Applications of Linear Equations
Linear equations in one variable are powerful tools for solving real-world problems. Any situation that involves an unknown quantity and a relationship that can be expressed as a first-degree equation can be solved using linear equations.
Common applications include age problems, number problems, perimeter problems, work-rate problems, digit problems, and money problems. The key skill is translating the word problem into an equation.
This topic covers a wide variety of application problems from NCERT Class 8, with step-by-step solutions.
What is Applications of Linear Equations?
Problem-solving strategy:
- Read the problem carefully and identify the unknown.
- Let the unknown be x (or any variable).
- Translate the given conditions into an equation.
- Solve the equation for x.
- Verify the answer against the original problem.
Applications of Linear Equations Formula
Common translations:
- "A number increased by 5" → x + 5
- "Twice a number" → 2x
- "3 less than a number" → x − 3
- "The sum of two consecutive numbers" → x + (x + 1) = 2x + 1
- "Perimeter of rectangle" → 2(l + b)
- "Father is 3 times as old as son" → F = 3S
Types and Properties
Types of application problems:
- Number problems: Find a number given conditions about its parts or multiples.
- Age problems: Find present ages given conditions about past or future ages.
- Perimeter and geometry: Find dimensions given perimeter or angle conditions.
- Digit problems: Find a two-digit number given conditions about its digits.
- Money and coins: Find the number of coins of each denomination.
- Work and rate: Find time or rate given work conditions.
- Distribution: Divide a quantity into parts satisfying given conditions.
Solved Examples
Example 1: Example 1: Number problem
Problem: The sum of three consecutive integers is 72. Find the integers.
Solution:
Let the three consecutive integers be x, x+1, x+2.
- x + (x+1) + (x+2) = 72
- 3x + 3 = 72
- 3x = 69
- x = 23
The integers: 23, 24, 25.
Verification: 23 + 24 + 25 = 72 ✓
Answer: The integers are 23, 24, and 25.
Example 2: Example 2: Age problem
Problem: A father is 30 years older than his son. In 5 years, the father's age will be 3 times the son's age. Find their present ages.
Solution:
Let son's present age = x years. Father's present age = x + 30.
After 5 years:
- Son's age = x + 5
- Father's age = x + 35
- x + 35 = 3(x + 5)
- x + 35 = 3x + 15
- 35 − 15 = 3x − x
- 20 = 2x
- x = 10
Verification: Son = 10, Father = 40. After 5 years: 15 and 45. Is 45 = 3 × 15? Yes ✓
Answer: Son is 10 years, Father is 40 years.
Example 3: Example 3: Perimeter problem
Problem: The length of a rectangle is 5 cm more than its breadth. If the perimeter is 50 cm, find the dimensions.
Solution:
Let breadth = x cm. Length = x + 5 cm.
- Perimeter = 2(l + b) = 50
- 2(x + 5 + x) = 50
- 2(2x + 5) = 50
- 4x + 10 = 50
- 4x = 40
- x = 10
Verification: Breadth = 10, Length = 15. Perimeter = 2(15+10) = 50 ✓
Answer: Breadth = 10 cm, Length = 15 cm.
Example 4: Example 4: Digit problem
Problem: A two-digit number is 7 times the sum of its digits. If 27 is subtracted from the number, the digits are reversed. Find the number.
Solution:
Let tens digit = x, units digit = y.
Number = 10x + y. Reversed number = 10y + x.
- 10x + y = 7(x + y) → 10x + y = 7x + 7y → 3x = 6y → x = 2y ... (i)
- 10x + y − 27 = 10y + x → 9x − 9y = 27 → x − y = 3 ... (ii)
From (i): x = 2y. Substitute in (ii):
- 2y − y = 3 → y = 3
- x = 2(3) = 6
Verification: Number = 63. Sum of digits = 9. 7 × 9 = 63 ✓. 63 − 27 = 36 (reversed) ✓
Answer: The number is 63.
Example 5: Example 5: Money problem
Problem: A man has 25 coins in ₹5 and ₹2 denominations. The total value is ₹95. How many coins of each type does he have?
Solution:
Let number of ₹5 coins = x. Then ₹2 coins = 25 − x.
- 5x + 2(25 − x) = 95
- 5x + 50 − 2x = 95
- 3x = 45
- x = 15
Verification: 15 coins of ₹5 = ₹75. 10 coins of ₹2 = ₹20. Total = ₹95 ✓
Answer: 15 coins of ₹5 and 10 coins of ₹2.
Example 6: Example 6: Distribution problem
Problem: Divide 54 into two parts such that one part is 2/7 of the other.
Solution:
Let one part = x. Other part = 54 − x.
- x = (2/7)(54 − x)
- 7x = 2(54 − x)
- 7x = 108 − 2x
- 9x = 108
- x = 12
Other part = 54 − 12 = 42.
Verification: Is 12 = 2/7 of 42? 2/7 × 42 = 12 ✓
Answer: The two parts are 12 and 42.
Example 7: Example 7: Angle problem
Problem: Two supplementary angles differ by 40°. Find the angles.
Solution:
Let one angle = x°. Other angle = (180 − x)°.
- x − (180 − x) = 40
- x − 180 + x = 40
- 2x = 220
- x = 110°
Other angle = 180 − 110 = 70°.
Verification: 110 + 70 = 180 (supplementary) ✓. 110 − 70 = 40 ✓
Answer: The angles are 110° and 70°.
Example 8: Example 8: Speed and distance
Problem: A car travels at 60 km/h for some time and then at 80 km/h for the remaining time. If the total distance is 280 km and total time is 4 hours, find the time at each speed.
Solution:
Let time at 60 km/h = t hours. Time at 80 km/h = (4 − t) hours.
- 60t + 80(4 − t) = 280
- 60t + 320 − 80t = 280
- −20t = −40
- t = 2
Verification: 60(2) + 80(2) = 120 + 160 = 280 ✓
Answer: 2 hours at 60 km/h and 2 hours at 80 km/h.
Example 9: Example 9: Consecutive even numbers
Problem: The sum of three consecutive even numbers is 78. Find them.
Solution:
Let the numbers be x, x+2, x+4.
- x + x + 2 + x + 4 = 78
- 3x + 6 = 78
- 3x = 72
- x = 24
Answer: The numbers are 24, 26, and 28.
Example 10: Example 10: Work problem
Problem: A can do a piece of work in 12 days, B can do it in 15 days. In how many days can they finish the work together?
Solution:
A's one day work = 1/12. B's one day work = 1/15.
- Together in one day = 1/12 + 1/15
- = (5 + 4)/60 = 9/60 = 3/20
- Total days = 20/3 = 6⅔ days
Answer: They can finish the work together in 6⅔ days (or 6 days and 16 hours).
Real-World Applications
Where linear equations are applied:
- Finance: Profit-loss calculations, cost-revenue analysis.
- Geometry: Finding unknown dimensions from perimeter/area conditions.
- Physics: Speed-distance-time problems, temperature conversions.
- Daily life: Splitting bills, distributing items, planning schedules.
- Business: Break-even analysis, pricing strategies.
Key Points to Remember
- Always define the variable clearly ("Let x = ...").
- Translate each condition in the problem into an algebraic expression or equation.
- For age problems: note the time difference ("5 years ago" means subtract 5).
- For digit problems: a two-digit number with digits a and b = 10a + b.
- For consecutive numbers: use x, x+1, x+2 (odd: x, x+2, x+4).
- Always verify your answer in the original problem statement.
- If the answer is negative or non-integer and the problem demands a positive integer, recheck your equation.
- Units matter — ensure time, distance, and speed use consistent units.
Practice Problems
- The sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
- A mother is 25 years older than her daughter. After 8 years, she will be twice her daughter's age. Find their present ages.
- The perimeter of a triangle is 60 cm. The longest side is 4 cm more than the middle side, and the shortest is 8 cm less than the longest. Find all sides.
- A two-digit number is 4 times the sum of its digits. If 18 is added, the digits are reversed. Find the number.
- Ravi has ₹200 in ₹10 and ₹5 notes. He has 25 notes in all. How many of each?
- Three angles of a triangle are in the ratio 2:3:4. Find each angle.
- A train travels at 50 km/h for 3 hours and then at 70 km/h for 2 hours. Find the average speed.
- Divide 100 into two parts such that one-fourth of one part equals one-fifth of the other.
Frequently Asked Questions
Q1. How do you convert a word problem into an equation?
Identify the unknown (let it be x). Express each condition as a mathematical relationship. Set up the equation by connecting these relationships.
Q2. What are consecutive integers?
Integers that follow one after another: x, x+1, x+2. For consecutive even or odd integers: x, x+2, x+4.
Q3. How do you set up age problems?
Let present age = x. For 'n years ago', use (x − n). For 'n years later', use (x + n). Write the condition as an equation.
Q4. How do you represent a two-digit number algebraically?
If tens digit = a and units digit = b, the number = 10a + b. The reversed number = 10b + a.
Q5. Why is verification important?
Verification ensures the answer satisfies ALL given conditions. It catches errors in equation setup or solving.
Q6. What if I get a negative answer for an age problem?
A negative age is not possible. Re-read the problem and check if you set up the equation correctly. You may have swapped the relationship.
Q7. Can a word problem have no solution?
Yes. If the conditions are contradictory, the equation may have no solution. For example, two supplementary angles cannot both be greater than 100°.
Q8. What is a work-rate problem?
If A does a job in 'a' days, A's rate = 1/a per day. Combined rate of A and B = 1/a + 1/b. Time together = 1/(1/a + 1/b) = ab/(a+b) days.
