term added to the 'x' term.
" represents one of the inequality symbols: , ≤, or ≥.
'c' is the constant value on the right-hand side of the inequality.
The goal is to isolate 'x' to determine the range of values that satisfy the inequality. The process involves algebraic manipulation similar to solving linear equations, with one crucial difference: if you multiply or divide both sides by a negative number, you must reverse the direction of the inequality sign.
Step-by-step derivation:
Start with the inequality: ax + b c
Subtract 'b' from both sides: ax c – b
Now, consider the coefficient 'a':
If a > 0: Divide both sides by 'a'. The inequality sign remains the same. x (c – b) / a
If a < 0: Divide both sides by 'a'. The inequality sign must be reversed. x (c – b) / a
If a = 0: The inequality becomes 0*x + b c, which simplifies to b c. This is either always true (if b c is valid) or always false (if b c is invalid), and 'x' can be any real number.
The result gives the solution set for 'x'.
Variables Table
Variables in the linear inequality ax + b c
Variable
Meaning
Unit
Typical Range
a
Coefficient of x
Unitless
Any real number (often non-zero for a unique solution range)
b
Constant term
Unitless
Any real number
Inequality Operator
Symbol
, ≤, ≥
c
Right-hand side value
Unitless
Any real number
x
The variable being solved for
Unitless
Real numbers (satisfying the derived inequality)
Practical Examples (Real-World Use Cases)
Example 1: Budget Constraint
Suppose you have a budget of $50 for snacks. Apples cost $2 each, and bananas cost $1 each. You decide you want at least 5 apples. How many bananas can you buy?
Let 'a' be the number of apples and 'b' be the number of bananas.
The cost inequality is: 2a + 1b ≤ 50.
You decide to buy exactly 5 apples (a=5).
Substitute a=5 into the inequality:
2(5) + b ≤ 50
10 + b ≤ 50
Subtract 10 from both sides:
b ≤ 40
Interpretation: You can buy 40 or fewer bananas if you buy 5 apples, given your budget.
Example 2: Production Limit
A factory produces two types of widgets, Type A and Type B. Type A requires 3 hours of labor, and Type B requires 2 hours. The factory has a maximum of 120 labor hours available per week. If they plan to produce at least 20 units of Type A, what is the maximum number of Type B widgets they can produce?
Let 'a' be the number of Type A widgets and 'b' be the number of Type B widgets.
The labor constraint is: 3a + 2b ≤ 120.
They plan to produce at least 20 units of Type A (a ≥ 20).
To find the maximum 'b' for a given 'a', we rearrange the inequality:
2b ≤ 120 – 3a
b ≤ (120 – 3a) / 2
If they produce exactly 20 units of Type A (a=20):
b ≤ (120 – 3*20) / 2
b ≤ (120 – 60) / 2
b ≤ 60 / 2
b ≤ 30
Interpretation: If the factory produces 20 units of Type A, they can produce a maximum of 30 units of Type B without exceeding the labor hour limit.
How to Use This Linear Inequality Calculator
Our Linear Inequality Calculator is designed for simplicity and accuracy. Follow these steps to get your solution:
Input Coefficients: Enter the value for the coefficient of 'x' (denoted as 'a') and the constant term added to 'x' (denoted as 'b') in the respective fields.
Select Operator: Choose the correct inequality symbol (, ≤, ≥) from the dropdown menu that represents your specific linear inequality.
Input Right-Hand Value: Enter the constant value on the right side of the inequality (denoted as 'c').
Calculate: Click the "Calculate Solution" button.
Review Results: The calculator will display the primary solution for 'x' (e.g., x > 2.5), key intermediate values used in the calculation, and the formula applied.
Interpret the Solution: The primary result tells you the range of values 'x' can take. For example, "x > 5" means any number strictly greater than 5 is a solution. "x ≤ 10" means 10 and any number less than 10 are solutions.
Visualize (Chart): The generated chart visually represents the solution set on a number line, showing the boundary point and whether it's included (solid circle) or excluded (hollow circle), and the direction of the solution region.
Use the Table: The table provides a structured breakdown of the input values and the derived solution.
Reset: If you need to start over or try a new inequality, click the "Reset" button to clear all fields and return to default values.
Copy Results: Use the "Copy Results" button to easily transfer the calculated solution and key details to another document or application.
Decision-Making Guidance: Use the results to understand the boundaries and possibilities within a given constraint. For instance, if 'x' represents the number of items you can produce, a result like "x ≤ 50" tells you the maximum you can produce. If 'x' represents a cost, "x ≥ $100" indicates the minimum you must spend.
Key Factors That Affect Linear Inequality Results
While the calculation itself is straightforward algebra, several underlying factors influence the context and interpretation of linear inequality results:
Coefficient of x ('a'): The sign and magnitude of 'a' are critical. A positive 'a' means 'x' increases as the right side increases (or decreases as the right side decreases). A negative 'a' reverses this relationship, and crucially, requires flipping the inequality sign when isolating 'x'. If 'a' is zero, the inequality might be universally true, universally false, or dependent only on 'b' and 'c'.
Constant Terms ('b' and 'c'): These values set the baseline and the target. Changing 'b' shifts the boundary point, while changing 'c' directly alters the target value. Their interaction determines the final boundary for 'x'.
Inequality Operator (, ≤, ≥): This symbol dictates whether the boundary value itself is included in the solution set. " exclude the boundary (open interval), while '≤' and '≥' include it (closed interval). This distinction is vital in applications where the boundary condition is permissible or strictly forbidden.
Context of the Variable 'x': Is 'x' a quantity, a cost, a time, a rate? The interpretation of "x > 5" changes drastically depending on what 'x' represents. For example, "x > 5 units" is different from "x > $5".
Real-World Constraints (Implicit): Often, variables must be non-negative (e.g., you can't produce negative items). While not explicitly part of the ax + b c formula, these implicit constraints (like x ≥ 0) must be considered alongside the calculated inequality to define a realistic solution set.
Units of Measurement: Ensure all terms (a, b, c) are in compatible units. If 'a' is in hours/widget, 'b' is in hours, and 'c' is in hours, the resulting 'x' will be in widgets. Mismatched units lead to nonsensical results.
Integer vs. Real Solutions: The calculator provides real number solutions. However, in many practical scenarios (like producing discrete items), only integer solutions are valid. You may need to round or select integers within the calculated range.
Multiple Inequalities (Systems): This calculator handles one inequality at a time. In real-world problems (like linear programming), you often face multiple constraints simultaneously, defining a feasible region where all inequalities are satisfied.
Frequently Asked Questions (FAQ)
Q1: What's the difference between a linear inequality and a linear equation?
A linear equation (e.g., 2x + 5 = 15) typically has one specific solution (x=5). A linear inequality (e.g., 2x + 5 < 15) represents a range of solutions (x < 5), often an infinite set.
Q2: When do I flip the inequality sign?
You must flip the inequality sign (, ≥ becomes ≤, etc.) only when you multiply or divide both sides of the inequality by a negative number.
Q3: Can 'a' (the coefficient of x) be zero?
Yes. If 'a' is zero, the inequality simplifies. For example, 0x + 3 < 7 becomes 3 < 7, which is always true. If it was 0x + 8 < 7, it becomes 8 < 7, which is always false. In these cases, 'x' can be any real number (if true) or no real number (if false).
Q4: How do I represent the solution on a number line?
For x > c, draw a number line, place an open circle at 'c', and shade to the right. For x ≥ c, use a closed circle at 'c' and shade to the right. For x < c, use an open circle at 'c' and shade to the left. For x ≤ c, use a closed circle at 'c' and shade to the left.
Q5: What does the calculator's "primary result" mean?
The primary result is the simplified form of the inequality, isolating 'x'. For example, if you input 2x + 4 > 10, the primary result might be "x > 3", indicating all values greater than 3 satisfy the original inequality.
Q6: Does the calculator handle inequalities with 'x' on both sides?
This specific calculator is designed for the form ax + b c. For inequalities like 3x + 2 < x + 8, you would first rearrange it algebraically to the standard form before using the calculator.
Q7: What if my inequality involves multiplication or division by a variable?
Inequalities involving multiplication or division by a variable (e.g., x/2 12) are generally handled by the same algebraic principles. However, if the variable could be negative, you must consider the case where the inequality sign flips. This calculator assumes a standard linear form.
Q8: Can this calculator solve systems of linear inequalities?
No, this calculator solves a single linear inequality. Systems of inequalities, which define a feasible region, require graphical methods or more advanced solvers.
var coefficientAInput = document.getElementById('coefficientA');
var constantBInput = document.getElementById('constantB');
var operatorSelect = document.getElementById('operator');
var valueCInput = document.getElementById('valueC');
var errorA = document.getElementById('errorA');
var errorB = document.getElementById('errorB');
var errorC = document.getElementById('errorC');
var resultsContainer = document.getElementById('results-container');
var primaryResultDiv = document.getElementById('primary-result');
var intermediateADiv = document.getElementById('intermediate-a');
var intermediateBDiv = document.getElementById('intermediate-b');
var intermediateCDiv = document.getElementById('intermediate-c');
var intermediateOperatorDiv = document.getElementById('intermediate-operator');
var formulaExplanationDiv = document.getElementById('formula-explanation');
var tableContainer = document.getElementById('table-container');
var inequalityChart = document.getElementById('inequalityChart');
var chartLegend = document.getElementById('chart-legend');
var ctx = inequalityChart.getContext('2d');
var chartInstance = null;
function validateInput(value, errorElement, fieldName) {
if (value === ") {
errorElement.textContent = fieldName + ' cannot be empty.';
errorElement.style.display = 'block';
return false;
}
if (isNaN(value)) {
errorElement.textContent = 'Please enter a valid number.';
errorElement.style.display = 'block';
return false;
}
errorElement.textContent = ";
errorElement.style.display = 'none';
return true;
}
function calculateInequality() {
var a = parseFloat(coefficientAInput.value);
var b = parseFloat(constantBInput.value);
var operator = operatorSelect.value;
var c = parseFloat(valueCInput.value);
var isValidA = validateInput(coefficientAInput.value, errorA, 'Coefficient a');
var isValidB = validateInput(constantBInput.value, errorB, 'Constant b');
var isValidC = validateInput(valueCInput.value, errorC, 'Value c');
if (!isValidA || !isValidB || !isValidC) {
resultsContainer.classList.add('hidden');
return;
}
var inequalityString = coefficientAInput.value + 'x + ' + constantBInput.value + ' ' + operator + ' ' + valueCInput.value;
var solutionX = ";
var solutionOperator = ";
var formula = ";
var boundaryValue = ";
var inequalityType = "; // ", '='
if (operator === '<') inequalityType = '<';
else if (operator === '<=') inequalityType = '') inequalityType = '>';
else if (operator === '>=') inequalityType = '>=';
if (a > 0) {
boundaryValue = (c – b) / a;
solutionOperator = operator;
formula = 'ax + b ' + operator + ' c => ax ' + operator + ' c – b => x ' + operator + ' (c – b) / a';
} else if (a < 0) {
boundaryValue = (c – b) / a;
if (operator === '';
else if (operator === '=';
else if (operator === '>') solutionOperator = '=') solutionOperator = ' ax ' + operator + ' c – b => x ' + solutionOperator + ' (c – b) / a (sign flipped)';
} else { // a === 0
if (operator === '<') {
if (b < c) { solutionX = 'Any real number'; formula = '0x + b b < c (True)'; }
else { solutionX = 'No solution'; formula = '0x + b b < c (False)'; }
} else if (operator === '<=') {
if (b <= c) { solutionX = 'Any real number'; formula = '0x + b b <= c (True)'; }
else { solutionX = 'No solution'; formula = '0x + b b ') {
if (b > c) { solutionX = 'Any real number'; formula = '0x + b > c => b > c (True)'; }
else { solutionX = 'No solution'; formula = '0x + b > c => b > c (False)'; }
} else if (operator === '>=') {
if (b >= c) { solutionX = 'Any real number'; formula = '0x + b >= c => b >= c (True)'; }
else { solutionX = 'No solution'; formula = '0x + b >= c => b >= c (False)'; }
}
}
if (a !== 0) {
solutionX = 'x ' + solutionOperator + ' ' + boundaryValue.toFixed(4);
}
primaryResultDiv.textContent = solutionX;
intermediateADiv.innerHTML = 'Coefficient a: ' + a;
intermediateBDiv.innerHTML = 'Constant b: ' + b;
intermediateCDiv.innerHTML = 'Value c: ' + c;
intermediateOperatorDiv.innerHTML = 'Operator: ' + operator;
formulaExplanationDiv.textContent = 'Formula Used: ' + formula;
generateTable(a, b, operator, c, boundaryValue, solutionOperator, inequalityType);
updateChart(boundaryValue, solutionOperator, inequalityType);
resultsContainer.classList.remove('hidden');
}
function generateTable(a, b, operator, c, boundaryValue, solutionOperator, inequalityType) {
var tableHTML = '
' + ( (operator === '<' && b < c) || (operator === '<=' && b ' && b > c) || (operator === '>=' && b >= c) ? 'True (Any x)' : 'False (No solution)' ) + '
';
}
tableHTML += '
';
tableContainer.innerHTML = tableHTML;
}
function updateChart(boundaryValue, solutionOperator, inequalityType) {
if (chartInstance) {
chartInstance.destroy();
}
var dataPoints = [];
var labels = [];
var minValue = Math.min(-10, boundaryValue – 5);
var maxValue = Math.max(10, boundaryValue + 5);
var step = (maxValue – minValue) / 100;
for (var i = 0; i <= 100; i++) {
var xVal = minValue + i * step;
labels.push(xVal.toFixed(2));
var satisfies = false;
if (inequalityType === '<') satisfies = xVal < boundaryValue;
else if (inequalityType === '<=') satisfies = xVal ') satisfies = xVal > boundaryValue;
else if (inequalityType === '>=') satisfies = xVal >= boundaryValue;
dataPoints.push(satisfies ? xVal : NaN); // Use NaN to break the line
}
var boundaryLineValue = boundaryValue;
var boundaryLineX = [boundaryLineValue, boundaryValue];
var boundaryLineY = [-1, 1]; // Arbitrary Y range for visibility
var chartData = {
datasets: [{
label: 'Solution Region (Satisfies Inequality)',
data: dataPoints.map(function(val, index) { return {x: minValue + index * step, y: val === NaN ? NaN : 0}; }), // Plot on y=0
borderColor: 'rgba(40, 167, 69, 0.8)',
backgroundColor: 'rgba(40, 167, 69, 0.2)',
fill: true,
pointRadius: 0,
showLine: true
}, {
label: 'Boundary Line',
data: [{x: boundaryLineValue, y: -1}, {x: boundaryLineValue, y: 1}],
borderColor: 'rgba(0, 74, 153, 1)',
borderWidth: 2,
borderDash: [5, 5],
pointRadius: 0,
showLine: true
}]
};
chartInstance = new Chart(ctx, {
type: 'scatter',
data: chartData,
options: {
responsive: true,
maintainAspectRatio: false,
scales: {
x: {
type: 'linear',
position: 'bottom',
title: {
display: true,
text: 'Value of x'
},
min: minValue,
max: maxValue,
ticks: {
callback: function(value, index, ticks) {
return value.toFixed(1);
}
}
},
y: {
display: false // Hide Y-axis as it's just for visualization
}
},
plugins: {
legend: {
display: true,
position: 'top',
},
tooltip: {
callbacks: {
label: function(context) {
if (context.dataset.label === 'Boundary Line') {
return 'Boundary: x = ' + context.parsed.x.toFixed(4);
}
if (!isNaN(context.raw.y)) {
return 'Solution: x = ' + context.raw.x.toFixed(4);
}
return null;
}
}
}
}
}
});
var legendHTML = 'Chart Legend:';
legendHTML += '■ Solution Region ';
legendHTML += ' Boundary Line (x = ' + boundaryValue.toFixed(4) + ')';
chartLegend.innerHTML = legendHTML;
}
function resetCalculator() {
coefficientAInput.value = '2';
constantBInput.value = '5';
operatorSelect.value = '<';
valueCInput.value = '10';
resultsContainer.classList.add('hidden');
errorA.textContent = ''; errorA.style.display = 'none';
errorB.textContent = ''; errorB.style.display = 'none';
errorC.textContent = ''; errorC.style.display = 'none';
if (chartInstance) {
chartInstance.destroy();
chartInstance = null;
}
tableContainer.innerHTML = '';
chartLegend.innerHTML = '';
}
function copyResults() {
var resultsText = "Linear Inequality Calculation:\n\n";
resultsText += "Inequality: " + coefficientAInput.value + "x + " + constantBInput.value + " " + operatorSelect.value + " " + valueCInput.value + "\n\n";
resultsText += "Primary Solution: " + primaryResultDiv.textContent + "\n\n";
resultsText += "Intermediate Values:\n";
resultsText += "- Coefficient a: " + intermediateADiv.textContent.replace('Coefficient a: ', '') + "\n";
resultsText += "- Constant b: " + intermediateBDiv.textContent.replace('Constant b: ', '') + "\n";
resultsText += "- Value c: " + intermediateCDiv.textContent.replace('Value c: ', '') + "\n";
resultsText += "- Operator: " + intermediateOperatorDiv.textContent.replace('Operator: ', '') + "\n\n";
resultsText += "Formula Used: " + formulaExplanationDiv.textContent.replace('Formula Used: ', '') + "\n\n";
var tempTextArea = document.createElement("textarea");
tempTextArea.value = resultsText;
document.body.appendChild(tempTextArea);
tempTextArea.select();
document.execCommand("copy");
document.body.removeChild(tempTextArea);
// Provide visual feedback
var copyButton = document.querySelector('button.btn-success');
var originalText = copyButton.textContent;
copyButton.textContent = 'Copied!';
setTimeout(function() {
copyButton.textContent = originalText;
}, 1500);
}
// Initial calculation on load if values are present
document.addEventListener('DOMContentLoaded', function() {
calculateInequality();
});