Mathematics notes, chapter 6

Linear equations

Textbook Mathematics Form 1 (DLP), pages 122–146

What this chapter covers

Identify linear equations in one variable and describe their characteristics (6.1)
Form linear equations in one variable from a statement or a situation, and write a situation for an equation (6.1)
Solve linear equations in one variable by trial and improvement, the equality concept and backtracking (6.1)
Solve problems involving linear equations in one variable (6.1)
Identify linear equations in two variables and write them in the general form ax + by = c (6.2)
Form linear equations in two variables from a statement or a situation, and vice versa (6.2)
Find possible solutions of a linear equation in two variables and draw its straight line graph (6.2)
Form simultaneous linear equations from daily situations, draw their graphs and state the type of solution (6.3)
Solve simultaneous linear equations by the graphical, substitution and elimination methods (6.3)
Solve problems involving simultaneous linear equations in two variables (6.3)

A class of 30 students is to be split into groups. There are 6 more boys than girls. How many girls are there? If we let the number of girls be x, the boys are x + 6 and so x + (x + 6) = 30. This is a linear equation. Solving it gives x = 12 girls and 18 boys. In this chapter you will learn how to form equations like this from a situation and how to solve them.

English	Bahasa Melayu
linear equation	persamaan linear
variable	pemboleh ubah
solution, root of an equation	penyelesaian, punca persamaan
trial and improvement	cuba jaya
backtracking (working backwards)	pematahbalikan
simultaneous linear equations	persamaan linear serentak
substitution, elimination	penggantian, penghapusan

6.1Linear equations in one variable

What is an equation?

An equation is a mathematical sentence with an equals sign "=". The value on the left hand side is equal to the value on the right hand side. For example, x + 2 = 5 and y − 7 = 11. An expression such as 5a − 4 has no "=" sign, so it is not an equation.

A linear equation in one variable has

only one variable (one letter), and
the power of the variable is 1 (no x², no 1x, no xy).

The general form is ax + b = c.

Example 1: is it a linear equation in one variable? (textbook p. 125)

Equation	Answer	Reason
3x + 2 = 5	Yes	One variable, x, and its power is 1
p − 4q = 6	No	Two variables, p and q
2(k − 7) = k	Yes	One variable, k, and its power is 1 (the brackets do not matter)
y² + 3y = 1	No	The highest power of y is 2

x appearing on both sides is still one variable. 3n + 1 = 6n is a linear equation in one variable.

Which are linear equations in one variable? a) m + 5 = 12 b) 9x + 8y = 10 c) k² − 5k = 4 d) 3(p − 2) = −7

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a) Yes b) No, two variables c) No, power of k is 2 d) Yes

Forming linear equations

Choose a letter for the unknown, then turn the words into a mathematical sentence.

Words	Maths
sum of, more than, added to, increased by	+
difference, less than, subtract … from, reduced by	−
times, twice, thrice, product of	×
quotient of, shared equally, half of	÷
is, the result is, gives, equals	=

Example 2 (textbook p. 126)

Subtract 8 from a number, the remainder is 2. Let the number be x: x − 8 = 2
Peck Chin bought five pens at y sen each and an exercise book for RM3. She paid RM7 in total.
Change everything to sen: RM3 = 300 sen, RM7 = 700 sen. 5y + 300 = 700

Every term must have the same unit. Writing 5y + 3 = 7 when y is in sen is wrong. Change RM to sen first (or sen to RM).

Example 3: from an equation to a situation (textbook p. 126)

2x + 5 = 19, where x is a number: "Twice a number added to 5 gives 19."
y − 2 = 8, where y is Rajes's age now: "Rajes was 8 years old 2 years ago."

Form a linear equation. a) The quotient of a number and 6 is 12. b) Rozita bought 5 kg of chicken at RMy per kg and paid RM40. c) Wei Ling bought 3 pens at x sen each and a ruler for RM2. She paid RM5 in total.

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a) x6 = 12
b) 5y = 40
c) 3x + 200 = 500 (in sen). Solving gives x = 100, so each pen costs 100 sen = RM1.

Solving linear equations in one variable

Two blocks of x kg and 6 kg are on a scale that reads 10 kg, so x + 6 = 10. A 4 kg block makes the scale read 10 kg, so x = 4 satisfies the equation.

An x kg block and a 6 kg block on a weighing scale that reads 10 kg — x + 6 = 10, so x = 4

The solution (or root) of an equation is the value of the variable that makes both sides equal. A linear equation in one variable has only one solution.

There are three methods. Example 4 solves 2x + 1 = 7 with each one.

Example 4: solve 2x + 1 = 7 (textbook pp. 127–128)

(a) Trial and improvement. Try a value, check the left hand side, then improve the guess.

Try x =	2x + 1	Compare with 7
1	2(1) + 1 = 3	3 is less than 7. Try a bigger value.
2	2(2) + 1 = 5	5 is closer, still less than 7.
3	2(3) + 1 = 7	Equal to 7. x = 3

(b) Equality concept. An equation is like a balance. Whatever you do to one side, do the same to the other side and it stays level.

Balance: two x boxes and 1 ball balance 7 balls; remove 1 ball from each side: two x boxes balance 6 balls; halve both sides: one x box balances 3 balls — Take 1 from both sides, then divide both sides by 2

2x + 1 − 1 = 7 − 1 → 2x = 6 → 2x2 = 62 → x = 3

(c) Backtracking. List what happens to x, then undo the operations in reverse order starting from the answer.

Forward: x times 2 plus 1 gives 7. Backwards: 7 minus 1 is 6, divided by 2 is 3 — 7 − 1 = 6, then 6 ÷ 2 = 3

Always check your answer: put it back into the equation. For x = 3: 2(3) + 1 = 7 ✓

In backtracking, the order matters. 2x + 1 = 7 means "× 2 then + 1", so undo "+ 1" first. 7 ÷ 2 − 1 = 2.5 is wrong.

Harder equations: fractions, brackets and x on both sides

Use the equality concept. Write a short reason for each step.

Example 5 (textbook p. 129)

(a) 2x + 135 = 7

Multiply both sides by 5: 2x + 13 = 35
Subtract 13 from both sides: 2x = 22
Divide both sides by 2: x = 11

(b) 5(x − 4) = x + 16

Expand the brackets: 5x − 20 = x + 16
Add 20 to both sides: 5x = x + 36
Subtract x from both sides: 4x = 36
Divide both sides by 4: x = 9

Check: 5(9 − 4) = 25 and 9 + 16 = 25 ✓

Multiply every term inside the brackets. 5(x − 4) = 5x − 20, not 5x − 4.

When a term moves to the other side, its sign changes because you add or subtract it on both sides: 5x − 20 = x + 16 gives 5x − x = 16 + 20, not 5x + x = 16 − 20.

Solve. a) x − 10 = 3 b) 4x − 1 = 7 c) x6 + 3 = 5 d) x + 82 = 9 e) 3(x + 2) = 5x

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a) x = 3 + 10 = 13
b) 4x = 8, x = 2
c) x6 = 2, x = 12
d) x + 8 = 18, x = 10
e) 3x + 6 = 5x, 6 = 2x, x = 3

Solving problems

Four steps: understand the problem, let a letter stand for the unknown, form and solve the equation, then check the answer in the story.

Example 6: Jalil's age (textbook p. 130)

After 10 years, Jalil's age will be three times his age this year. How old is Jalil?

Let Jalil's age now be x years. After 10 years he is x + 10.
x + 10 = 3x
Subtract x from both sides: 10 = 2x
x = 5. Jalil is 5 years old.

Check: 5 + 10 = 15 = 3 × 5 ✓

Example 7: a rectangle (textbook p. 130)

The perimeter of the rectangle is 66 cm. Find its area.

Rectangle with length (2x + 1) cm and width (x + 5) cm, perimeter 66 cm

Perimeter = 2(length + width): 2[(2x + 1) + (x + 5)] = 66
2(3x + 6) = 66 → 6x + 12 = 66 → 6x = 54 → x = 9
Length = 2(9) + 1 = 19 cm, width = 9 + 5 = 14 cm
Area = 19 × 14 = 266 cm²

Finding x = 9 is not the end. Read the question again: it asks for the area.

a) The sum of three consecutive odd numbers is 63. Find the numbers. b) Azmah scored 17 marks more than Yazid. Suzana scored twice Yazid's marks. Their total is 161. Find Azmah's score.

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a) x + (x + 2) + (x + 4) = 63, 3x + 6 = 63, 3x = 57, x = 19. The numbers are 19, 21 and 23.
b) Let Yazid's score be y. y + (y + 17) + 2y = 161, 4y + 17 = 161, 4y = 144, y = 36. Azmah = 36 + 17 = 53.

6.2Linear equations in two variables

What is a linear equation in two variables?

A linear equation in two variables has two variables and the power of each variable is 1. Its general form is ax + by = c, where a and b are not zero. Example: 6p + 4q = 9.

Example 8 (textbook p. 132)

Equation	Two variables, linear?	Reason
3p − q = 6	Yes	Variables p and q, each with power 1
7x² + 5y = 9	No	The power of x is 2
2(k + 8) = 5k	No	Only one variable
y6 + 4 = −2x	Yes	Variables x and y, each with power 1

xy + 2y = 3 is not linear: in the term xy two variables are multiplied together. 1x + 1y = 5 is not linear either.

Forming linear equations in two variables

Example 9 (textbook p. 133)

The difference between two numbers is 18: p − q = 18
Ahmad saved RMx and Norita saved RMy. Together they saved RM600: x + y = 600
A bus holds m passengers and a van holds n. Two buses and five vans carry 100 passengers: 2m + 5n = 100
Going the other way: 4x + 5y = 35, where x and y are the prices in RM of a plate of nasi lemak and a plate of fried noodles: "4 plates of nasi lemak and 5 plates of fried noodles cost RM35."

Form a linear equation in two variables. a) Eight adult tickets at RMa and five child tickets at RMc cost RM265. b) A test has Section A questions worth 1 mark and Section B questions worth 2 marks. There are x questions in A and y in B, and the total is 40 marks.

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a) 8a + 5c = 265
b) x + 2y = 40

Possible solutions

A linear equation in two variables has many pairs of solutions. Choose a value for one variable, substitute it, and solve for the other. Write each pair as an ordered pair (x, y).

Example 10: three solutions of 2x + y = 6 (textbook p. 134)

Choose	Substitute	Solve	Ordered pair
x = 0	2(0) + y = 6	y = 6	(0, 6)
x = 1	2(1) + y = 6	y = 4	(1, 4)
x = 2	2(2) + y = 6	y = 2	(2, 2)

Example 11: Osman's jerseys (textbook p. 135)

Osman bought 5 jerseys for Harimau Team (x) and Kancil Team (y): x + y = 5. The possible pairs are (1, 4), (2, 3), (3, 2) and (4, 1). In a real situation, only whole numbers make sense: you cannot buy 2.5 jerseys.

In an ordered pair the x-value comes first. (1, 4) means x = 1, y = 4, which is a different point from (4, 1).

a) Write three pairs of solutions for x − y = 7. b) Sani won 4 medals, all gold or bronze, with at least one of each. List the possible numbers.

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a) For example (7, 0), (8, 1), (10, 3). Any pair where x is 7 more than y.
b) Gold + bronze = 4: (1 gold, 3 bronze), (2, 2), (3, 1).

Graphs of linear equations in two variables

Make a table of values, plot the ordered pairs on a grid and join them. The points always lie on a straight line, which is why these equations are called linear.

Example 12: draw the graph of x − y = −2 (textbook p. 136)

Rewrite as y = x + 2, then fill in the table.

x	0	1	2	3	4	5
y	2	3	4	5	6	7

Grid with the points (0, 2), (1, 3), (2, 4), (3, 5), (4, 6) and (5, 7) joined by a straight line labelled x − y = −2 — All the points lie on one straight line

Every point on the line is a solution of the equation, and every solution is a point on the line. A point that is not on the line is not a solution.

a) Is (3, 5) a solution of x − y = −2? Is (4, 2)? b) Complete the table for 2x + y = 2 when y = 10, 8, 6, 4, 2, 0.

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a) 3 − 5 = −2 ✓ so yes. 4 − 2 = 2 ≠ −2, so (4, 2) is not.
b) x = 2 − y2: x = −4, −3, −2, −1, 0, 1

6.3Simultaneous linear equations in two variables

Forming simultaneous linear equations

Faizah rears 7 chickens and ducks. A chicken costs RM2 a week and a duck RM1. The total cost is RM12 a week. Let x = number of chickens and y = number of ducks.

x + y = 7 (number of animals) 2x + y = 12 (cost)

Simultaneous linear equations are two linear equations with the same two variables. Their solution is the pair (x, y) that satisfies both equations at the same time. On a graph it is the point of intersection of the two lines. For Faizah, x = 5 and y = 2: 5 + 2 = 7 ✓ and 2(5) + 2 = 12 ✓.

Three panels: two intersecting lines (unique solution), two parallel lines (no solution), two overlapping lines (infinite solutions) — Three possible cases (textbook p. 139)

The two straight lines	Type of solution	Example
Intersect at one point	Unique solution	x + y = 7 and 2x + y = 12
Are parallel	No solution	2x − 3y = −6 and 4x − 6y = 18
Overlap (the same line)	Infinite solutions	x + y = 4 and 5x + 5y = 20

5x + 5y = 20 is just x + y = 4 multiplied by 5, so the two lines are the same line.

Method 1: graphical method

Example 13: x + y = 6 and 2x + y = 8 (textbook p. 140)

Graph of x + y = 6 from (0, 6) to (6, 0) and 2x + y = 8 from (0, 8) to (4, 0), meeting at (2, 4)

Draw both lines on the same axes. They meet at (2, 4), so x = 2, y = 4.

Method 2: substitution method

Example 14: x − 3y = 7 and 5x + 2y = 1 (textbook p. 140)

x − 3y = 7 …… ① 5x + 2y = 1 …… ②

From ①, write x in terms of y: x = 7 + 3y …… ③
Substitute ③ into ②: 5(7 + 3y) + 2y = 1
35 + 15y + 2y = 1 → 17y = −34 → y = −2
Substitute y = −2 into ③: x = 7 + 3(−2) = 1

x = 1, y = −2. Check in ②: 5(1) + 2(−2) = 1 ✓

Method 3: elimination method

Make the coefficients of one variable the same, then add (if the signs are different) or subtract (if the signs are the same) to remove it.

Example 15 (textbook p. 141)

(a) x + 2y = 9 …… ① 3x − 2y = 15 …… ②

+2y and −2y have different signs, so add: ① + ②: 4x = 24 → x = 6
Substitute into ①: 6 + 2y = 9 → 2y = 3 → y = 32

(b) 2x + 5y = 14 …… ① 3x + 4y = 7 …… ②

The LCM of 2 and 3 is 6. ① × 3: 6x + 15y = 42 …… ③
② × 2: 6x + 8y = 14 …… ④
Same signs, so subtract: ③ − ④: 7y = 28 → y = 4
Substitute into ①: 2x + 20 = 14 → 2x = −6 → x = −3

x = −3, y = 4

When you multiply an equation, multiply every term, including the number on the right. ① × 3 is 6x + 15y = 42, not 6x + 15y = 14.

Subtracting a negative: (3x − 2y) − (x + 2y) gives −4y, so subtraction would not remove y in Example 15(a). Look at the signs before you choose to add or subtract.

Solve. a) y = 3x + 1 and x + 2y = 16 b) x + y = 5 and 2x − y = 22 c) 4x + 3y = 8 and x − 3y = 2

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a) Substitute: x + 2(3x + 1) = 16, 7x + 2 = 16, x = 2, y = 3(2) + 1 = 7
b) Add: 3x = 27, x = 9, y = 5 − 9 = −4
c) Add: 5x = 10, x = 2; 4(2) + 3y = 8, y = 0

Solving problems

Example 16: Zoo Negara tickets (textbook p. 142)

2 adults and 3 children paid RM97. 4 adults and 1 child paid RM139. Find the price of each ticket.

Let an adult ticket be RMx and a child ticket be RMy.
2x + 3y = 97 …… ① 4x + y = 139 …… ②
① × 2: 4x + 6y = 194 …… ③
③ − ②: 5y = 55 → y = 11
Substitute into ①: 2x + 33 = 97 → 2x = 64 → x = 32

An adult ticket costs RM32 and a child ticket costs RM11. Check: 2(32) + 3(11) = 97 ✓, 4(32) + 11 = 139 ✓

a) 800 coupon booklets were sold at RM30 and RM50 each, collecting RM30 000. How many of each were sold? b) A pool is p m long and q m wide. The length is twice the width and the perimeter is 150 m. Find p and q.

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a) x + y = 800 and 30x + 50y = 30 000. Multiply the first by 30: 30x + 30y = 24 000. Subtract: 20y = 6 000, y = 300, x = 500. So 500 RM30 booklets and 300 RM50 booklets.
b) p = 2q and 2p + 2q = 150. Substitute: 4q + 2q = 150, 6q = 150, q = 25, p = 50.

Summary

	Linear equation in one variable	Linear equation in two variables	Simultaneous linear equations
Form	ax + b = c	ax + by = c	Two equations in the same x and y
Example	2x + 1 = 7	x + y = 7	x + y = 7 and 2x + y = 12
Solutions	Only one (x = 3)	Many pairs; they form a straight line	Unique, none (parallel) or infinite (overlapping)
Methods	Trial and improvement, equality concept, backtracking	Choose x, substitute, solve for y	Graphical, substitution, elimination

Remember	Key fact
Balance rule	Do the same thing to both sides
Brackets	Multiply every term inside: 3(x − 2) = 3x − 6
Units	Change RM and sen (or m and cm) to the same unit before forming the equation
Backtracking	Undo the operations in reverse order
Elimination	Different signs: add. Same signs: subtract.
Check	Substitute your answer into the original equation(s)