Linear Equations in Real Life

Q: How do you identify the two unknowns in a word problem?

Read the problem and identify the two quantities you are asked to find. These become x and y. For example, if the problem asks for ages of father and son, let x = father's age and y = son's age.

Q: What if the equations are not linear at first?

Some problems (especially speed-distance and work problems) lead to non-linear equations initially. Use substitution to convert them to linear form. For example, let u = 1/(x−y) and v = 1/(x+y) in boat problems.

Q: How do you check if the answer is correct?

Substitute both values back into both original word conditions (not just the equations). If both conditions are satisfied, the answer is correct.

Q: What does an inconsistent system mean in real life?

It means the conditions described in the problem contradict each other. For example, '2 apples and 1 orange cost Rs. 100' and '4 apples and 2 oranges cost Rs. 250' are contradictory (the second should be double the first).

Q: Which method is best for solving real-life problems?

Elimination is best when coefficients can be easily matched. Substitution is best when one equation is simple. Cross-multiplication works for all cases but is mechanical.

Q: Can linear equations model all real-life situations?

No. Linear equations model only situations where the relationship between quantities is constant (proportional). Situations involving squares, cubes, or exponential growth need non-linear equations.

Class 10Pair of Linear Equations in Two Variables

Pair of linear equations can model a wide variety of real-life situations — from ticket pricing and age puzzles to speed-distance and work problems. The key skill is translating English sentences into mathematical equations.

Once the equations are formed, they can be solved using substitution, elimination, cross-multiplication, or graphical methods. This topic is covered in Class 10 NCERT under "Pair of Linear Equations in Two Variables".

The ability to model real situations with linear equations is one of the most practical outcomes of Class 10 algebra.

What is Linear Equations in Real Life?

Definition: A real-life linear equation problem describes a situation involving two unknown quantities that are related by two linear conditions. The task is to:

Identify the two unknowns and assign variables (x, y).
Translate the conditions into two linear equations.
Solve the system to find the values of x and y.
Interpret the result in the context of the problem.

Linear Equations in Real Life Formula

Methods of solving:

Substitution: Express one variable in terms of the other from one equation, substitute into the second.
Elimination: Multiply equations to make coefficients equal, then add or subtract.
Cross-multiplication: For a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0:

x/(b₁c₂ − b₂c₁) = y/(c₁a₂ − c₂a₁) = 1/(a₁b₂ − a₂b₁)

Types and Properties

Common categories of real-life problems:

Age Problems: "Father is 3 times as old as son. After 5 years..."
Number Problems: "Sum of two numbers is 50. If one is doubled..."
Cost/Price Problems: "3 books and 5 pens cost Rs. 250..."
Speed-Distance-Time: "Upstream and downstream boat problems."
Fixed Charge + Rate: "Taxi fare = fixed + per-km charge."
Fraction Problems: "If numerator is increased by 2 and denominator by 1..."
Work Problems: "A and B together complete work in 12 days..."
Geometry Problems: "Perimeter of a rectangle is 40, length is 4 more than breadth..."

Solved Examples

Example 1: Age Problem

Problem: The sum of the ages of two friends is 20 years. Four years ago, the product of their ages was 48. Find their present ages.

Solution:

Let: Ages = x and y, where x + y = 20

Four years ago: (x − 4)(y − 4) = 48

From first equation: y = 20 − x
Substitute: (x − 4)(20 − x − 4) = 48
(x − 4)(16 − x) = 48
16x − x² − 64 + 4x = 48
−x² + 20x − 112 = 0
x² − 20x + 112 = 0
x = (20 ± √(400 − 448))/2 → discriminant is negative

Since the discriminant is negative, this problem has no real solution with these exact numbers. Let us adjust: if the product was 48 → try (x−4)(y−4) = 48 with x+y = 20. Since discriminant < 0, there is no solution.

Revised Problem: Sum = 20, four years ago product = 36.

(x−4)(16−x) = 36 → x² − 20x + 100 = 0 → (x−10)² = 0 → x = 10, y = 10.

Answer: Both friends are 10 years old.

Example 2: Cost of Items

Problem: 2 kg of apples and 1 kg of grapes cost Rs. 160. 4 kg of apples and 2 kg of grapes cost Rs. 300. Find the cost per kg of each.

Solution:

Let: Cost of 1 kg apples = x, Cost of 1 kg grapes = y

Equations:

2x + y = 160 ... (i)
4x + 2y = 300 ... (ii)

Check consistency: a₁/a₂ = 1/2, b₁/b₂ = 1/2, c₁/c₂ = 160/300 = 8/15. Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel — no solution.

Answer: The given conditions are inconsistent. Such a pricing situation is not possible.

Example 3: Taxi Fare Problem

Problem: A taxi charges consist of a fixed charge plus a charge per km. For a journey of 10 km the charge is Rs. 75 and for 15 km it is Rs. 110. Find the fixed charge and the rate per km.

Solution:

Let: Fixed charge = x, Per-km rate = y

Equations:

x + 10y = 75 ... (i)
x + 15y = 110 ... (ii)

Subtract (i) from (ii):

5y = 35 → y = 7
x = 75 − 70 = 5

Answer: Fixed charge = Rs. 5, Rate per km = Rs. 7.

Example 4: Fraction Problem

Problem: A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to the denominator. Find the fraction.

Solution:

Let: Numerator = x, Denominator = y

Equations:

(x − 1)/y = 1/3 → 3x − 3 = y → 3x − y = 3 ... (i)
x/(y + 8) = 1/4 → 4x = y + 8 → 4x − y = 8 ... (ii)

Subtract (i) from (ii):

x = 5
y = 3(5) − 3 = 12

Verification: (5−1)/12 = 4/12 = 1/3 ✓. 5/(12+8) = 5/20 = 1/4 ✓.

Answer: The fraction is 5/12.

Example 5: Speed-Distance (Boat) Problem

Problem: A boat goes 30 km upstream and 44 km downstream in 10 hours. It can go 40 km upstream and 55 km downstream in 13 hours. Find the speed of the boat in still water and the speed of the stream.

Solution:

Let: Speed of boat = x km/h, Speed of stream = y km/h

Upstream speed = (x − y), Downstream speed = (x + y)

Equations (time = distance/speed):

30/(x−y) + 44/(x+y) = 10
40/(x−y) + 55/(x+y) = 13

Let u = 1/(x−y), v = 1/(x+y):

30u + 44v = 10 ... (i)
40u + 55v = 13 ... (ii)

Solve: Multiply (i) by 4 and (ii) by 3:

120u + 176v = 40
120u + 165v = 39
Subtract: 11v = 1 → v = 1/11 → x + y = 11
30u + 4 = 10 → u = 1/5 → x − y = 5

x = 8, y = 3.

Answer: Speed of boat = 8 km/h, Speed of stream = 3 km/h.

Example 6: Rectangle Problem

Problem: The length of a rectangle is 5 m more than its breadth. If the perimeter is 50 m, find the length and breadth.

Solution:

Let: Breadth = x, Length = y

Equations:

y = x + 5 ... (i)
2(x + y) = 50 → x + y = 25 ... (ii)

Substitute (i) in (ii):

x + (x + 5) = 25 → 2x = 20 → x = 10
y = 15

Answer: Breadth = 10 m, Length = 15 m.

Example 7: Digit Problem

Problem: A two-digit number is such that the sum of its digits is 9. If the digits are reversed, the new number is 27 more than the original. Find the number.

Solution:

Let: Tens digit = x, Units digit = y

Original number = 10x + y, Reversed number = 10y + x

Equations:

x + y = 9 ... (i)
(10y + x) − (10x + y) = 27 → 9y − 9x = 27 → y − x = 3 ... (ii)

Adding (i) and (ii):

2y = 12 → y = 6, x = 3

Answer: The number is 36.

Example 8: Investment Problem

Problem: A person invests a total of Rs. 50,000 in two schemes at 8% and 10% annual interest. If the total annual interest is Rs. 4,400, find the amount in each scheme.

Solution:

Let: Amount at 8% = x, Amount at 10% = y

Equations:

x + y = 50000 ... (i)
0.08x + 0.10y = 4400 → 8x + 10y = 440000 ... (ii)

From (i): x = 50000 − y. Substitute in (ii):

8(50000 − y) + 10y = 440000
400000 − 8y + 10y = 440000
2y = 40000 → y = 20000
x = 30000

Answer: Rs. 30,000 at 8% and Rs. 20,000 at 10%.

Real-World Applications

Real-life linear equations are used in:

Finance: Budget allocation, loan calculations, investment splitting.
Transport: Calculating speeds, distances, and travel times.
Commerce: Pricing, discount calculations, profit-loss problems.
Planning: Resource allocation, workforce distribution.
Science: Mixing solutions of different concentrations.

Key Points to Remember

Always define what x and y represent before forming equations.
Each condition in the problem gives one equation — you need exactly two conditions for two unknowns.
Check consistency: if a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the system has no solution.
For speed-time-distance problems, use substitution (u = 1/speed) to linearise.
For digit problems, the number = 10 × (tens digit) + (units digit).
Always verify the answer in both original conditions.
Reject negative values when they don't make sense (e.g., negative age, negative cost).
For fraction problems, set up equations for numerator and denominator separately.
Substitution is often easier for problems where one equation is simple (like y = x + 5).
Elimination is faster when coefficients are convenient to match.

Practice Problems

The sum of two numbers is 35. If the larger number is divided by the smaller, the quotient is 2 and remainder is 5. Find the numbers.
A two-digit number is 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.
5 years hence, a father's age will be 3 times his son's age. 5 years ago, the father was 7 times as old as his son. Find their present ages.
A boat travels 24 km upstream and 28 km downstream in 6 hours. It travels 30 km upstream and 21 km downstream in 6.5 hours. Find the speed of the boat and the current.
Rs. 900 was divided among two people such that 5% of the first amount + 4% of the second amount = Rs. 40. Find each person's share.
The cost of 4 chairs and 3 tables is Rs. 2100. The cost of 5 chairs and 2 tables is Rs. 1750. Find the cost of each.

Frequently Asked Questions

Q1. How do you identify the two unknowns in a word problem?

Read the problem and identify the two quantities you are asked to find. These become x and y. For example, if the problem asks for ages of father and son, let x = father's age and y = son's age.

Q2. What if the equations are not linear at first?

Some problems (especially speed-distance and work problems) lead to non-linear equations initially. Use substitution to convert them to linear form. For example, let u = 1/(x−y) and v = 1/(x+y) in boat problems.

Q3. How do you check if the answer is correct?

Substitute both values back into both original word conditions (not just the equations). If both conditions are satisfied, the answer is correct.

Q4. What does an inconsistent system mean in real life?

It means the conditions described in the problem contradict each other. For example, '2 apples and 1 orange cost Rs. 100' and '4 apples and 2 oranges cost Rs. 250' are contradictory (the second should be double the first).

Q5. Which method is best for solving real-life problems?

Elimination is best when coefficients can be easily matched. Substitution is best when one equation is simple. Cross-multiplication works for all cases but is mechanical.

Q6. Can linear equations model all real-life situations?

No. Linear equations model only situations where the relationship between quantities is constant (proportional). Situations involving squares, cubes, or exponential growth need non-linear equations.

Linear Equations in Real Life

What is Linear Equations in Real Life?

Linear Equations in Real Life Formula

Types and Properties

Solved Examples

Example 1: Age Problem

Example 2: Cost of Items

Example 3: Taxi Fare Problem

Example 4: Fraction Problem

Example 5: Speed-Distance (Boat) Problem

Example 6: Rectangle Problem

Example 7: Digit Problem

Example 8: Investment Problem

Real-World Applications

Key Points to Remember

Practice Problems

Frequently Asked Questions

Q1. How do you identify the two unknowns in a word problem?

Q2. What if the equations are not linear at first?

Q3. How do you check if the answer is correct?

Q4. What does an inconsistent system mean in real life?

Q5. Which method is best for solving real-life problems?

Q6. Can linear equations model all real-life situations?

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