Substitution Method
The substitution method is an algebraic technique for solving a pair of linear equations in two variables. It is part of Chapter 3 of the NCERT Class 10 Mathematics syllabus.
The basic idea is straightforward:
- Express one variable in terms of the other from one equation.
- Substitute that expression into the second equation.
- This converts the system into a single equation in one variable, which is solved directly.
The substitution method gives an exact algebraic solution, unlike the graphical method which may involve estimation. It is especially convenient when one equation already has a variable with coefficient 1 or −1.
What is Substitution Method — Solving Pair of Linear Equations in Two Variables?
Definition: The substitution method solves a system of two linear equations by isolating one variable from one equation and replacing it in the other equation.
A pair of linear equations in two variables x and y has the general form:
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
where a₁, a₂, b₁, b₂, c₁, c₂ are real numbers, and a₁² + b₁² ≠ 0, a₂² + b₂² ≠ 0.
What can happen during substitution:
- If you get a unique value for the variable, the system has exactly one solution (intersecting lines).
- If the variable cancels out and you get a true statement (like 0 = 0), the system has infinitely many solutions (coincident lines).
- If the variable cancels out and you get a false statement (like 0 = 5), the system has no solution (parallel lines).
Substitution Method Formula
Substitution Method — General Procedure:
From a₁x + b₁y = c₁, get y = (c₁ − a₁x) / b₁
Substitute into a₂x + b₂y = c₂ and solve for x
Consistency conditions:
| Condition | Type of System | Graphical Meaning |
|---|---|---|
| a₁/a₂ ≠ b₁/b₂ | Unique solution | Lines intersect at one point |
| a₁/a₂ = b₁/b₂ = c₁/c₂ | Infinitely many solutions | Lines are coincident |
| a₁/a₂ = b₁/b₂ ≠ c₁/c₂ | No solution | Lines are parallel |
Derivation and Proof
Steps of the Substitution Method:
- Choose an equation and a variable that is easiest to isolate (preferably with coefficient 1 or −1).
- Express the chosen variable in terms of the other. For example, from x + 2y = 8, get x = 8 − 2y.
- Substitute this expression into the other equation. This gives a single equation in one variable.
- Solve the resulting equation to find the value of that variable.
- Back-substitute the value into the expression from Step 2 to find the second variable.
- Write the solution as an ordered pair (x, y).
- Verify by substituting both values into both original equations.
Why it works:
- If y = (c₁ − a₁x)/b₁ is true for the solution, replacing y in the second equation preserves the solution set.
- The transformation is logically equivalent — any (x, y) satisfying both original equations also satisfies the substituted equation, and vice versa.
Important:
- Always substitute into the other equation, not the same equation you derived the expression from.
- Substituting into the same equation gives a tautology (always true), not a useful result.
Types and Properties
Types of problems solved using the Substitution Method:
Type 1: One equation already solved for a variable
- Example: y = 3x + 2 and 5x + 2y = 12.
- Substitute y = 3x + 2 directly into the second equation.
Type 2: One variable has coefficient 1 or −1
- Example: x + 3y = 7 and 2x − y = 4.
- From the first equation, x = 7 − 3y. Substitute into the second.
Type 3: Equations with fractions
- Example: x/2 + y/3 = 4 and x/3 − y/2 = 1.
- First clear fractions by multiplying by the LCM, then apply substitution.
Type 4: Reducible to linear form
- Example: 2/x + 3/y = 13 and 5/x − 4/y = −2.
- Let u = 1/x and v = 1/y. Solve for u, v, then find x = 1/u and y = 1/v.
Type 5: Word problems
- Form two equations from the given conditions.
- Solve using substitution.
Methods
Choosing Substitution vs. Other Methods:
- Use substitution when one variable is already isolated or has coefficient 1 or −1.
- Use elimination when both variables have larger coefficients and fractions would arise from isolation.
- Use cross-multiplication for direct formula-based solutions.
Common mistakes to avoid:
- Sign errors — be careful when distributing a negative through parentheses.
- Substituting into the wrong equation — always use the equation you did NOT derive the expression from.
- Forgetting to verify — always check the solution in both original equations.
- Dropping fractions — when clearing fractions, multiply every term, not just some.
Tips for efficiency:
- Isolate the variable with the smallest coefficient to minimize fractions.
- If both equations look equally complex, either variable from either equation will work — the method still gives the correct answer.
- Keep the work organized: label each step clearly.
Solved Examples
Example 1: Basic Substitution — Variable Already Isolated
Problem: Solve by substitution: y = 2x − 1 and 3x + y = 9.
Solution:
Given:
- Equation (i): y = 2x − 1
- Equation (ii): 3x + y = 9
Steps:
- y is already expressed: y = 2x − 1
- Substitute into (ii): 3x + (2x − 1) = 9
- 5x − 1 = 9
- 5x = 10, so x = 2
- Back-substitute: y = 2(2) − 1 = 3
Verification:
- (i): y = 2(2) − 1 = 3 ✓
- (ii): 3(2) + 3 = 9 ✓
Answer: x = 2, y = 3.
Example 2: Isolating a Variable with Coefficient 1
Problem: Solve: x + 2y = 8 and 3x − y = 3.
Solution:
Given:
- Equation (i): x + 2y = 8
- Equation (ii): 3x − y = 3
Steps:
- From (i): x = 8 − 2y (coefficient of x is 1)
- Substitute into (ii): 3(8 − 2y) − y = 3
- 24 − 6y − y = 3
- 24 − 7y = 3
- −7y = −21, so y = 3
- x = 8 − 2(3) = 2
Verification:
- (i): 2 + 2(3) = 8 ✓
- (ii): 3(2) − 3 = 3 ✓
Answer: x = 2, y = 3.
Example 3: Substitution with Fractional Coefficients
Problem: Solve: x/2 + 2y/3 = −1 and x − y/3 = 3.
Solution:
Given:
- Equation (i): x/2 + 2y/3 = −1
- Equation (ii): x − y/3 = 3
Steps:
- Multiply (i) by 6: 3x + 4y = −6 ... (iii)
- Multiply (ii) by 3: 3x − y = 9 ... (iv)
- From (iv): y = 3x − 9
- Substitute into (iii): 3x + 4(3x − 9) = −6
- 3x + 12x − 36 = −6
- 15x = 30, so x = 2
- y = 3(2) − 9 = −3
Verification:
- (i): 2/2 + 2(−3)/3 = 1 − 2 = −1 ✓
- (ii): 2 − (−3)/3 = 2 + 1 = 3 ✓
Answer: x = 2, y = −3.
Example 4: System with No Solution (Parallel Lines)
Problem: Solve: 2x + 4y = 10 and x + 2y = 8.
Solution:
Given:
- Equation (i): 2x + 4y = 10
- Equation (ii): x + 2y = 8
Steps:
- From (ii): x = 8 − 2y
- Substitute into (i): 2(8 − 2y) + 4y = 10
- 16 − 4y + 4y = 10
- 16 = 10 — a false statement
Interpretation:
- The variable y has been eliminated, leaving a contradiction.
- a₁/a₂ = 2/1 = 2, b₁/b₂ = 4/2 = 2, c₁/c₂ = 10/8 = 5/4.
- Since a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel.
Answer: No solution. The system is inconsistent.
Example 5: System with Infinitely Many Solutions (Coincident Lines)
Problem: Solve: 2x + 3y = 12 and 4x + 6y = 24.
Solution:
Given:
- Equation (i): 2x + 3y = 12
- Equation (ii): 4x + 6y = 24
Steps:
- From (i): 2x = 12 − 3y, so x = (12 − 3y)/2
- Substitute into (ii): 4 × (12 − 3y)/2 + 6y = 24
- 2(12 − 3y) + 6y = 24
- 24 − 6y + 6y = 24
- 24 = 24 — a true statement
Interpretation:
- The variable y has been eliminated, leaving a tautology.
- a₁/a₂ = 2/4 = 1/2, b₁/b₂ = 3/6 = 1/2, c₁/c₂ = 12/24 = 1/2.
- Since a₁/a₂ = b₁/b₂ = c₁/c₂, the lines are coincident.
Answer: Infinitely many solutions. Every point on 2x + 3y = 12 is a solution.
Example 6: Reducible to Linear Form — Using 1/x and 1/y
Problem: Solve: 2/x + 3/y = 13 and 5/x − 4/y = −2, where x ≠ 0, y ≠ 0.
Solution:
Given:
- Equation (i): 2/x + 3/y = 13
- Equation (ii): 5/x − 4/y = −2
Steps:
- Let u = 1/x and v = 1/y
- Equations become: 2u + 3v = 13 ... (iii) and 5u − 4v = −2 ... (iv)
- From (iii): 2u = 13 − 3v, so u = (13 − 3v)/2
- Substitute into (iv): 5 × (13 − 3v)/2 − 4v = −2
- Multiply by 2: 5(13 − 3v) − 8v = −4
- 65 − 15v − 8v = −4
- −23v = −69, so v = 3
- u = (13 − 9)/2 = 2
- x = 1/u = 1/2, y = 1/v = 1/3
Verification:
- (i): 2/(1/2) + 3/(1/3) = 4 + 9 = 13 ✓
- (ii): 5/(1/2) − 4/(1/3) = 10 − 12 = −2 ✓
Answer: x = 1/2, y = 1/3.
Example 7: Word Problem — Ages
Problem: Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. Find their present ages.
Solution:
Given:
- Let Nuri's present age = x years, Sonu's present age = y years.
Forming equations:
- Five years ago: x − 5 = 3(y − 5) → x − 3y = −10 ... (i)
- Ten years later: x + 10 = 2(y + 10) → x − 2y = 10 ... (ii)
Steps:
- From (i): x = 3y − 10
- Substitute into (ii): (3y − 10) − 2y = 10
- y − 10 = 10
- y = 20
- x = 3(20) − 10 = 50
Verification:
- Five years ago: 45 = 3 × 15 ✓
- Ten years later: 60 = 2 × 30 ✓
Answer: Nuri is 50 years old, Sonu is 20 years old.
Example 8: Word Problem — Numbers
Problem: The sum of two numbers is 80. The larger number exceeds four times the smaller by 5. Find the numbers.
Solution:
Given:
- Let the smaller number = x, larger number = y.
Forming equations:
- x + y = 80 ... (i)
- y = 4x + 5 ... (ii)
Steps:
- Equation (ii) already gives y in terms of x.
- Substitute into (i): x + (4x + 5) = 80
- 5x + 5 = 80
- 5x = 75, so x = 15
- y = 4(15) + 5 = 65
Verification:
- 15 + 65 = 80 ✓
- 65 = 4(15) + 5 = 65 ✓
Answer: The numbers are 15 and 65.
Example 9: Negative Coefficients
Problem: Solve: 7x − 2y = 5 and 3x + y = 11.
Solution:
Given:
- Equation (i): 7x − 2y = 5
- Equation (ii): 3x + y = 11
Steps:
- From (ii): y = 11 − 3x
- Substitute into (i): 7x − 2(11 − 3x) = 5
- 7x − 22 + 6x = 5
- 13x = 27
- x = 27/13
- y = 11 − 3(27/13) = (143 − 81)/13 = 62/13
Verification:
- (i): 7(27/13) − 2(62/13) = (189 − 124)/13 = 65/13 = 5 ✓
- (ii): 3(27/13) + 62/13 = (81 + 62)/13 = 143/13 = 11 ✓
Answer: x = 27/13, y = 62/13.
Example 10: Word Problem — Fixed and Variable Charges
Problem: A taxi charges consist of a fixed amount plus a per-kilometre rate. For 10 km, the charge is ₹105. For 15 km, the charge is ₹155. Find the fixed charge and per-km rate.
Solution:
Given:
- Let fixed charge = ₹x, rate per km = ₹y.
Forming equations:
- x + 10y = 105 ... (i)
- x + 15y = 155 ... (ii)
Steps:
- From (i): x = 105 − 10y
- Substitute into (ii): (105 − 10y) + 15y = 155
- 105 + 5y = 155
- 5y = 50, so y = 10
- x = 105 − 10(10) = 5
Verification:
- 10 km: 5 + 10(10) = 105 ✓
- 15 km: 5 + 15(10) = 155 ✓
Answer: Fixed charge = ₹5, rate per km = ₹10.
Real-World Applications
Applications of the Substitution Method:
- Age problems — relate present, past, and future ages with two unknowns.
- Number problems — find two numbers given their sum, difference, or a relationship.
- Cost and pricing — determine fixed and variable costs from total cost data.
- Speed-distance-time — find upstream/downstream speeds, or speeds of two vehicles.
- Geometry — find dimensions of rectangles given perimeter and area conditions.
- Mixtures — determine quantities of two ingredients in a blend.
- Business and economics — break-even analysis, profit-cost models with two unknowns.
- Physics — solving simultaneous equations in circuit analysis (Kirchhoff's laws), force equilibrium.
Key Points to Remember
- The substitution method converts a system of two equations in two variables into a single equation in one variable.
- Always isolate the variable with coefficient 1 or −1 first to avoid fractions.
- Substitute into the other equation, never back into the same equation.
- If the variable cancels and you get a true statement (e.g., 0 = 0), the system has infinitely many solutions.
- If the variable cancels and you get a false statement (e.g., 0 = 5), the system has no solution.
- Always verify the solution in both original equations.
- For equations with fractions, clear fractions first by multiplying through by the LCM of the denominators.
- For equations of the form a/x + b/y = c, substitute u = 1/x, v = 1/y to reduce to linear form.
- The substitution method works for all consistent systems and is equivalent to the elimination method — both give the same solution.
- In CBSE board exams, substitution method problems carry 3–4 marks. Show all steps clearly.
Practice Problems
- Solve by substitution: 2x + y = 7 and x − y = 2.
- Solve: x + 4y = 14 and 7x − 3y = 5.
- Solve: 3x − 5y = −1 and x − y = −1.
- Solve: 5x + 3y = 35 and 2x + 4y = 28.
- A fraction becomes 9/11 if 2 is added to both numerator and denominator. If 3 is added to both, it becomes 5/6. Find the fraction.
- The sum of a two-digit number and the number formed by reversing its digits is 66. If the digits differ by 2, find the number.
- Solve: 3/x + 2/y = 12 and 2/x + 3/y = 13 (x ≠ 0, y ≠ 0).
- The length of a rectangle exceeds its breadth by 7 cm. If the perimeter is 50 cm, find the dimensions.
Frequently Asked Questions
Q1. What is the substitution method?
The substitution method is an algebraic technique to solve a pair of linear equations. One variable is expressed in terms of the other from one equation, and this expression is substituted into the second equation. This converts the system to a single equation in one variable.
Q2. When should I use the substitution method instead of elimination?
Use substitution when one equation already has a variable isolated (e.g., y = 3x + 5) or when a variable has coefficient 1 or −1, making isolation easy. Use elimination when both variables have larger coefficients and isolation would create fractions.
Q3. How do I know if a system has no solution using substitution?
If, during substitution, the variable cancels out and you get a false statement like 16 = 10 or 0 = 5, the system has no solution. The lines are parallel.
Q4. How do I know if a system has infinitely many solutions?
If the variable cancels out and you get a true statement like 0 = 0 or 24 = 24, the system has infinitely many solutions. The lines are coincident (identical).
Q5. Does it matter which variable I isolate?
No. You can isolate x or y from either equation. The final answer is the same. However, choosing the variable with the simplest coefficient (1 or −1) avoids fractions and makes arithmetic easier.
Q6. Can substitution solve non-linear equations?
The substitution method for Class 10 is designed for linear equations. However, the substitution technique also applies to non-linear systems (quadratic, etc.) in higher classes.
Q7. Why must I substitute into the OTHER equation?
If you substitute back into the same equation you derived the expression from, every value of the variable satisfies it (tautology). You need both equations to determine the unique solution.
Q8. Is the substitution method expected in CBSE board exams?
Yes. CBSE board exams frequently ask students to solve a pair of linear equations by the substitution method. Such questions carry 3–4 marks. Show each step: isolate, substitute, solve, back-substitute, and verify.
Related Topics
- Elimination Method
- Cross-Multiplication Method
- Pair of Linear Equations in Two Variables
- Graphical Method for Linear Equations
- Consistency of Linear Equations
- Word Problems on Pair of Linear Equations
- Word Problems Solved Graphically
- Reducing Equations to Linear Form
- Conditions for Solvability
- Linear Equations in Real Life
