8 min readPathway sheetPart III (5 marks), often as a word problem; sometimes a 1-mark MCQ

Linear equations in three variables

Three equations in x, y and z are solved by elimination: remove one variable twice, solve the pair, then substitute back.

General formax + by + cz + d = 0, where a, b, c are not all zero

PictureEach equation is a plane in space

MethodElimination, then back substitution

CheckPut the values of x, y, z in all three equations

Method of elimination

Number the equations (1), (2) and (3). Then follow these steps.

1
Choose the variable that is easiest to remove. Often it has coefficient 1 or −1.
2★
Use one pair, such as (1) and (2), to remove it. Call the new equation (4).
3★
Use a different pair, such as (2) and (3), to remove the same variable. Call it (5).
4
Solve (4) and (5). They now have only two variables.
5
Put both values into any one of (1), (2) or (3) to find the third variable.
6
Check the three values in an equation you did not use last.

★ key step

What the elimination tells you

What you get at the end	Meaning	Picture
Values such as x = 1, y = 2, z = 3	Unique solution	The three planes meet at one point
A true statement such as 0 = 0	Infinitely many solutions	The planes meet along a line, or are the same plane
A false statement such as 0 = 5	No solution	No point lies on all three planes

Worked example 1: solve by elimination

Problem

Solve x + y + z = 6, 2x − y + z = 3 and x + 2y − z = 2.

Working

Number them: x + y + z = 6 ... (1), 2x − y + z = 3 ... (2), x + 2y − z = 2 ... (3).
Remove z using (1) + (3): 2x + 3y = 8 ... (4).
Remove z again using (2) + (3): 3x + y = 5 ... (5).
Multiply (5) by 3: 9x + 3y = 15 ... (6).
(6) − (4): 7x = 7, so x = 1.
Put x = 1 in (5): 3 + y = 5, so y = 2.
Put x = 1 and y = 2 in (1): 1 + 2 + z = 6, so z = 3.
Check in (2): 2(1) − 2 + 3 = 3. True.

Answer

x = 1, y = 2, z = 3

Worked example 2: a stationery shop word problem

Problem

At a stationery shop in Madurai, Kavin pays ₹ 190 for 2 notebooks, 3 pens and 1 geometry box. Meena pays ₹ 220 for 1 notebook, 2 pens and 2 geometry boxes. Arun pays ₹ 210 for 3 notebooks, 1 pen and 1 geometry box. Find the price of each item. (These are example prices.)

Working

Let a notebook cost ₹ x, a pen ₹ y and a geometry box ₹ z.
Kavin: 2x + 3y + z = 190 ... (1). Meena: x + 2y + 2z = 220 ... (2). Arun: 3x + y + z = 210 ... (3).
Remove z using (3) − (1): x − 2y = 20 ... (4).
Remove z again using 2 × (1) − (2). That is (4x + 6y + 2z) − (x + 2y + 2z) = 380 − 220.
So 3x + 4y = 160 ... (5).
Multiply (4) by 2: 2x − 4y = 40 ... (6).
(5) + (6): 5x = 200, so x = 40.
Put x = 40 in (4): 40 − 2y = 20, so y = 10.
Put x = 40 and y = 10 in (3): 120 + 10 + z = 210, so z = 80.
Check in (2): 40 + 20 + 160 = 220. True.

Answer

Notebook ₹ 40, pen ₹ 10, geometry box ₹ 80

In real life

Three bills with three unknown prices are a common puzzle at a shop or a canteen. Write one equation for each bill, then eliminate to find every price.

Exam angle

1 marks: “While solving three linear equations, elimination ends in 0 = 0. What does this mean?” The system has **infinitely many solutions**.

5 marks: “Solve: 3x − 2y + z = 11, x + y + z = 4, 2x + 3y − z = −2.” Add (1) and (3): 5x + y = 9. Add (2) and (3): 3x + 4y = 2. Solve the pair: x = 2, y = −1. Then z = 4 − x − y = 3. Answer: x = 2, y = −1, z = 3.

5 marks: “Divya's marks in Tamil, English and Maths add up to 240. Her Maths mark is 20 more than her Tamil mark. English and Maths together come to 170. Find each mark.” Let the marks be T, E, M. T + E + M = 240, M − T = 20, E + M = 170. Subtract the third from the first: T = 70. Then M = 90 and E = 80.

Remember it

Same variable, two pairs: remove one variable from two different pairs, solve the pair, substitute back, then check.