Elimination Method
The elimination method (also called the method of addition or subtraction) is a powerful algebraic technique for solving a pair of linear equations in two variables. While the substitution method works by expressing one variable in terms of the other, the elimination method takes a different approach: it manipulates the equations so that one variable can be completely removed (eliminated) by adding or subtracting the equations. Once one variable is eliminated, the resulting single-variable equation is solved, and the value is substituted back to find the other variable. This method is often faster than substitution, especially when both variables have larger coefficients that would produce fractions under substitution. The elimination method is particularly elegant when the coefficients of one variable are already equal (or negatives of each other) in the two equations. In Class 10, this is one of the most commonly used methods for solving systems of equations and is widely applicable in word problems.
What is Elimination Method for Solving Linear Equations?
The elimination method (or method of elimination by equating coefficients) is a technique for solving a pair of linear equations by manipulating the equations so that the addition or subtraction of the two equations eliminates one of the variables.
Core idea: If we can make the coefficients of one variable (say y) equal in both equations but with opposite signs, then adding the equations will eliminate y. If the coefficients are already equal with the same sign, subtracting the equations will eliminate y.
The procedure involves:
1. Multiplying one or both equations by suitable non-zero constants so that the coefficients of one variable become equal in magnitude.
2. Adding or subtracting the equations to eliminate that variable.
3. Solving the resulting single-variable equation.
4. Substituting back to find the other variable.
What can happen during elimination:
- If one variable is eliminated and a valid equation in the other variable remains, the system has a unique solution.
- If both variables are eliminated and the result is a true statement (like 0 = 0), the system has infinitely many solutions.
- If both variables are eliminated and the result is a false statement (like 0 = 7), the system has no solution.
Elimination Method Formula
Given the system:
a1*x + b1*y = c1 ... (i)
a2*x + b2*y = c2 ... (ii)
To eliminate y:
Multiply (i) by b2: a1*b2*x + b1*b2*y = c1*b2
Multiply (ii) by b1: a2*b1*x + b1*b2*y = c2*b1
Subtract: (a1*b2 - a2*b1)*x = c1*b2 - c2*b1
x = (c1*b2 - c2*b1) / (a1*b2 - a2*b1), provided a1*b2 - a2*b1 =/= 0.
To eliminate x:
Multiply (i) by a2: a1*a2*x + b1*a2*y = c1*a2
Multiply (ii) by a1: a1*a2*x + b2*a1*y = c2*a1
Subtract: (b1*a2 - b2*a1)*y = c1*a2 - c2*a1
y = (c1*a2 - c2*a1) / (b1*a2 - b2*a1), provided b1*a2 - b2*a1 =/= 0.
Note: The denominator a1*b2 - a2*b1 is the same (up to sign) for both formulas. If it equals zero, the system is either inconsistent or dependent.
Derivation and Proof
The elimination method is justified by the properties of equality and the ability to perform the same operation on both sides of an equation.
Principle 1: Scaling invariance. If a1*x + b1*y = c1 is true, then multiplying both sides by any non-zero constant k gives k*a1*x + k*b1*y = k*c1, which is also true. The solution set is unchanged.
Principle 2: Addition property. If equation (i) holds and equation (ii) holds, then their sum (i) + (ii) also holds for the same values of x and y. Similarly, (i) - (ii) holds. This is because equals added to equals give equals.
Combining these principles:
Given: a1*x + b1*y = c1 ... (i) and a2*x + b2*y = c2 ... (ii).
Multiply (i) by b2 and (ii) by b1:
a1*b2*x + b1*b2*y = c1*b2 ... (iii)
a2*b1*x + b1*b2*y = c2*b1 ... (iv)
Subtract (iv) from (iii): (a1*b2 - a2*b1)*x = c1*b2 - c2*b1.
The y terms cancel because both have coefficient b1*b2*y. This is the essence of elimination — by careful choice of multipliers, we engineer the cancellation of one variable.
When a1*b2 - a2*b1 = 0: This means a1/a2 = b1/b2 (the lines have equal slopes). The elimination of one variable also eliminates the other, leaving either 0 = 0 (infinitely many solutions) or 0 = nonzero (no solution).
Connection to determinants: The expression a1*b2 - a2*b1 is the determinant of the coefficient matrix. In linear algebra, this is denoted as det(A) = |a1 b1; a2 b2| = a1*b2 - a2*b1. The system has a unique solution if and only if this determinant is non-zero. This connection foreshadows matrix methods studied in higher mathematics.
Types and Properties
Different types of elimination scenarios arise based on the coefficients:
Type 1: Coefficients already equal (opposite signs)
Example: 3x + 2y = 11 and 3x - 2y = 5. The y coefficients are +2 and -2. Simply add the equations: 6x = 16, x = 8/3. This is the easiest case.
Type 2: Coefficients already equal (same sign)
Example: 5x + 3y = 11 and 2x + 3y = 5. The y coefficients are both +3. Subtract the equations: 3x = 6, x = 2.
Type 3: One coefficient is a multiple of the other
Example: 2x + 3y = 7 and 4x + y = 13. Multiply the first equation by 2: 4x + 6y = 14. Now subtract from 4x + y = 13 (or vice versa): (4x + 6y) - (4x + y) = 14 - 13, giving 5y = 1.
Type 4: No simple multiple — multiply both equations
Example: 3x + 4y = 10 and 5x + 6y = 16. To eliminate x, multiply first by 5 and second by 3: 15x + 20y = 50 and 15x + 18y = 48. Subtract: 2y = 2, y = 1.
Type 5: Equations need simplification first
Example: 0.2x + 0.3y = 1.3 and 0.4x + 0.5y = 2.3. Multiply by 10 to clear decimals: 2x + 3y = 13 and 4x + 5y = 23. Then apply standard elimination.
Methods
Step-by-Step Procedure for the Elimination Method:
Step 1: Write both equations in standard form. Make sure both equations are in the form ax + by = c with like terms aligned.
Step 2: Choose which variable to eliminate. Look at the coefficients and decide which variable is easier to eliminate. If one variable already has equal (or opposite) coefficients, choose that one.
Step 3: Make the coefficients equal. Multiply one or both equations by appropriate constants so that the coefficients of the chosen variable have the same absolute value.
To eliminate x: Make coefficients of x equal. Multiply equation (i) by |a2| and equation (ii) by |a1| (or use the LCM of a1 and a2).
To eliminate y: Make coefficients of y equal. Multiply equation (i) by |b2| and equation (ii) by |b1| (or use the LCM of b1 and b2).
Step 4: Add or subtract the equations. If the equal coefficients have opposite signs, add the equations. If they have the same sign, subtract one from the other.
Step 5: Solve the resulting equation. After elimination, you have a single equation in one variable. Solve it.
Step 6: Substitute back. Plug the value from Step 5 into either original equation to find the other variable.
Step 7: Verify. Substitute both values into both original equations to confirm.
Efficient choices:
- If one coefficient is already 1, multiply only the other equation.
- Use the LCM of the coefficients (rather than the product) to keep numbers small.
- When coefficients are already opposites, just add — no multiplication needed.
Solved Examples
Example 1: Example 1: Basic elimination (opposite coefficients)
Problem: Solve: x + y = 10 and x - y = 4.
Solution:
The y coefficients are +1 and -1 (opposite signs). Add the equations:
(x + y) + (x - y) = 10 + 4. 2x = 14. x = 7.
Substitute x = 7 in equation 1: 7 + y = 10. y = 3.
Verification: 7 + 3 = 10. Correct. 7 - 3 = 4. Correct.
Answer: x = 7, y = 3.
Example 2: Example 2: Same sign coefficients (subtraction)
Problem: Solve: 3x + 4y = 25 and 3x + 2y = 17.
Solution:
The x coefficients are both 3. Subtract equation 2 from equation 1:
(3x + 4y) - (3x + 2y) = 25 - 17. 2y = 8. y = 4.
Substitute y = 4 in equation 2: 3x + 8 = 17. 3x = 9. x = 3.
Verification: 3(3) + 4(4) = 9 + 16 = 25. Correct. 3(3) + 2(4) = 9 + 8 = 17. Correct.
Answer: x = 3, y = 4.
Example 3: Example 3: Multiply one equation
Problem: Solve: 2x + 3y = 11 and x + 2y = 7.
Solution:
Multiply equation 2 by 2: 2x + 4y = 14.
Subtract equation 1 from this: (2x + 4y) - (2x + 3y) = 14 - 11. y = 3.
Substitute y = 3 in equation 2: x + 6 = 7. x = 1.
Verification: 2(1) + 3(3) = 2 + 9 = 11. Correct. 1 + 2(3) = 7. Correct.
Answer: x = 1, y = 3.
Example 4: Example 4: Multiply both equations
Problem: Solve: 3x + 4y = 10 and 5x - 3y = 7.
Solution:
To eliminate y, make y-coefficients equal: LCM of 4 and 3 is 12.
Multiply equation 1 by 3: 9x + 12y = 30.
Multiply equation 2 by 4: 20x - 12y = 28.
Add (since coefficients of y are +12 and -12): 29x = 58. x = 2.
Substitute x = 2 in equation 1: 6 + 4y = 10. 4y = 4. y = 1.
Verification: 3(2) + 4(1) = 10. Correct. 5(2) - 3(1) = 7. Correct.
Answer: x = 2, y = 1.
Example 5: Example 5: System with no solution
Problem: Solve: 2x + 3y = 7 and 4x + 6y = 18.
Solution:
Multiply equation 1 by 2: 4x + 6y = 14.
Subtract from equation 2: (4x + 6y) - (4x + 6y) = 18 - 14. 0 = 4.
This is a false statement. Both variables have been eliminated, leaving a contradiction.
Interpretation: The system has no solution. The lines are parallel.
Check ratios: a1/a2 = 2/4 = 1/2, b1/b2 = 3/6 = 1/2, c1/c2 = 7/18. Since 1/2 = 1/2 but 1/2 =/= 7/18, confirmed parallel.
Answer: No solution. The system is inconsistent.
Example 6: Example 6: System with infinitely many solutions
Problem: Solve by elimination: 4x - 6y = 10 and 6x - 9y = 15.
Solution:
Multiply equation 1 by 3: 12x - 18y = 30.
Multiply equation 2 by 2: 12x - 18y = 30.
Subtract: 0 = 0. This is always true.
Interpretation: The system has infinitely many solutions.
Note: Equation 2 is (3/2) times equation 1. They represent the same line.
General solution: From 4x - 6y = 10, we get 2x - 3y = 5, or x = (5 + 3y)/2. For any value of y, x is determined.
Answer: Infinitely many solutions. Both equations represent the line 2x - 3y = 5.
Example 7: Example 7: Word problem — fixed charges and per-unit cost
Problem: A taxi charges a fixed amount plus a per-kilometre rate. For 10 km the charge is Rs 105 and for 15 km it is Rs 155. Find the fixed charge and the per-km rate.
Solution:
Let fixed charge = f Rs and rate = r Rs/km.
Equation 1: f + 10r = 105.
Equation 2: f + 15r = 155.
Subtract equation 1 from equation 2: 5r = 50. r = 10.
Substitute in equation 1: f + 100 = 105. f = 5.
Verification: 5 + 10(10) = 105. Correct. 5 + 15(10) = 155. Correct.
Answer: Fixed charge = Rs 5, per-km rate = Rs 10.
Example 8: Example 8: Equations with decimal coefficients
Problem: Solve: 0.4x + 0.3y = 1.7 and 0.7x - 0.2y = 0.8.
Solution:
Multiply both equations by 10 to clear decimals:
4x + 3y = 17 ... (i)
7x - 2y = 8 ... (ii)
To eliminate y: Multiply (i) by 2 and (ii) by 3.
8x + 6y = 34 and 21x - 6y = 24.
Add: 29x = 58. x = 2.
Substitute in (i): 8 + 3y = 17. 3y = 9. y = 3.
Verification: 0.4(2) + 0.3(3) = 0.8 + 0.9 = 1.7. Correct. 0.7(2) - 0.2(3) = 1.4 - 0.6 = 0.8. Correct.
Answer: x = 2, y = 3.
Example 9: Example 9: Investment problem
Problem: A man invested Rs 12,000 in two schemes. Scheme A offers 10% per annum and Scheme B offers 15% per annum. If the total interest after one year is Rs 1,500, find the amount invested in each scheme.
Solution:
Let amount in Scheme A = x Rs and amount in Scheme B = y Rs.
Equation 1: x + y = 12000.
Equation 2: 0.10x + 0.15y = 1500. Multiply by 100: 10x + 15y = 150000. Simplify by dividing by 5: 2x + 3y = 30000.
To eliminate x: Multiply equation 1 by 2: 2x + 2y = 24000.
Subtract from equation 2: (2x + 3y) - (2x + 2y) = 30000 - 24000. y = 6000.
Substitute: x = 12000 - 6000 = 6000.
Verification: 6000 + 6000 = 12000. Correct. 0.10(6000) + 0.15(6000) = 600 + 900 = 1500. Correct.
Answer: Rs 6,000 in Scheme A and Rs 6,000 in Scheme B.
Example 10: Example 10: Large coefficients — efficient elimination
Problem: Solve: 37x + 43y = 123 and 43x + 37y = 117.
Solution:
Instead of directly eliminating with large multipliers, observe a pattern.
Add both equations: 80x + 80y = 240. Divide by 80: x + y = 3 ... (iii).
Subtract equation 2 from equation 1: -6x + 6y = 6. Divide by 6: -x + y = 1 ... (iv).
Add (iii) and (iv): 2y = 4. y = 2.
From (iii): x = 3 - 2 = 1.
Verification: 37(1) + 43(2) = 37 + 86 = 123. Correct. 43(1) + 37(2) = 43 + 74 = 117. Correct.
Observation: When coefficients are symmetric (a1 = b2 and a2 = b1), adding and subtracting the equations is a very efficient strategy.
Answer: x = 1, y = 2.
Real-World Applications
The elimination method is used extensively in mathematics, science, and practical problem-solving.
Engineering: Electrical circuits with multiple loops lead to systems of equations (Kirchhoff's laws). The elimination method is used to solve for unknown currents and voltages in these circuits.
Business and Finance: Problems involving two types of investments, two products, or two cost components are modelled as pairs of linear equations and solved using elimination to find the individual values.
Chemistry: Balancing complex chemical equations sometimes requires solving systems of linear equations. The elimination method helps determine the coefficients of reactants and products.
Sports Analytics: When two players' statistics satisfy certain total and difference conditions, the elimination method can be used to find individual statistics.
Traffic Flow: In urban planning, traffic at intersections is modelled using systems of equations. The flow rates on different roads can be determined using elimination.
Linear Algebra: The elimination method is the foundation of Gaussian elimination, a systematic procedure used to solve systems of any number of equations with any number of unknowns. This is one of the most important algorithms in computational mathematics.
Key Points to Remember
- The elimination method removes one variable by adding or subtracting suitably multiplied equations.
- Multiply one or both equations so that the coefficients of one variable become equal in magnitude.
- If the equal coefficients have the same sign, subtract the equations. If they have opposite signs, add them.
- Use the LCM of the coefficients (not the product) for efficiency — it keeps numbers smaller.
- If both variables are eliminated and the result is 0 = 0, the system has infinitely many solutions.
- If both variables are eliminated and the result is a false statement (like 0 = 7), the system has no solution.
- For equations with decimals or fractions, clear them first by multiplying by appropriate constants.
- When coefficients are symmetric (a1 = b2, a2 = b1), adding and subtracting the equations is a clever shortcut.
- Always substitute back to find the second variable and verify in both original equations.
- The elimination method extends to Gaussian elimination for larger systems in higher mathematics.
Practice Problems
- Solve by elimination: 3x + 2y = 12 and 5x - 2y = 4.
- Solve by elimination: 2x + 5y = 1 and 3x + 7y = 1.
- Solve: 99x + 101y = 499 and 101x + 99y = 501. (Hint: add and subtract the equations.)
- The sum of a two-digit number and the number obtained by reversing the digits is 66. If the digits differ by 2, find the number using elimination.
- Solve: x/6 + y/3 = 5 and x/3 - y/6 = 1.
- 8 men and 12 women can finish a piece of work in 10 days while 6 men and 8 women can finish it in 14 days. Find the time taken by one man alone and one woman alone.
- Use the elimination method to find the values of p and q if the system px + qy = 5 and qx + py = 7 has the solution x = 1 and y = 1.
Frequently Asked Questions
Q1. What is the elimination method?
The elimination method is a technique for solving a pair of linear equations by multiplying the equations by suitable constants so that when they are added or subtracted, one variable is eliminated. This leaves a single equation in one variable, which can be solved directly.
Q2. When should you add and when should you subtract?
Add the equations when the coefficients of the variable you want to eliminate have opposite signs (one positive, one negative). Subtract when they have the same sign. The goal is to make the coefficients cancel out.
Q3. How do you choose which variable to eliminate?
Choose the variable whose coefficients are easier to make equal. If one variable already has equal coefficients in both equations, choose that one. If the coefficients of one variable have a smaller LCM, that variable is usually easier to eliminate.
Q4. Is the elimination method better than substitution?
Neither is universally better. Elimination is often more efficient when both variables have larger coefficients (to avoid fractions). Substitution is simpler when one variable is already isolated or has coefficient 1. For symmetric coefficient patterns, elimination with adding and subtracting is very elegant.
Q5. What if both variables get eliminated?
If both variables are eliminated and you get a true statement (like 0 = 0), the system has infinitely many solutions (the equations represent the same line). If you get a false statement (like 0 = 5), the system has no solution (parallel lines).
Q6. Can the elimination method be used for three variables?
Yes, the method extends to systems with more variables. For three variables, you would eliminate one variable from two pairs of equations to get two equations in two variables, then apply elimination again. This is the basis of Gaussian elimination in linear algebra.
Q7. How do you handle equations with fractions or decimals?
First, multiply each equation by the LCM of its denominators (for fractions) or by a power of 10 (for decimals) to get integer coefficients. Then apply the standard elimination procedure.
Q8. What is the relationship between elimination and matrices?
The elimination method is closely related to row operations on the augmented matrix of the system. Multiplying an equation by a constant corresponds to scaling a row, and adding/subtracting equations corresponds to adding/subtracting rows. This connection leads to the matrix method (Gaussian elimination) studied in higher classes.
Q9. Can elimination give an approximate answer?
No, the elimination method gives exact answers (unlike the graphical method). The only source of error is arithmetic mistakes. If you perform the calculations correctly, the answer is precise.
Q10. How is the elimination method tested in CBSE exams?
CBSE Class 10 exams ask students to solve systems of equations specifically using the elimination method. Common formats include direct equation solving (2-3 marks), word problems requiring formation and solving of equations (4-5 marks), and questions asking for values of parameters (like k) that make the system consistent or inconsistent.
Related Topics
- Substitution 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










