Linear Equations In One Variable Class 8 Chapter Notes

Key Definitions

Variable

A symbol, usually a letter (like x, y, t), that represents an unknown quantity which can take various numerical values.

Constant

A quantity that has a fixed numerical value. For example, 5, -10, 3/4.

Algebraic Expression

A combination of constants and variables connected by mathematical operators (+, -, ×, ÷), but without an equals sign. Example: 2x + 7.

Equation

A statement of equality between two algebraic expressions, which involves an equals sign (=). It holds true only for a specific value of the variable. Example: 2x + 7 = 15.

Linear Equation

An equation in which the highest power of the variable is 1. When graphed, it forms a straight line.

Linear Equation in One Variable

A linear equation that contains only one variable. Its standard form is ax + b = c, where a, b, and c are real numbers and a ≠ 0. Example: 5y - 3 = 12.

Solution or Root

The specific value of the variable that makes the equation true (i.e., makes the Left Hand Side (LHS) equal to the Right Hand Side (RHS)).

Understanding What Makes an Equation 'Linear in One Variable'

An equation acts like a balanced weighing scale. The equals sign (=) is the pivot point. For an equation to be a 'Linear Equation in One Variable', it must satisfy two conditions. First, it must be 'linear', which means the highest power of the variable in the equation is 1. You will see terms like x, 3y, or (1/2)z, but you will NOT see x², y³, or √z. For example, 4x + 5 = 13 is linear, but 4x² + 5 = 13 is not (it's a quadratic equation). Second, it must be 'in one variable', which means there is only one type of unknown in the entire equation. For example, 2x - 10 = 5x + 2 is an equation in one variable (the variable is x). However, 2x + 3y = 10 is an equation in two variables (x and y). So, for this chapter, we focus on simple equations like ax + b = c, where our goal is to find the single unique value of x that keeps the scale balanced.

Must Remember

• An equation has two sides: Left Hand Side (LHS) and Right Hand Side (RHS), separated by the equals (=) sign.
• The primary rule of solving an equation is to keep it balanced. Whatever operation you perform on one side, you must perform the same on the other.
• The main goal is to isolate the variable (e.g., 'x') on one side of the equation.
• Transposing a term means moving it to the other side of the equation. When you transpose, the sign of the term changes: + becomes -, and - becomes +.
• A number that multiplies a variable on one side will divide on the other side when transposed, and vice versa.
• Always verify your solution by substituting the found value of the variable back into the original equation. If LHS = RHS, your solution is correct.
• For equations with brackets, use the distributive property to open the brackets first.
• For equations with fractions, you can simplify by multiplying the entire equation by the LCM of the denominators.

Step-by-Step Method to Solve Linear Equations

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Solved Examples

• {"title":"Example 1: Variable on One Side","bodyMarkdown":"Problem: Solve 3p - 7 = 11\n\nSolution:\n1. We need to isolate p. First, move the constant -7 to the RHS.\n 3p = 11 + 7\n2. Simplify the RHS.\n 3p = 18\n3. The 3 is multiplying p. Transpose it to the RHS, where it will divide.\n p = 18 / 3\n4. Solve for p.\n p = 6\n\nCheck: 3(6) - 7 = 18 - 7 = 11. LHS = RHS. The solution is correct."}
• {"title":"Example 2: Variable on Both Sides","bodyMarkdown":"Problem: Solve 5x + 9 = 5 + 3x\n\nSolution:\n1. Collect variable terms on the LHS. Move 3x from RHS to LHS.\n 5x - 3x + 9 = 5\n2. Collect constant terms on the RHS. Move +9 from LHS to RHS.\n 5x - 3x = 5 - 9\n3. Combine like terms on both sides.\n 2x = -4\n4. Isolate x by dividing by its coefficient, 2.\n x = -4 / 2\n x = -2\n\nCheck: LHS: 5(-2) + 9 = -10 + 9 = -1. RHS: 5 + 3(-2) = 5 - 6 = -1. LHS = RHS. The solution is correct."}
• {"title":"Example 3: Forming an Equation from a Word Problem","bodyMarkdown":"Problem: If you subtract 1/2 from a number and multiply the result by 1/2, you get 1/8. What is the number?\n\nSolution:\n1. Let the unknown number be x.\n2. Subtract 1/2 from the number: x - 1/2\n3. Multiply the result by 1/2: (1/2) (x - 1/2)\n4. This is equal to 1/8. So, the equation is: (1/2)(x - 1/2) = 1/8\n5. Solve the equation: Multiply both sides by 2 to remove the outer 1/2.\n x - 1/2 = (1/8) 2\n x - 1/2 = 2/8 = 1/4\n6. Isolate x by transposing -1/2.\n x = 1/4 + 1/2\n7. Find the common denominator to add the fractions.\n x = 1/4 + 2/4 = 3/4\n\nAnswer: The number is 3/4."}

Exam Tip

In exams, especially for word problems, marks are often allocated for correctly forming the equation and for the method of solving, not just the final answer. Always write down the equation you've formed before you start solving it. Be extremely careful with signs (+ and -) during transposition. A single sign error is the most common mistake and will lead to an incorrect answer. After finding a solution, do a quick mental check by substituting the value back into the equation; it takes just a few seconds and can save you from silly mistakes.

Practice Questions with Solutions

• Q: Solve for x: 2x - 3 = x + 2 A: Transposing x to LHS and -3 to RHS gives 2x - x = 2 + 3, so x = 5.
• Q: Find the value of y: y/5 + 1 = 2 A: y/5 = 2 - 1 => y/5 = 1 => y = 5.
• Q: Solve: 8x = 20 + 3x A: 8x - 3x = 20 => 5x = 20 => x = 4.
• Q: The sum of three consecutive integers is 51. What are the integers? (Hint: Let them be x, x+1, x+2) A: Equation: x + (x+1) + (x+2) = 51 => 3x + 3 = 51 => 3x = 48 => x = 16. The integers are 16, 17, and 18.

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