Pair of Linear Equations in Two Variables · Lesson 3 of 4
Algebraic Method 2: The Elimination Method
“The mathematical equivalent of 'this town ain't big enough for the both of us', where we aggressively delete one variable just to make the other one confess its value.”
• Explain how eliminating one variable reduces a pair of linear equations to a single-variable equation. • Choose suitable multipliers to make the coefficients of one variable equal or opposite. • Solve pairs of linear equations accurately using addition or subtraction. • Interpret complete cancellation to identify systems with no solution or infinitely many solutions.
Introduction
The elimination method solves a pair of linear equations by removing one variable completely. This is done by making the coefficients of one variable numerically equal in both equations and then adding or subtracting the equations.
Make the coefficients of either x or y equal. Then add or subtract the equations so that the chosen variable cancels out.
Standard Step-by-Step Procedure
Multiply one or both equations by suitable non-zero constants so that the coefficients of one variable become numerically equal.
Add or subtract the modified equations. The selected variable will cancel, leaving an equation in only one variable.
Substitute the value obtained into either original equation to calculate the value of the second variable.
Complete Worked Example
Problem
The ratio of monthly incomes of two people is 9:7 and the ratio of their monthly expenditures is 4:3. Each person saves ₹2,000 per month. Find their exact monthly incomes.
- 1.Represent the incomes and expenditures using variables.
- 2.Form two linear equations using Income − Expenditure = Savings.
- 3.Make the coefficients of one variable equal.
- 4.Subtract the equations to eliminate that variable.
- 5.Substitute back to find the remaining value.
- 6.Calculate the actual monthly incomes.
Formulate the Equations
Let the monthly incomes of the two people be 9x and 7x. Let their monthly expenditures be 4y and 3y.
Step 1: Match the Coefficients of y
The least common multiple of 4 and 3 is 12. Multiply Equation (1) by 3 and Equation (2) by 4.
Step 2: Eliminate y
Subtract Equation (3) from Equation (4). Since both equations contain −12y, the y-terms cancel.
The common income unit is x = 2000.
Step 3: Substitute Back to Find y
Substitute x = 2000 into Equation (1).
Calculate the Final Incomes
The monthly income of the first person is ₹18,000 and the monthly income of the second person is ₹14,000.
Special Algebraic Edge Cases: No Solution and Infinitely Many Solutions
Sometimes, while solving a pair of equations, both variables cancel completely. The numerical statement that remains tells us whether the system has no solution or infinitely many solutions.
Case 1: A False Numerical Statement
The variables cancel, but the remaining statement is impossible. Therefore, the equations have no common solution. Their graphs are parallel lines, and the system is inconsistent.
Case 2: A True Numerical Statement
The variables cancel, and the remaining statement is always true. Therefore, both equations represent the same line. The system has infinitely many solutions and is dependent and consistent.
A false statement means no solution. A true statement means infinitely many solutions.
Choose the variable whose coefficients can be made equal using the smallest and simplest multipliers. This reduces arithmetic and lowers the chance of calculation errors.
Multiply every term on both sides of an equation, handle negative signs carefully while subtracting, and always substitute back to find the second variable.
Quiz
What should generally be done when one variable has equal coefficients with the same sign in both equations?
What is the solution of x + y = 9 and x − y = 3?
Which operation eliminates y from 2x + 3y = 12 and 5x − 3y = 9?
What does the result 0 = 0 after eliminating both variables indicate?
What does an impossible result such as 0 = 5 indicate after elimination?
Practice Problems
- Solve x + y = 11 and x − y = 3 using the elimination method.
- Solve 2x + 3y = 12 and 2x − y = 4 using the elimination method.
- Solve 2x + 3y = 13 and 5x + 2y = 16 by choosing suitable multipliers and eliminating one variable.
- Use elimination to determine whether the equations 2x + 3y = 7 and 4x + 6y = 20 have one solution, no solution or infinitely many solutions.
- At a school event, 18 adult and child tickets were sold. An adult ticket cost ₹80 and a child ticket cost ₹60. The total amount collected was ₹1,260. Form a pair of linear equations and use elimination to find the number of each type of ticket sold.
Key Takeaways
• Use the least common multiple of coefficients when choosing multipliers. • Subtract the equations when equal coefficients have the same sign. • Add the equations when equal coefficients have opposite signs. • Always multiply every term in the equation. • Interpret complete cancellation carefully.