Linear Equations in One Variable
Linear equations in one variable are among the most fundamental concepts in algebra, forming the backbone of equation-solving skills taught in Class 8 mathematics. A linear equation is an equation in which the highest power of the variable is 1 — there are no squared terms, cubed terms, or any higher powers. The phrase 'one variable' means the equation contains only one unknown quantity, typically represented by a letter like x, y, or z. For example, 3x + 5 = 14 is a linear equation in one variable (x). Solving such an equation means finding the value of the variable that makes the equation true — in this case, x = 3 because 3(3) + 5 = 9 + 5 = 14. Linear equations appear everywhere in daily life. When a shopkeeper calculates how many items you can buy with a given budget, when an engineer determines the required length of material, or when a scientist converts temperatures between Celsius and Fahrenheit — all these involve solving linear equations. In Class 8, students move beyond simple one-step and two-step equations to tackle equations with variables on both sides, equations involving fractions and decimals, and word problems that require translating everyday situations into algebraic equations. Mastering linear equations in one variable is essential preparation for systems of linear equations, quadratic equations, and more advanced algebra in Classes 9 and 10.
What is Linear Equations in One Variable?
A linear equation in one variable is an equation that can be written in the standard form:
ax + b = 0
where a and b are real numbers (constants) and a is not equal to 0, and x is the variable (the unknown quantity we want to find).
The word 'linear' comes from the fact that the graph of such an equation is a straight line. The word 'one variable' means there is exactly one unknown in the equation.
Key terminology:
Variable: The unknown quantity, usually denoted by x, y, z, or another letter. In the equation 5x - 7 = 18, the variable is x.
Coefficient: The number multiplied by the variable. In 5x - 7 = 18, the coefficient of x is 5.
Constant: A number without a variable. In 5x - 7 = 18, the constants are -7 and 18.
Left-Hand Side (LHS): The expression on the left of the equals sign. In 5x - 7 = 18, the LHS is 5x - 7.
Right-Hand Side (RHS): The expression on the right of the equals sign. In 5x - 7 = 18, the RHS is 18.
Solution (or root): The value of the variable that makes the LHS equal to the RHS. For 5x - 7 = 18, the solution is x = 5 because 5(5) - 7 = 25 - 7 = 18.
A linear equation in one variable has exactly one solution (one unique value of the variable that satisfies it), unless it is an identity (true for all values) or a contradiction (true for no values).
Methods
There are several methods and techniques for solving linear equations in one variable:
Method 1: Transposition Method (most commonly used)
Transposition means moving a term from one side of the equation to the other by changing its sign. Steps:
1. Simplify both sides of the equation if needed (expand brackets, combine like terms).
2. Move all variable terms to the LHS and all constant terms to the RHS (or vice versa) by transposing.
3. Combine like terms on each side.
4. Divide both sides by the coefficient of the variable to find its value.
Key rule: When a term is transposed (moved to the other side), its sign changes. A positive term becomes negative, and vice versa. Multiplication becomes division, and vice versa.
Method 2: Balancing Method
In this method, you perform the same operation on both sides of the equation to maintain the balance:
1. Add or subtract the same number from both sides to isolate the variable term.
2. Multiply or divide both sides by the same number to solve for the variable.
This method is conceptually clearer because it emphasises that an equation is like a balanced scale — whatever you do to one side, you must do to the other.
Method 3: Cross-Multiplication (for equations with fractions)
When the equation is in the form a/b = c/d, cross multiply to get ad = bc, which eliminates the fractions and gives a simpler equation to solve.
Method 4: LCM Method (for equations with multiple fractions)
Multiply every term on both sides by the LCM of all denominators. This eliminates all fractions at once and gives a simpler equation to work with.
Solved Examples
Example 1: Example 1: Simple one-step equation
Problem: Solve: x + 7 = 15
Solution:
Transpose 7 to the RHS (change its sign):
x = 15 - 7
x = 8
Check: LHS = 8 + 7 = 15 = RHS. Correct!
Answer: x = 8
Example 2: Example 2: Two-step equation
Problem: Solve: 3x - 5 = 16
Solution:
Step 1: Transpose -5 to the RHS.
3x = 16 + 5
3x = 21
Step 2: Divide both sides by 3.
x = 21 / 3
x = 7
Check: LHS = 3(7) - 5 = 21 - 5 = 16 = RHS. Correct!
Answer: x = 7
Example 3: Example 3: Equation with variables on both sides
Problem: Solve: 5x + 9 = 2x + 30
Solution:
Step 1: Transpose the variable term 2x from RHS to LHS (it becomes -2x).
5x - 2x + 9 = 30
Step 2: Transpose the constant 9 from LHS to RHS (it becomes -9).
5x - 2x = 30 - 9
Step 3: Simplify both sides.
3x = 21
Step 4: Divide both sides by 3.
x = 7
Check: LHS = 5(7) + 9 = 35 + 9 = 44. RHS = 2(7) + 30 = 14 + 30 = 44. LHS = RHS. Correct!
Answer: x = 7
Example 4: Example 4: Equation with brackets
Problem: Solve: 4(x - 3) = 2(x + 5)
Solution:
Step 1: Expand the brackets.
4x - 12 = 2x + 10
Step 2: Transpose 2x to the LHS.
4x - 2x - 12 = 10
Step 3: Transpose -12 to the RHS.
4x - 2x = 10 + 12
Step 4: Simplify.
2x = 22
Step 5: Divide by 2.
x = 11
Check: LHS = 4(11 - 3) = 4(8) = 32. RHS = 2(11 + 5) = 2(16) = 32. Correct!
Answer: x = 11
Example 5: Example 5: Equation with fractions
Problem: Solve: (2x + 1) / 3 = (3x - 2) / 5
Solution:
Step 1: Cross multiply.
5(2x + 1) = 3(3x - 2)
Step 2: Expand.
10x + 5 = 9x - 6
Step 3: Transpose 9x to the LHS and 5 to the RHS.
10x - 9x = -6 - 5
Step 4: Simplify.
x = -11
Check: LHS = (2(-11) + 1) / 3 = (-22 + 1) / 3 = -21 / 3 = -7. RHS = (3(-11) - 2) / 5 = (-33 - 2) / 5 = -35 / 5 = -7. LHS = RHS. Correct!
Answer: x = -11
Example 6: Example 6: Equation with multiple fractions using LCM
Problem: Solve: x/2 + x/3 = 5
Solution:
Step 1: Find the LCM of the denominators 2 and 3. LCM = 6.
Step 2: Multiply every term by 6.
6 x (x/2) + 6 x (x/3) = 6 x 5
3x + 2x = 30
Step 3: Simplify.
5x = 30
Step 4: Divide by 5.
x = 6
Check: LHS = 6/2 + 6/3 = 3 + 2 = 5 = RHS. Correct!
Answer: x = 6
Example 7: Example 7: Word problem — age problem
Problem: Ravi's father is 4 times as old as Ravi. After 5 years, the father's age will be 3 times Ravi's age. Find their present ages.
Solution:
Step 1: Let Ravi's present age be x years.
Then father's present age = 4x years.
Step 2: After 5 years:
Ravi's age = x + 5
Father's age = 4x + 5
Step 3: According to the condition:
4x + 5 = 3(x + 5)
Step 4: Expand and solve.
4x + 5 = 3x + 15
4x - 3x = 15 - 5
x = 10
Step 5: Ravi's age = 10 years. Father's age = 4 x 10 = 40 years.
Check: After 5 years: Ravi = 15, Father = 45. Is 45 = 3 x 15? Yes, 45 = 45. Correct!
Answer: Ravi is 10 years old and his father is 40 years old.
Example 8: Example 8: Word problem — consecutive numbers
Problem: The sum of three consecutive even numbers is 78. Find the numbers.
Solution:
Step 1: Let the three consecutive even numbers be x, x + 2, and x + 4.
Step 2: Write the equation.
x + (x + 2) + (x + 4) = 78
Step 3: Simplify and solve.
3x + 6 = 78
3x = 78 - 6
3x = 72
x = 24
Step 4: The three numbers are 24, 26, and 28.
Check: 24 + 26 + 28 = 78. Correct!
Answer: The three consecutive even numbers are 24, 26, and 28.
Example 9: Example 9: Word problem — geometry
Problem: The perimeter of a rectangle is 46 cm. If the length is 5 cm more than the width, find the dimensions of the rectangle.
Solution:
Step 1: Let the width = x cm.
Then the length = (x + 5) cm.
Step 2: Perimeter of rectangle = 2(length + width).
46 = 2(x + 5 + x)
46 = 2(2x + 5)
Step 3: Expand and solve.
46 = 4x + 10
4x = 46 - 10
4x = 36
x = 9
Step 4: Width = 9 cm, Length = 9 + 5 = 14 cm.
Check: Perimeter = 2(14 + 9) = 2(23) = 46 cm. Correct!
Answer: The width is 9 cm and the length is 14 cm.
Example 10: Example 10: Word problem — money and pricing
Problem: A notebook costs Rs 5 more than a pen. If 3 notebooks and 4 pens together cost Rs 110, find the cost of each.
Solution:
Step 1: Let the cost of a pen = Rs x.
Then the cost of a notebook = Rs (x + 5).
Step 2: Write the equation.
3(x + 5) + 4x = 110
Step 3: Expand and solve.
3x + 15 + 4x = 110
7x + 15 = 110
7x = 110 - 15
7x = 95
x = 95 / 7
x = 13.57 (approximately)
Let us try with a cleaner approach. Perhaps the cost of pen = Rs x.
7x + 15 = 110
7x = 95
x = 95/7 which is not a whole number.
Let us recheck the problem — if 3 notebooks and 4 pens cost Rs 110:
3(x + 5) + 4x = 110
3x + 15 + 4x = 110
7x = 95
x = 95/7 = Rs 13.57 (approximately)
Cost of pen = Rs 95/7 (approximately Rs 13.57)
Cost of notebook = Rs 95/7 + 5 = Rs 130/7 (approximately Rs 18.57)
Check: 3(130/7) + 4(95/7) = 390/7 + 380/7 = 770/7 = 110. Correct!
Answer: The cost of a pen is Rs 95/7 (approximately Rs 13.57) and the cost of a notebook is Rs 130/7 (approximately Rs 18.57).
Real-World Applications
Linear equations in one variable have countless practical applications:
Everyday Arithmetic and Shopping: When you know the total cost and the price per item, setting up a linear equation helps find how many items you can buy. For example, if each pencil costs Rs 8 and you have Rs 100, the equation 8x = 100 gives x = 12.5, meaning you can buy at most 12 pencils.
Age Problems: Problems involving present ages, future ages, and past ages are naturally modelled by linear equations. These are very common in school mathematics and competitive exams.
Geometry and Measurement: Finding unknown side lengths, angles, or dimensions when given the perimeter, area, or other geometric constraints often leads to linear equations.
Speed, Distance, and Time: The relationship distance = speed x time is a linear equation. If you know any two of these quantities, you can find the third.
Temperature Conversion: The formula F = (9/5)C + 32 is a linear equation connecting Fahrenheit and Celsius temperatures. Given one temperature scale, you can solve for the other.
Profit and Loss: Business calculations involving cost price, selling price, profit percentage, and discount percentage involve linear equations.
Mixture Problems: When mixing solutions of different concentrations or combining items at different prices, linear equations help find the required quantities.
Science: Ohm's law (V = IR), Newton's laws, and many other physical relationships are expressed as linear equations that scientists solve daily.
Key Points to Remember
- A linear equation in one variable has the form ax + b = 0, where a is not 0 and x is the variable with highest power 1.
- A linear equation in one variable has exactly one solution (one value of x that satisfies it).
- The transposition method involves moving terms from one side to the other by changing their signs.
- The balancing method involves performing the same operation on both sides to maintain equality.
- When an equation has fractions, cross-multiplication or the LCM method can be used to clear the fractions.
- Always expand brackets and simplify before attempting to isolate the variable.
- For equations with variables on both sides, collect all variable terms on one side and all constant terms on the other.
- Always verify your solution by substituting it back into the original equation.
- Word problems require translating English statements into algebraic equations before solving.
- Be careful with negative signs, especially when distributing negatives across brackets.
Practice Problems
- Solve: 7x - 3 = 2x + 22
- Solve: 3(2x - 1) = 5(x + 2) - 4
- Solve: (x + 3)/4 = (2x - 1)/6
- Solve: x/3 + x/4 + x/5 = 47
- The sum of two numbers is 95. If one exceeds the other by 15, find the two numbers.
- A number is 7 more than half of itself. Find the number.
- Anu is 4 years older than Binu. Ten years ago, Anu was twice as old as Binu. Find their present ages.
- The length of a rectangular field is twice its width. If the perimeter is 180 m, find the dimensions and area of the field.
Frequently Asked Questions
Q1. What is a linear equation in one variable?
A linear equation in one variable is an equation with one unknown (variable) where the highest power of the variable is 1. It has the general form ax + b = 0, where a and b are constants and a is not zero. For example, 2x + 5 = 11 is a linear equation in one variable (x).
Q2. How many solutions does a linear equation in one variable have?
A standard linear equation in one variable has exactly one solution. There is only one value of the variable that makes the equation true. For example, 3x + 6 = 15 has only one solution: x = 3. (The exception is if the equation simplifies to 0 = 0, making it an identity with infinitely many solutions, or to a false statement like 0 = 5, meaning no solution exists.)
Q3. What is transposition in solving equations?
Transposition is the process of moving a term from one side of an equation to the other side by changing its sign. For example, in 3x + 5 = 20, transposing +5 to the right side gives 3x = 20 - 5 = 15. This is mathematically equivalent to subtracting 5 from both sides but is a quicker shorthand.
Q4. How do you solve equations with fractions?
There are two main approaches: (1) Cross-multiplication — if the equation is in the form a/b = c/d, cross multiply to get ad = bc. (2) LCM method — multiply every term on both sides by the LCM of all denominators to eliminate fractions entirely. Both methods convert the equation into a simpler form without fractions.
Q5. What is the difference between an equation and an expression?
An expression is a mathematical phrase that combines numbers, variables, and operations (like 3x + 5 or 2y - 7). It does not have an equals sign. An equation has an equals sign and states that two expressions are equal (like 3x + 5 = 20). You can solve an equation to find the value of the variable, but you cannot 'solve' an expression.
Q6. How do you check if your solution is correct?
Substitute the value you found for the variable back into the original equation. Calculate the LHS and RHS separately. If LHS equals RHS, your solution is correct. For example, if you found x = 4 for the equation 2x + 3 = 11, check: LHS = 2(4) + 3 = 11, RHS = 11. Since LHS = RHS, x = 4 is correct.
Q7. What are the common types of word problems involving linear equations?
The most common types are: (1) Age problems — finding present, past, or future ages. (2) Number problems — finding unknown numbers given relationships between them. (3) Geometry problems — finding dimensions given perimeter or area conditions. (4) Money problems — finding costs, quantities, or rates. (5) Speed-distance-time problems. (6) Consecutive number problems.
Q8. Can a linear equation have no solution?
Yes. If simplifying the equation leads to a false statement like 0 = 5, then the equation has no solution. This happens when the variable terms cancel out but the constants do not match. For example, 2x + 3 = 2x + 8 simplifies to 3 = 8, which is false, so there is no value of x that satisfies this equation.
Q9. What is the difference between a linear equation and a quadratic equation?
In a linear equation, the highest power of the variable is 1 (like 3x + 2 = 0). In a quadratic equation, the highest power is 2 (like x squared + 3x + 2 = 0). Linear equations have exactly one solution, while quadratic equations can have zero, one, or two solutions. Linear equations are studied in Class 8, while quadratic equations are typically introduced in Class 10.
Q10. Why is it important to learn linear equations?
Linear equations are the foundation of algebra. Almost every branch of mathematics and science uses them. They teach logical thinking and problem-solving skills. They are essential for understanding systems of equations, coordinate geometry, and calculus in higher classes. In everyday life, they help with budgeting, planning, measurement, and decision-making.
