Solve linear inequalities of the form Ax + B C, Ax + B = C.
< (Less Than)
> (Greater Than)
≤ (Less Than or Equal To)
≥ (Greater Than or Equal To)
Solution
Enter values to see the solution.
Understanding and Solving Linear Inequalities
Linear inequalities are mathematical statements that compare two expressions using inequality symbols. They describe a range of values for a variable that make the statement true, unlike equations which typically have specific solutions. The general form of a linear inequality in one variable is:
Ax + B < C
Where 'A' is the coefficient of the variable 'x', 'B' is a constant term, 'C' is another constant, and the symbol (, ≤, ≥) indicates the type of relationship between the two sides.
How to Solve Them
Solving a linear inequality involves isolating the variable (x) on one side of the inequality, similar to solving a linear equation. The key difference lies in how multiplication and division affect the inequality sign:
Adding or Subtracting Terms: You can add or subtract any number from both sides of an inequality without changing the direction of the inequality sign.
Multiplying or Dividing by a Positive Number: If you multiply or divide both sides by a positive number, the inequality sign remains the same.
Multiplying or Dividing by a Negative Number: If you multiply or divide both sides by a negative number, you MUST reverse the direction of the inequality sign.
Steps to Solve Ax + B < C (and similar inequalities):
Subtract B from both sides: This isolates the term with x.
Ax < C - B
Divide both sides by A:
If A is positive: x < (C - B) / A
If A is negative: x > (C - B) / A (Note the sign flip!)
If A is zero: This leads to a special case. If 0*x + B < C, then it's true if B < C, and false otherwise. It does not depend on x.
The same principles apply to other inequality symbols (>, ≤, ≥).
Use Cases
Linear inequalities are used in various fields:
Resource Allocation: Determining how many units of a product can be produced given constraints on labor, materials, or time.
Budgeting: Ensuring spending stays below a certain limit.
Physics: Describing conditions where one force is greater than another, or a system's state is within a certain range.
Optimization Problems: Forming the constraints in linear programming.
General Decision Making: Evaluating conditions where one quantity must be greater than, less than, or equal to another.
Example Calculation
Let's solve the inequality: 3x + 5 ≥ 20
Here, A = 3, B = 5, C = 20, and the operator is ≥.
Subtract B (5) from both sides:
3x ≥ 20 - 5 3x ≥ 15
Divide both sides by A (3). Since 3 is positive, the sign stays the same:
x ≥ 15 / 3 x ≥ 5
Solution: x must be greater than or equal to 5.
Another Example (with negative coefficient)
Let's solve the inequality: -2x + 10 < 4
Here, A = -2, B = 10, C = 4, and the operator is <.
Subtract B (10) from both sides:
-2x < 4 - 10 -2x < -6
Divide both sides by A (-2). Since -2 is negative, we MUST reverse the inequality sign:
x > -6 / -2 x > 3
Solution: x must be strictly greater than 3.
function solveInequality() {
var a = parseFloat(document.getElementById("coefficientA").value);
var b = parseFloat(document.getElementById("constantB").value);
var c = parseFloat(document.getElementById("constantC").value);
var comparison = document.getElementById("comparison").value;
var solutionOutput = document.getElementById("solutionOutput");
// Clear previous results and error messages
solutionOutput.textContent = "";
solutionOutput.style.color = "#28a745"; // Reset to success green
// Input validation
if (isNaN(a) || isNaN(b) || isNaN(c)) {
solutionOutput.textContent = "Error: Please enter valid numbers for all fields.";
solutionOutput.style.color = "red";
return;
}
var simplifiedInequality = "";
var inequalitySign = "";
var resultValue = "";
var isTrueForAllX = false;
var isFalseForAllX = false;
// Step 1: Simplify to Ax C-B etc.
var rhs = c – b;
// Define inequality symbols for display
var symbols = {
"less_than": "",
"less_than_equal": "≤",
"greater_than_equal": "≥"
};
// Determine the operator for the simplified inequality
switch(comparison) {
case "less_than": inequalitySign = ""; break;
case "less_than_equal": inequalitySign = "≤"; break;
case "greater_than_equal": inequalitySign = "≥"; break;
}
// Step 2: Solve for x based on the sign of 'a'
if (a === 0) {
// Case: Coefficient of x is zero
if (comparison === "less_than") {
isTrueForAllX = (b c);
} else if (comparison === "less_than_equal") {
isTrueForAllX = (b = c);
}
if (isTrueForAllX) {
simplifiedInequality = "0x " + symbols[comparison] + " " + rhs;
solutionOutput.textContent = "The inequality is true for all real numbers (x ∈ ℝ).";
solutionOutput.style.color = "#28a745";
} else {
simplifiedInequality = "0x " + symbols[comparison] + " " + rhs;
solutionOutput.textContent = "The inequality is false for all real numbers (no solution).";
solutionOutput.style.color = "red";
}
} else if (a > 0) {
// Case: Coefficient of x is positive
resultValue = rhs / a;
simplifiedInequality = "x " + inequalitySign + " " + resultValue;
solutionOutput.innerHTML = "The solution is: x " + inequalitySign + " " + resultValue + "";
} else { // a "; break;
case "greater_than": flippedSign = "<"; break;
case "less_than_equal": flippedSign = "≥"; break;
case "greater_than_equal": flippedSign = "≤"; break;
}
simplifiedInequality = "x " + flippedSign + " " + resultValue;
solutionOutput.innerHTML = "The solution is: x " + flippedSign + " " + resultValue + "";
}
// Display the simplified form if it's not a special case
if (a !== 0 && !isTrueForAllX && !isFalseForAllX) {
var originalInequality = "" + a + "x + " + b + " " + symbols[comparison] + " " + c + "";
var simplifiedFormDisplay = "x " + (a < 0 ? (flippedSign || "?") : (symbols[comparison] || "?")) + " " + resultValue + "";
var stepsText = originalInequality + " simplifies to " + simplifiedFormDisplay;
// Additional text for context
if (a < 0) {
stepsText += " (Note: Inequality sign flipped due to division by a negative number).";
}
// You could potentially add more descriptive text here about intervals or number lines if needed
}
}