Enter the coefficients and constants for a linear equation in the form Ax + B = C or A(x + B) = C to solve for x.
Enter values to see the solution
Understanding Algebra One Equation Solving
Algebra One is a foundational course in mathematics that introduces students to the manipulation of variables and the solving of equations. A key skill developed in Algebra One is the ability to solve linear equations, which are equations where the highest power of the variable is one. These equations typically take forms like Ax + B = C or variations that can be simplified to this form.
The Standard Form: Ax + B = C
The equation Ax + B = C represents a linear relationship where:
x is the variable we want to solve for.
A is the coefficient of the variable x. It tells us how many x's we have.
B is a constant term added to the variable term.
C is the constant term on the other side of the equation.
The goal is to isolate x on one side of the equation. We achieve this by using inverse operations, ensuring that whatever we do to one side of the equation, we must also do to the other side to maintain equality.
Steps to Solve Ax + B = C:
Isolate the term with x: Subtract B from both sides of the equation.
Ax + B - B = C - B This simplifies to: Ax = C - B
Solve for x: Divide both sides of the equation by A (assuming A is not zero).
(Ax) / A = (C - B) / A This gives the solution: x = (C - B) / A
Handling Other Forms: A(x + B) = C
Sometimes, equations are presented with parentheses, such as A(x + B) = C. Before applying the steps above, we first need to simplify this form:
Distribute A: Multiply A by each term inside the parentheses.
A * x + A * B = C This results in: Ax + AB = C
Identify New Constants: Notice that AB is now a single constant term. Let's call this new constant B', where B' = AB. The equation becomes Ax + B' = C.
Solve as Standard Form: Now, you can solve this equation using the same steps as the Ax + B = C form, where the constant term is B' instead of B. The solution would be x = (C - AB) / A.
Why is this Important?
Solving linear equations is a fundamental skill that forms the basis for more complex mathematical concepts. It's used extensively in:
Science and Engineering: Modeling physical phenomena, calculating rates, and solving problems involving motion or forces.
Finance: Calculating simple interest, analyzing costs, and budgeting.
Everyday Problem Solving: Determining quantities, proportions, and solving practical scenarios that can be represented linearly.
This calculator is designed to help students and enthusiasts quickly verify their solutions or understand the process for solving basic linear equations encountered in Algebra One.
Example Calculation
Let's solve the equation: 5(x + 3) = 18
Here, A = 5, and within the parentheses, we have x + 3. So, we can identify:
A = 5
B = 3 (from the parenthesis)
C = 18
Using the calculator:
Coefficient A: 5
Constant B: 3
Constant C: 18
The calculator will first distribute A: 5x + 15 = 18. Then it will solve:
Subtract 15 from both sides: 5x = 18 - 15 => 5x = 3
Divide by 5: x = 3 / 5
The result displayed would be x = 0.6.
function solveEquation() {
var a = parseFloat(document.getElementById("coefficientA").value);
var b = parseFloat(document.getElementById("constantB").value);
var c = parseFloat(document.getElementById("constantC").value);
var resultDiv = document.getElementById("result");
resultDiv.classList.remove("error"); // Remove error class by default
// Input validation
if (isNaN(a) || isNaN(b) || isNaN(c)) {
resultDiv.innerHTML = "Please enter valid numbers for all fields.";
resultDiv.classList.add("error");
return;
}
// Check if A is zero, as division by zero is undefined for Ax = D
// However, if A=0, the equation becomes B = C, which is either true or false,
// and not a solvable linear equation for x in the standard sense.
// For Ax + B = C form:
if (a === 0) {
if (b === c) {
resultDiv.innerHTML = "Infinite solutions (0 = 0)";
} else {
resultDiv.innerHTML = "No solution (" + b + " != " + c + ")";
}
resultDiv.classList.add("error");
return;
}
// Calculate x for Ax + B = C
var numerator = c – b;
var x = numerator / a;
// Format the output
var resultHTML;
if (Number.isInteger(x)) {
resultHTML = "x = " + x;
} else {
// Display as fraction if possible and cleaner, otherwise decimal
var gcdValue = gcd(Math.abs(numerator), Math.abs(a));
var simplifiedNumerator = numerator / gcdValue;
var simplifiedDenominator = a / gcdValue;
if (simplifiedDenominator < 0) {
simplifiedNumerator *= -1;
simplifiedDenominator *= -1;
}
if (simplifiedDenominator === 1) {
resultHTML = "x = " + simplifiedNumerator;
} else {
resultHTML = "x = " + simplifiedNumerator + " / " + simplifiedDenominator;
}
// Optionally, show decimal approximation as well if needed
// resultHTML += " (≈ " + x.toFixed(4) + ")";
}
resultDiv.innerHTML = resultHTML;
}
// Helper function to calculate Greatest Common Divisor (GCD) for fraction simplification
function gcd(a, b) {
while (b) {
var temp = b;
b = a % b;
a = temp;
}
return a;
}