Simultaneous Linear Equations
Senior 2 · Mathematics
Simultaneous Linear Equations
Lesson 2 of 4
22 min
Graphs show the idea, but reading an intersection point from a graph is rarely exact. Algebraic methods give the exact answer every time. This lesson covers two of them.
Both methods turn two equations in two unknowns into one equation in one unknown, which you already know how to solve from S1.
Worked example 1
Solve y = 2x − 1 and 3x + 2y = 12.
y is already the subject. Substitute into the second equation: 3x + 2(2x − 1) = 12, so 3x + 4x − 2 = 12, 7x = 14 and x = 2. Then y = 2(2) − 1 = 3.
Check: 3(2) + 2(3) = 12 ✓. Solution: (2, 3).
Worked example 2
Solve x + 3y = 7 and 2x − y = 7.
From the first equation, x = 7 − 3y. Substitute: 2(7 − 3y) − y = 7, so 14 − 6y − y = 7, −7y = −7 and y = 1. Then x = 7 − 3 = 4. Check: 2(4) − 1 = 7 ✓. Solution: (4, 1).
Make the same variable the subject of both equations. Since both expressions equal the same thing, they are equal to each other.
Worked example 3
Solve y = 3x − 5 and y = −x + 7.
Compare: 3x − 5 = −x + 7, so 4x = 12 and x = 3. Then y = 3(3) − 5 = 4. Check: −3 + 7 = 4 ✓. Solution: (3, 4).
Worked example 4
Solve 2x + y = 8 and x − y = 1.
Make y the subject of both: y = 8 − 2x and y = x − 1. Compare: 8 − 2x = x − 1, so 9 = 3x and x = 3. Then y = 3 − 1 = 2. Solution: (3, 2).
Phone plan A costs 500 Frw a day plus 20 Frw per minute of calls. Plan B costs 200 Frw a day plus 30 Frw per minute. For how many minutes do the plans cost the same?
Let m be the number of minutes and C the cost in Frw. Plan A: C = 500 + 20m. Plan B: C = 200 + 30m. Compare: 500 + 20m = 200 + 30m, so 300 = 10m and m = 30.
At 30 minutes both plans cost 500 + 600 = 1,100 Frw. Below 30 minutes plan B is cheaper; above 30 minutes plan A is cheaper.
Tip
Choosing a method. Use substitution when one variable already has a coefficient of 1 or −1 (like x + 3y = 7), because making it the subject creates no fractions. Use comparison when both equations are already, or easily, in the form y = … . Put brackets around the expression you substitute: 2(7 − 3y), not 2 × 7 − 3y.
Solve by substitution: y = 11 − 4x and 2x + 3y = 13.
Solution: 2x + 3(11 − 4x) = 13, so 2x + 33 − 12x = 13, −10x = −20 and x = 2. Then y = 11 − 8 = 3. Solution: (2, 3).
Solve by substitution: x = 2y + 1 and 3x − 4y = 7.
Solution: 3(2y + 1) − 4y = 7, so 6y + 3 − 4y = 7, 2y = 4 and y = 2. Then x = 5. Solution: (5, 2).
Solve by comparison: y = 2x + 3 and y = 5x − 9.
Solution: 2x + 3 = 5x − 9, so 12 = 3x and x = 4. Then y = 11. Solution: (4, 11).
A taxi-moto company charges 400 Frw plus 150 Frw per km; another charges 700 Frw plus 100 Frw per km. For what distance are the fares equal?
Solution: 400 + 150d = 700 + 100d, so 50d = 300 and d = 6 km. Both fares are then 1,300 Frw.
Remember
Sources and credits
The Rwandan worked examples, the practice items and the exercises were written by Ganzaa.
