Simultaneous Linear Equations
Senior 2 · Mathematics
Simultaneous Linear Equations
Lesson 3 of 4
25 min
When neither equation has a variable with coefficient 1, substitution creates fractions. The elimination method avoids them by adding or subtracting whole equations. Cramer's rule goes further: it gives the answer by a formula, which is how calculators and computers solve systems.
If you add equal quantities to both sides of an equation, it stays balanced. So you may add (or subtract) two equations side by side. Choose it so that one variable cancels.
Worked example 1
Solve 3x + 2y = 16 and 5x − 2y = 16.
The y-terms, +2y and −2y, are opposites. Add the equations: 8x = 32, so x = 4. Substitute into the first: 12 + 2y = 16, so y = 2. Solution: (4, 2).
When no coefficients match, multiply first.
Worked example 2
Solve 2x + 3y = 12 and 3x − 4y = 1.
To eliminate y, make its coefficients 12 and −12. Multiply the first equation by 4 and the second by 3:
8x + 12y = 48
9x − 12y = 3
Add: 17x = 51, so x = 3. Substitute: 6 + 3y = 12, so y = 2. Solution: (3, 2).
A 2 × 2 determinant is a number calculated from four numbers arranged in a square:
the determinant with top row a, b and bottom row c, d equals ad − bc.
For example, with top row 5, 2 and bottom row 3, 4 the determinant is 5 × 4 − 2 × 3 = 14.
For the system a1x + b1y = c1 and a2x + b2y = c2, form three determinants:
If D ≠ 0, the system has exactly one solution: x = Dx ÷ D and y = Dy ÷ D. If D = 0, the rule fails, and the system has either no solution or infinitely many (use another method to decide).
Worked example 3
Solve 2x + 3y = 12 and 3x − 4y = 1 by Cramer's rule.
| Determinant | Top row | Bottom row | Value |
|---|---|---|---|
| D | 2, 3 | 3, −4 | (2)( − 4) − (3)(3) = −17 |
| Dx | 12, 3 | 1, −4 | (12)( − 4) − (3)(1) = −51 |
| Dy | 2, 12 | 3, 1 | (2)(1) − (12)(3) = −34 |
x = −51 ÷ ( − 17) = 3 and y = −34 ÷ ( − 17) = 2. This matches worked example 2.
Tip
Exam technique. In elimination, "same signs subtract, different signs add": if the matching coefficients are +2y and +2y, subtract; if they are +2y and −2y, add. In Cramer's rule, keep every sign in brackets, such as (2)( − 4) − (3)(3). Most wrong answers come from one lost minus sign, so check your pair in both equations at the end.
Solve by elimination: x + y = 9 and x − y = 3.
Solution: Add: 2x = 12, so x = 6 and y = 3.
Solve by elimination: 4x + 3y = 18 and 2x − 5y = −4.
Solution: Multiply the second equation by 2: 4x − 10y = −8. Subtract it from the first: 13y = 26, so y = 2. Then 4x + 6 = 18 and x = 3. Solution: (3, 2).
Evaluate the determinant with top row 7, 3 and bottom row 2, 1.
Solution: 7 × 1 − 3 × 2 = 1.
Solve by Cramer's rule: 3x + y = 9 and x − 2y = −4.
Solution: D = (3)( − 2) − (1)(1) = −7. Dx = (9)( − 2) − (1)( − 4) = −14. Dy = (3)( − 4) − (1)(9) = −21. So x = 2 and y = 3.
Remember
Sources and credits
The Rwandan worked examples, the practice items and the exercises were written by Ganzaa.
