Simultaneous Linear Equations
Senior 2 · Mathematics
Simultaneous Linear Equations
Lesson 4 of 4
25 min
In real problems nobody hands you the equations. You are given a situation, such as prices at a market, tickets sold for a school match, or two numbers described in words, and you must build the equations yourself. This is called modelling, and it is what national exam problems test.
Uwase pays 4,900 Frw for 2 kg of beans and 3 kg of rice. Her neighbour pays 3,500 Frw for 3 kg of beans and 1 kg of rice. Find the price of 1 kg of each.
Step 1 Let b be the price of 1 kg of beans and r the price of 1 kg of rice, both in Frw.
Step 2 2b + 3r = 4900 and 3b + r = 3500.
Step 3 From the second equation, r = 3500 − 3b. Substitute: 2b + 3(3500 − 3b) = 4900, so 2b + 10500 − 9b = 4900, −7b = −5600 and b = 800. Then r = 3500 − 2400 = 1100.
Step 4 Check: 2(800) + 3(1100) = 1600 + 3300 = 4900 ✓ and 3(800) + 1100 = 3500 ✓. Beans cost 800 Frw per kg and rice costs 1,100 Frw per kg.
For an inter-school football match, adult tickets cost 1,000 Frw and student tickets cost 500 Frw. 150 tickets were sold for a total of 100,000 Frw. How many of each were sold?
Let a be the number of adult tickets and s the number of student tickets. Then a + s = 150 and 1000a + 500s = 100000. Divide the second equation by 500: 2a + s = 200. Subtract the first equation: a = 50. Then s = 100.
Check: 50 + 100 = 150 ✓ and 50,000 + 50,000 = 100,000 ✓. 50 adult tickets and 100 student tickets.
The sum of two numbers is 42 and their difference is 8. Find them.
x + y = 42 and x − y = 8. Add: 2x = 50, so x = 25 and y = 17.
Tip
Exam technique. Write "Let x = …" with units before any algebra; examiners award a mark for it. Keep the units consistent: if prices are in Frw, every term must be in Frw. Finish with a sentence that answers the question asked. "x = 50" alone is not an answer to "how many adult tickets?".
3 pens and 2 exercise books cost 2,600 Frw. 1 pen and 4 exercise books cost 3,200 Frw. Find the price of each.
Solution: 3p + 2b = 2600 and p + 4b = 3200. From the second equation, p = 3200 − 4b. Substitute: 9600 − 12b + 2b = 2600, so 10b = 7000 and b = 700. Then p = 400. A pen costs 400 Frw and a book costs 700 Frw.
A farmer in Rwamagana keeps goats and chickens. There are 20 animals with 56 legs in total. How many goats are there?
Solution: g + c = 20 and 4g + 2c = 56. Double the first equation: 2g + 2c = 40. Subtract: 2g = 16, so g = 8 goats (and 12 chickens). Check: 32 + 24 = 56 ✓.
The difference of two numbers is 5 and their sum is 31. Find the numbers.
Solution: x + y = 31 and x − y = 5. Add: 2x = 36, so x = 18 and y = 13.
Remember
Sources and credits
The Rwandan worked examples, the practice items and the exercises were written by Ganzaa.
