Simultaneous Linear Equations
Introduction
A pair of linear equations in two variables is called simultaneous linear equations. Solving them means finding values of the variables that satisfy both equations simultaneously.
Method 1: Substitution Method
- Express one variable in terms of the other from one equation
- Substitute this expression into the second equation
- Solve for the first variable
- Substitute back to find the second variable
Method 2: Elimination Method
- Multiply equations (if needed) to make coefficients of one variable equal
- Add or subtract to eliminate that variable
- Solve for the remaining variable
- Substitute back to find the other
Multiply (i) by 3: 9x + 6y = 39 Multiply (ii) by 2: 10x - 6y = 18 Adding: 19x = 57 x = 3
From (i): 3(3) + 2y = 13 9 + 2y = 13 2y = 4 y = 2
Solution: x = 3, y = 2 </Solution> </ICSEExample>
Method 3: Cross-Multiplication Method
For equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0:
x/(b1c2 - b2c1) = y/(c1a2 - c2a1) = 1/(a1b2 - a2b1)
<ICSEExample title="Solve by Cross-Multiplication"> Solve: 2x - 3y + 5 = 0, 3x + 4y - 1 = 0 <Solution> a1 = 2, b1 = -3, c1 = 5 a2 = 3, b2 = 4, c2 = -1x/(b1c2 - b2c1) = x/[(-3)(-1) - (4)(5)] = x/(3 - 20) = x/(-17) y/(c1a2 - c2a1) = y/[(5)(3) - (-1)(2)] = y/(15 + 2) = y/17 1/(a1b2 - a2b1) = 1/[(2)(4) - (3)(-3)] = 1/(8 + 9) = 1/17
x/(-17) = 1/17 implies x = -1 y/17 = 1/17 implies y = 1 Solution: x = -1, y = 1 </Solution> </ICSEExample>
Word Problems
Type 1: Age Problems
<ICSEExample title="Age Problem"> Five years ago, a father was 3 times as old as his son. After 10 years, he will be twice as old as his son. Find their present ages. <Solution> Let fathers present age = x years, sons present age = y years Five years ago: x - 5 = 3(y - 5) x - 5 = 3y - 15 x - 3y = -10 ... (i)After 10 years: x + 10 = 2(y + 10) x + 10 = 2y + 20 x - 2y = 10 ... (ii)
Subtract (i) from (ii): y = 20 From (ii): x - 2(20) = 10 x = 50
Thus father is 50 years and son is 20 years. </Solution> </ICSEExample>
Type 2: Digit Problems
<ICSEExample title="Digit Problem"> A two-digit number is 7 times the sum of its digits. The number formed by reversing the digits is 27 less than the original number. Find the number. <Solution> Let tens digit = x, units digit = y Original number = 10x + y10x + y = 7(x + y) 10x + y = 7x + 7y 3x - 6y = 0 x = 2y ... (i)
Reversed number = 10y + x 10x + y - (10y + x) = 27 9x - 9y = 27 x - y = 3 ... (ii)
From (i) and (ii): 2y - y = 3 y = 3, x = 6
The original number is 63. </Solution> </ICSEExample>
Type 3: Fraction Problems
<ICSEExample title="Fraction Problem"> A fraction becomes 2/3 when 1 is added to both numerator and denominator. It becomes 1/2 when 1 is subtracted from both. Find the fraction. <Solution> Let fraction = x/y (x + 1)/(y + 1) = 2/3 3(x + 1) = 2(y + 1) 3x + 3 = 2y + 2 3x - 2y = -1 ... (i)(x - 1)/(y - 1) = 1/2 2(x - 1) = y - 1 2x - 2 = y - 1 2x - y = 1 ... (ii)
From (ii): y = 2x - 1 Substitute in (i): 3x - 2(2x - 1) = -1 3x - 4x + 2 = -1 -x = -3 x = 3
y = 2(3) - 1 = 5
The fraction is 3/5. </Solution> </ICSEExample>
Common Mistakes With Fixes
| Mistake | Correction |
|---|---|
| Sign errors in cross-multiplication | Write equations in standard form ax + by + c = 0 first |
| Forgetting to check the solution | Substitute back into BOTH original equations |
| Incorrect variable assignment in word problems | Define variables clearly before solving |
| Losing negative signs in elimination | Carefully handle signs when subtracting equations |
ICSE Exam Focus
| Topic | Marks (approx.) | Frequency |
|---|---|---|
| Solving by substitution/elimination | 4 marks | Very common |
| Cross-multiplication method | 4 marks | Common |
| Word problems (age, digits, fraction) | 4-5 marks | Very common |
| Speed and work problems | 4-5 marks | Frequently asked |
Self-Test
Q1: Solve: 2x + 3y = 12, 4x - y = 10
Q2: Solve by cross-multiplication: 3x - 4y = 7, 5x + 2y = 3
Q3: The sum of two numbers is 25 and their difference is 3. Find the numbers.
Q4: A two-digit number is 4 times the sum of its digits. The number formed by reversing the digits is 18 more than the original. Find the number.
Q5: 5 years from now, Ravi's age will be twice Shyam's age. 5 years ago, Ravi was 3 times as old as Shyam. Find their present ages.
